Constant Acceleration in 1D: past paper questions & worked solutions

Let the tension at D be $T$. The tension at C is double this, so it is $2T$.

The beam is in equilibrium, so the upward forces equal the weight: $2T+T=24$. Hence $3T=24$ and $T=8\ \mathrm{N}$.

The beam is uniform, so its weight acts at the midpoint, $3$ m from A. Also $AC=x$ and $AD=x+2$.

Clockwise and anticlockwise moments balance: $(2T)x+T(x+2)=24\cdot3$.

Substitute $T=8$: $16x+8(x+2)=72$. Thus $24x+16=72$, so $24x=56$ and $x=\dfrac{7}{3}$ m.

The velocity becomes zero at $t=2.5$ s. Since acceleration is $-g=-9.8\ \mathrm{m\,s^{-2}}$, the initial velocity is $u=9.8(2.5)$.

$u=24.5\ \mathrm{m\,s^{-1}}$.

From launch to $t=7$, the displacement is the area under the velocity-time graph. Algebraically, $s=ut-\dfrac12gt^2$ with $t=7$.

$s=24.5(7)-\dfrac12(9.8)(7^2)=171.5-240.1=-68.6$ m.

The negative sign shows the ground is below A, so the height of A above the ground is $68.6\ \mathrm{m}$.

Taking moments about A: $\sum M_A = 0$ Clockwise moments = Anticlockwise moments $P \times 3a = W \times d$ $3aP = Wd \quad \text{...(1)}$

Taking moments about C: $\sum M_C = 0$ Distance from C to COM = $(2a - d)$ Distance from C to B = $a$ $1.25P \times a = W \times (2a - d)$ $1.25Pa = W(2a - d)$ $1.25Pa = 2aW - Wd \quad \text{...(2)}$

From (1): $Wd = 3aP$ $1.25Pa = 2aW - 3aP$ $1.25Pa + 3aP = 2aW$ $4.25Pa = 2aW$ $P = \frac{2aW}{4.25a}$ $P = \frac{2W}{4.25}$ Substitute back into (1): $3a \times \frac{2W}{4.25} = Wd$ $\frac{6aW}{4.25} = Wd$ $d = \frac{6a}{4.25}$ $d = \frac{6a}{17/4}$ $d = \frac{24a}{17}$ $d = 1.41a \text{ (3 sf)}$ Or in exact form: $d = \frac{24a}{17}$

For P, take downward as positive. Its weight is $4mg$ and the tension in the string between P and Q is $3mg$ upward, so $4mg-3mg=4ma$.

The equation gives $mg=4ma$, hence $a=\dfrac{g}{4}$ downward.

For Q, downward forces are its weight $2mg$ and the lower string tension $3mg$. The upper string tension $T$ acts upward. Taking downward as positive: $2mg+3mg-T=2ma$.

$5mg-T=2m\left(\dfrac{g}{4}\right)=\dfrac{mg}{2}$.

$T=5mg-\dfrac{mg}{2}=\dfrac{9mg}{2}$.

Given $\tan\theta=\dfrac34$, use $\sin\theta=\dfrac35$ and $\cos\theta=\dfrac45$.

The horizontal force of $2$ N is balanced by the horizontal component of the tension in the sloping part of the string: $T\cos\theta=2$.

$T\cdot\dfrac45=2$, so $T=2.5\ \mathrm{N}$.

The ring is smooth, so the tension is the same in both parts of the string. Upward components are $T$ from the vertical part and $T\sin\theta$ from the sloping part. Therefore $T+T\sin\theta=Mg$.

$Mg=2.5+2.5\cdot\dfrac35=4.0$, so $M=\dfrac{4.0}{9.8}=0.408\ \mathrm{kg}$.

The rod length is $2a$. The centre of mass is at the midpoint, distance $a$ from A. Also $AC=\dfrac{2a}{5}$ and $AD=2a-\dfrac{3a}{5}=\dfrac{7a}{5}$.

For the largest possible mass at B, the string at C is just slack and the rod is about to tilt about D.

Let the particle mass be $X$. The distance from D to B is $\dfrac{3a}{5}$ and the distance from D to the rod's centre of mass is $\dfrac{2a}{5}$. Hence $Xg\left(\dfrac{3a}{5}\right)=Mg\left(\dfrac{2a}{5}\right)$.

Cancel $ga/5$: $3X=2M$, so $X=\dfrac{2M}{3}$.

Now the particle has mass $\dfrac12M$. Let the tensions at C and D be $T_C$ and $T_D$.

About D, $T_C(a)+\dfrac12Mg\left(\dfrac{3a}{5}\right)=Mg\left(\dfrac{2a}{5}\right)$.

$T_Ca+\dfrac{3}{10}Mga=\dfrac{2}{5}Mga$, so $T_C=\dfrac{1}{10}Mg$.

For the lift, the upward force is the cable tension $3616$ N. Downward forces on the lift are its own weight $250g$ and the contact forces from Alan and Bhavana, $565$ N and $226$ N.

Taking upward as positive: $3616-250g-565-226=250a$.

$3616-2450-565-226=250a$, so $375=250a$ and $a=1.5\ \mathrm{m\,s^{-2}}$.

Let Alan's mass be $m_A$. The upward force on Alan is $565$ N and his weight is $m_Ag$.

$565-m_Ag=m_A(1.5)$, so $565=m_A(9.8+1.5)=11.3m_A$. Hence $m_A=50\ \mathrm{kg}$.

The force $F$ is perpendicular to the string. The weight is $5$ N vertically downward, and the string makes $60^\circ$ with the vertical.

The component of the weight perpendicular to the string is $5\cos30^\circ$, so $F=5\cos30^\circ=\dfrac{5\sqrt3}{2}\ \mathrm{N}$.

The tension balances the component of the weight along the string. This component is $5\sin30^\circ$.

$T=5\sin30^\circ=\dfrac52\ \mathrm{N}$.

The bucket and cement have total mass $15+25=40$ kg and acceleration $0.2\ \mathrm{m\,s^{-2}}$ upward.

Taking upward as positive: $T-40g=40(0.2)$.

$T-392=8$, so $T=400\ \mathrm{N}$.

The bucket and tools are moving downward but decelerating, so their acceleration is upward with magnitude $0.1\ \mathrm{m\,s^{-2}}$.

For the $12$ kg box, upward force is the normal reaction $R$, and downward force is weight $12g$.

$R-12g=12(0.1)$. Thus $R=117.6+1.2=118.8\ \mathrm{N}$, about $119\ \mathrm{N}$.

The stone starts $5$ m above the ground. When it is $2$ m above the ground, its displacement from O is $-3$ m.

Taking upward as positive, $u=20$, $a=-9.8$ and $s=-3$. Use $v^2=u^2+2as$: $v^2=20^2+2(-9.8)(-3)=458.8$.

$v=21.4\ \mathrm{m\,s^{-1}}$ to 3 significant figures.

At the ground, displacement from O is $-5$ m. Use $s=ut+\dfrac12at^2$: $-5=20t-4.9t^2$.

$4.9t^2-20t-5=0$. The positive root is $t=4.3179\ldots$, so the total time is $4.32\ \mathrm{s}$.

The girder has length $6$ m, with C $1$ m from A and D $1$ m from B. So D is $5$ m from A.

When the $900$ N upward force is applied at A, the girder is about to tilt about D. Taking moments about D gives $900(5)=W(5-x)$.

When the $1500$ N upward force is applied at B, the girder is about to tilt about C. Taking moments about C gives $1500(5)=W(x-1)$.

The equations are $W(5-x)=4500$ and $W(x-1)=7500$. Add them: $W[(5-x)+(x-1)]=12000$, so $4W=12000$ and $W=3000$.

Use $W(x-1)=7500$: $3000(x-1)=7500$, so $x-1=2.5$ and $x=3.5$ m.

From the diagram: ; Distance AC = 2 m ; Distance CD = 8 - 2 - x = (6-x) m ; Distance from A to center = 4 m, so distance from C to center = 4 - 2 = 2 m

$\sum M_C = 0$ Clockwise moments = Anticlockwise moments $90 \times (6-x) = 240 \times 2$ $90(6-x) = 480$ $540 - 90x = 480$ $90x = 60$ $x = \frac{60}{90} = \frac{2}{3}$

$\sum F_{\text{vertical}} = 0$ $T_C + T_D = 240 + W$ $183 + T_D = 240 + W$ $T_D = 57 + W$

$\sum M_C = 0$ Clockwise moments = Anticlockwise moments Distance CD = $6 - \frac{2}{3} = \frac{16}{3}$ m $T_D \times \frac{16}{3} + W \times 2 = 240 \times 2$ $(57 + W) \times \frac{16}{3} + 2W = 480$ $\frac{16(57 + W)}{3} + 2W = 480$ $\frac{912 + 16W}{3} + 2W = 480$ Multiply through by 3: $912 + 16W + 6W = 1440$ $22W = 528$ $W = 24$

From $t=1$ to $t=9$, the displacement is $(10\mathbf{i}-3\mathbf{j})-(-2\mathbf{i}+5\mathbf{j})=12\mathbf{i}-8\mathbf{j}$.

The time interval is $8$ seconds, so $\mathbf{v}=\dfrac{12\mathbf{i}-8\mathbf{j}}{8}=1.5\mathbf{i}-\mathbf{j}\ \mathrm{m\,s^{-1}}$.

Time $t=7$ is 6 seconds after $t=1$. Therefore $\mathbf{r}_B(7)=(-2\mathbf{i}+5\mathbf{j})+6(1.5\mathbf{i}-\mathbf{j})=7\mathbf{i}-\mathbf{j}$.

The target has position vector $13\mathbf{i}-5\mathbf{j}$. The vector from the ball to the target is $(13\mathbf{i}-5\mathbf{j})-(7\mathbf{i}-\mathbf{j})=6\mathbf{i}-4\mathbf{j}$.

The distance is $\sqrt{6^2+(-4)^2}=\sqrt{52}=2\sqrt{13}\ \mathrm{m}$, which is about $7.2\ \mathrm{m}$.

Let the combined mass of boxes P and Q be $M$. The lift itself has mass $m$, so the total moving mass is $m+M$.

Taking upward as positive, $\dfrac{42}{5}mg-(m+M)g=(m+M)\dfrac{2g}{5}$.

Move the weight term to the right: $\dfrac{42}{5}mg=(m+M)\left(g+\dfrac{2g}{5}\right)=\dfrac{7g}{5}(m+M)$. Hence $42m=7(m+M)$ and $M=5m$.

For box P alone, the upward force from Q is $\dfrac{14}{5}mg$, while its weight is $pg$. The acceleration is still $\dfrac{2g}{5}$ upward.

$\dfrac{14}{5}mg-pg=p\dfrac{2g}{5}$.

Divide by $g$: $\dfrac{14}{5}m=p+\dfrac{2p}{5}=\dfrac{7p}{5}$. Therefore $p=2m$.

Let the speed at P be $u$. Over the $10$ seconds from P to Q, the speed changes uniformly from $u$ to $28$, and the distance is $220$ m.

$220=\dfrac{u+28}{2}\cdot10=5(u+28)$. Hence $u+28=44$ and $u=16\ \mathrm{m\,s^{-1}}$.

Use $v=u+at$: $28=16+10a$, so $a=1.2\ \mathrm{m\,s^{-2}}$.

$s=16t+\dfrac12(1.2)t^2=16t+0.6t^2$.

The fifth second is from $t=4$ to $t=5$. $s(5)=16(5)+0.6(25)=95$ and $s(4)=16(4)+0.6(16)=73.6$.

Distance in the fifth second $=95-73.6=21.4\ \mathrm{m}$.

Let the tension at A be $T$. The tension at C is then $T+20$. Since $CB=1.5$ m and $AB=6$ m, $AC=4.5$ m.

The upward tensions balance the weight: $T+(T+20)=W$, so $2T+20=W$.

The centre of mass is $2.625$ m from A. Taking moments about A gives $4.5(T+20)=2.625W$.

From the moments equation, $T+20=\dfrac{2.625}{4.5}W=\dfrac{7}{12}W$. Also $T+20=\dfrac{W+20}{2}$. Hence $\dfrac{W+20}{2}=\dfrac{7W}{12}$, so $6W+120=7W$ and $W=120$ N.

Modelling the beam as a rod means it stays straight and rigid, so the distances used in the moments equations do not change.

Taking upward as positive: ; $u = U$ m s$^{-1}$ (unknown, upward) ; $s = -12$ m (displacement from A to B, downward) ; $t = 4$ seconds ; $a = -9.8$ m s$^{-2}$ (gravity, downward)

$s = ut + \frac{1}{2}at^2$ $-12 = U(4) + \frac{1}{2}(-9.8)(4)^2$ $-12 = 4U + \frac{1}{2}(-9.8)(16)$ $-12 = 4U - 78.4$ $4U = -12 + 78.4$ $4U = 66.4$ $U = 16.6 \text{ m s}^{-1}$

$v^2 = u^2 + 2as$ $v^2 = (16.6)^2 + 2(-9.8)(-12)$ $v^2 = 275.56 + 235.2$ $v^2 = 510.76$ $v = \pm 22.6$ Since particle is moving downward at B, $v = -22.6$ m s$^{-1}$ (negative)

Speed $=|v|=22.6\ \mathrm{m\,s^{-1}}$, or about $23\ \mathrm{m\,s^{-1}}$ to 2 significant figures.

The particle: ; Starts at A with speed $U = 16.6$ m s$^{-1}$ upward ; Decelerates (speed decreases) as it goes up ; Reaches maximum height when speed = 0 ; Accelerates (speed increases) as it falls down ; Hits ground at B with speed 22.6 m s$^{-1}$ Time to reach maximum height: $v = u + at$ $0 = 16.6 - 9.8t$ $t = \frac{16.6}{9.8} = 1.69 \text{ seconds}$

Key Features: ; V-shaped graph (straight lines due to constant acceleration) ; First line: negative gradient (speed decreasing) ; Second line: positive gradient (speed increasing) ; Vertex on horizontal axis (speed = 0 at max height) ; Second line is longer and steeper (falls from greater height)

From rest, with downward displacement $1.5$ m and acceleration $g$, use $s=ut+\dfrac12at^2$.

$1.5=0+\dfrac12(9.8)t^2$, so $t^2=\dfrac{3}{9.8}$ and $t=0.553\ldots\ \mathrm{s}$.

Now the time is $0.6$ s. Again starting from rest, $1.5=\dfrac12a(0.6)^2=0.18a$, so $a=8.333\ldots\ \mathrm{m\,s^{-2}}$ downward.

Taking downward as positive, weight is $0.2g$ and resistance $R$ acts upward. Therefore $0.2g-R=0.2a$.

$1.96-R=0.2(8.333\ldots)=1.666\ldots$, so $R=0.293\ldots\ \mathrm{N}$.

The branch has length $1.5$ m. Since $AC=0.24$ m and $DB=0.36$ m, $CB=1.26$ m and $CD=0.90$ m. The centre of mass is $x$ m from C.

With the $150$ N upward force at B, the branch is on the point of tilting about C, so the support at D has zero reaction.

The upward force at B balances the weight moment: $150(1.26)=Wx$. Thus $Wx=189$.

With the $225$ N downward force at B, the branch is on the point of tilting about D, so the support at C has zero reaction.

The distance from D to the centre of mass is $0.90-x$. Hence $225(0.36)=W(0.90-x)$.

From the second equation, $81=0.90W-Wx$. Since $Wx=189$, $81=0.90W-189$, so $0.90W=270$ and $W=300\ \mathrm{N}$.

Using $Wx=189$, $300x=189$, so $x=0.63\ \mathrm{m}$.

At the highest point B, the velocity is $0$ and $AB=19.6$ m. Use $v^2=u^2+2as$ upward: $0=u^2-2g(19.6)$.

$u^2=2(9.8)(19.6)=384.16$, so $u=19.6\ \mathrm{m\,s^{-1}}$.

From A to the ground, taking upward as positive, displacement is $-24.5$ m. Use $s=ut+\dfrac12at^2$: $-24.5=19.6T-4.9T^2$.

$4.9T^2-19.6T-24.5=0$. Divide by $4.9$: $T^2-4T-5=0$, so $(T-5)(T+1)=0$. The positive time is $T=5$ s.

The speed starts at $19.6$, decreases linearly to $0$ at $t=2$, then increases linearly until the ball hits the ground at $t=5$. The final speed is $29.4$, so the second line is longer than the first.

$\mathbf{r}=(2-3)\mathbf{i}+(1-2(2))\mathbf{j}=-\mathbf{i}-3\mathbf{j}$.

The vector is in the third quadrant. The angle between $\mathbf{r}$ and the positive $\mathbf{j}$ direction satisfies $\tan\alpha=\dfrac{1}{3}$ from the vertical comparison.

$\alpha=18.4^\circ$ from the negative $\mathbf{j}$ direction, so the size of the angle with positive $\mathbf{j}$ is $180^\circ-18.4^\circ=161.6^\circ$, which is $162^\circ$.

Distance from O is $2.5$, so $(t-3)^2+(1-2t)^2=2.5^2$.

$t^2-6t+9+1-4t+4t^2=6.25$, so $5t^2-10t+3.75=0$.

Multiply by $4$: $20t^2-40t+15=0$, or $4t^2-8t+3=0$. Hence $(2t-1)(2t-3)=0$, so $t=\dfrac12$ or $t=\dfrac32$.

Using $s=ut+\dfrac12at^2$, the first $3$ seconds give $20=3u+\dfrac12a(3^2)=3u+\dfrac92a$.

The distance in the fourth second is $u+a(4-\dfrac12)=u+\dfrac72a$. Hence $10=u+\dfrac72a$.

From $10=u+\dfrac72a$, $u=10-\dfrac72a$. Substitute into $20=3u+\dfrac92a$: $20=30-\dfrac{21}{2}a+\dfrac92a=30-6a$. Thus $a=\dfrac53$.

$u=10-\dfrac72\cdot\dfrac53=10-\dfrac{35}{6}=\dfrac{25}{6}$.

At B, speed is $20$. Use $v=u+at$: $20=\dfrac{25}{6}+\dfrac53t$.

$\dfrac53t=20-\dfrac{25}{6}=\dfrac{95}{6}$, so $t=\dfrac{95}{10}=9.5\ \mathrm{s}$.

The beam is $5$ m long. Since $CB=1.5$ m, $AC=3.5$ m. The centre of mass is $1.75$ m from A, so it is also $1.75$ m from C.

For the largest possible $M$ at B, the rope at A is just slack, so the beam is supported at C only.

The particle at B is $1.5$ m to the right of C, and the beam's weight acts $1.75$ m to the left of C. Hence $Mg(1.5)=12g(1.75)$.

$M=\dfrac{12(1.75)}{1.5}=14$.

Let the tension at A be $T$. The tension at C is $2T$. Vertical equilibrium gives $T+2T=12g+15g=27g$, so $T=9g$ and the tension at C is $18g$.

For the new particle at $AD=x$, moments about A give $18g(3.5)=12g(1.75)+15gx$.

Cancel $g$: $63=21+15x$. Thus $15x=42$ and $x=2.8$ m.

The rod is 6 m long and C is 0.4 m from B, so $AC=5.6$ m. The rod is uniform, so its weight $30g$ acts at the midpoint, 3 m from A. The 20 kg particle at B has weight $20g$.

Moments about A remove the tension at A. Clockwise moments from the rod and particle equal the anticlockwise moment from the rope at C: $T_C(5.6)=30g(3)+20g(6)$.

$T_C(5.6)=210g$, so $T_C=37.5g=367.5\ \mathrm{N}$. To suitable accuracy, $T_C=368\ \mathrm{N}$, or $370\ \mathrm{N}$ to 2 significant figures.

Vertically, $T_A+T_C=30g+20g=50g$. Therefore $T_A=50g-37.5g=12.5g=122.5\ \mathrm{N}$, so $T_A=123\ \mathrm{N}$, or $120\ \mathrm{N}$ to 2 significant figures.

For the greatest possible mass at B, the rope at A is just slack, so $T_A=0$. Take moments about C: the clockwise moment of $Mg$ at B balances the anticlockwise moment of the rod weight.

The distance from C to B is $0.4$ m and the rod weight acts $5.6-3=2.6$ m to the left of C. Hence $Mg(0.4)=30g(2.6)$, so $M=\dfrac{78}{0.4}=195$.

At maximum height, velocity = 0 ; $u = 5$ m s$^{-1}$ (initial velocity) ; $v = 0$ m s$^{-1}$ (at maximum height) ; $a = -9.8$ m s$^{-2}$ (gravity, downward) ; $s = ?$ (displacement above A)

$v^2 = u^2 + 2as$ $0^2 = 5^2 + 2(-9.8)s$ $0 = 25 - 19.6s$ $19.6s = 25$ $s = \frac{25}{19.6}$ $s = 1.276...$ $s = 1.28 \text{ m (3 s.f.)}$

Taking upward as positive, with origin at ground level: $s = s_0 + ut + \frac{1}{2}at^2$ For P (starting at A, 10 m above ground): $s_P = 10 + 5t + \frac{1}{2}(-9.8)t^2$ $s_P = 10 + 5t - 4.9t^2$

Q starts at $t = 1$, so Q has been moving for $(t-1)$ seconds: For Q (starting at B, ground level): $s_Q = 0 + 7(t-1) + \frac{1}{2}(-9.8)(t-1)^2$ $s_Q = 7(t-1) - 4.9(t-1)^2$ $s_Q = 7t - 7 - 4.9(t^2 - 2t + 1)$ $s_Q = 7t - 7 - 4.9t^2 + 9.8t - 4.9$ $s_Q = -4.9t^2 + 16.8t - 11.9$

$10 + 5T - 4.9T^2 = -4.9T^2 + 16.8T - 11.9$ $10 + 5T = 16.8T - 11.9$ $10 + 11.9 = 16.8T - 5T$ $21.9 = 11.8T$ $T = \frac{21.9}{11.8}$ $T = 1.855...$ $T = 1.86 \text{ s (3 s.f.)}$

$v = u + at$ For particle P: ; $u = 5$ m s$^{-1}$ ; $a = -9.8$ m s$^{-2}$ ; $t = 1.86$ s $v_P = 5 + (-9.8)(1.86)$ $= 5 - 18.228$ $= -13.228...$ $= -13.2 \text{ m s}^{-1}$ The negative sign indicates P is moving downward.

Speed = $|v|$ $\text{Speed} = |-13.2|$ $= 13.2 \text{ m s}^{-1}$

From P to Q, the speed changes uniformly from $u$ to $2u$ over $12$ s. The average speed is $\dfrac{u+2u}{2}=\dfrac{3u}{2}$.

Distance $PQ=6x$, so $6x=\dfrac{3u}{2}\cdot12=18u$. Hence $x=3u$.

The distance $QR=5x=15u$. The speed changes from $2u$ to $3u$ with acceleration $1.5\ \mathrm{m\,s^{-2}}$.

Use $v^2=u^2+2as$: $(3u)^2=(2u)^2+2(1.5)(15u)$. Thus $9u^2=4u^2+45u$, so $5u^2=45u$ and $u=9$.

In the first $12$ seconds, the distance is $PQ=6x=18u=18(9)=162$ m.

From $t=12$ to $t=14$, the vehicle starts this part at speed $2u=18$ and accelerates at $1.5$. Distance $=18(2)+\dfrac12(1.5)(2^2)=36+3=39$ m.

Total distance in the first $14$ seconds is $162+39=201\ \mathrm{m}$.

The speed is less than $14.7\ \mathrm{m\,s^{-1}}$ near the highest point. From speed $14.7$ to speed $0$, use $v^2=u^2+2as$: $0=14.7^2-2gs$.

This gives $s=11.025$ m on the way up. The same distance occurs on the way down before the speed reaches $14.7$ again, so the total distance is $2s=22.05\ \mathrm{m}$, about $22.1\ \mathrm{m}$.

The ball has speed $29.4$ again when it passes A on the way down. It is faster than $29.4$ only after passing A downward until it hits the ground.

From A to the ground, taking downward as positive, $19.6=29.4t+\dfrac12gt^2$.

$19.6=29.4t+4.9t^2$, giving the positive root $t=0.606\ldots\ \mathrm{s}$.

The graph starts at speed $29.4$ when $t=0$, decreases linearly to speed $0$ at $t=3$, then increases linearly until the ball hits the ground. The final speed is greater than $29.4$, so the right-hand arm ends above the starting height on the speed axis.

For vertical equilibrium, $R+200\sin15^\circ+T\sin25^\circ=20g$.

For the smallest $T$, friction is limiting and balances the remaining horizontal tendency. Horizontally, $200\cos15^\circ-T\cos25^\circ-F=0$.

$F=0.3R$.

From the vertical equation, $R=20g-200\sin15^\circ-T\sin25^\circ$. Substitute into $F=0.3R$ and then into the horizontal equation.

$200\cos15^\circ-T\cos25^\circ-0.3(20g-200\sin15^\circ-T\sin25^\circ)=0$. Solving gives $T=192.3\ldots\ \mathrm{N}$, so the smallest value is about $192\ \mathrm{N}$.

The ball is released from rest and falls $40$ m. Use $s=ut+\dfrac12gt^2$: $40=\dfrac12gt_1^2$. Thus $t_1=\sqrt{\dfrac{80}{g}}=\dfrac{20}{7}$ s.

Using $v^2=u^2+2gs$, $v_1^2=2g(40)$, so $v_1=28\ \mathrm{m\,s^{-1}}$.

After the first bounce the speed is reduced by $50\%$, so the upward speed is $14\ \mathrm{m\,s^{-1}}$.

The time to go up and return to the ground is $\dfrac{2(14)}{g}=\dfrac{20}{7}$ s.

Total time to the second hit is $\dfrac{20}{7}+\dfrac{20}{7}=\dfrac{40}{7}=5.71$ s approximately.

After each bounce, the speed is halved. The upward speeds after the first, second and third bounces are $14$, $7$ and $3.5\ \mathrm{m\,s^{-1}}$.

At the top after the third bounce, $0=(3.5)^2-2gh$. Therefore $h=\dfrac{(3.5)^2}{2g}=0.625$ m.

Distance is the area under a speed-time graph. The three areas are a triangle from $0$ to $12$, a rectangle from $12$ to $31$, and a trapezium from $31$ to $41$.

$\dfrac12(12)(3U)+19(3U)+\dfrac12(3U+2U)(10)=600$. This gives $18U+57U+25U=600$, so $100U=600$ and $U=6$.

Since $U=6$, the final speed $6\ \mathrm{m\,s^{-1}}$ occurs during the first section. The acceleration there is $\dfrac{18-0}{12}=1.5\ \mathrm{m\,s^{-2}}$, so the time to reach $6$ is $4$ s. The distance is $\dfrac12(4)(6)=12\ \mathrm{m}$.

From $0$ to $12$, $a=\dfrac{18}{12}=1.5\ \mathrm{m\,s^{-2}}$. From $31$ to $41$, the speed falls from $18$ to $12$, so the deceleration is $\dfrac{18-12}{10}=0.6\ \mathrm{m\,s^{-2}}$.

The acceleration-time graph has three horizontal sections: $1.5$ from $0\leq t\leq 12$, $0$ from $12<t\leq31$, and $-0.6$ from $31<t\leq41$.

(i) Acceleration for $0 \leq t \leq 10$ $\text{Acceleration} = \frac{\Delta v}{\Delta t} = \text{gradient of v-t graph}$ $a = \frac{30 - 0}{10 - 0}$ $a = 3 \text{ m s}^{-2}$ (ii) Deceleration for $30 \leq t \leq 35$ $a = \frac{0 - 30}{35 - 30}$ $a = \frac{-30}{5}$ $a = -6 \text{ m s}^{-2}$ Deceleration = $|-6| = 6$ m s$^{-2}$

Analysis of each section: ; $0 \leq t \leq 10$: Constant acceleration = 3 m s$^{-2}$ ; $10 < t \leq 30$: Constant velocity, so acceleration = 0 m s$^{-2}$ ; $30 < t \leq 35$: Constant deceleration = $-6$ m s$^{-2}$

$\text{Distance} = \text{Area under v-t graph}$ The shape is a trapezium. We can split it into regions: Method 1: Trapezium formula $\text{Area} = \frac{1}{2}(a + b)h$ Where the parallel sides are the time intervals and height is velocity. Better to calculate as sum of shapes: Method 2: Sum of triangle + rectangle + triangle $\text{Area}_1 = \frac{1}{2} \times 10 \times 30 = 150 \text{ m}$ $\text{Area}_2 = 20 \times 30 = 600 \text{ m}$ $\text{Area}_3 = \frac{1}{2} \times 5 \times 30 = 75 \text{ m}$ $\text{Total distance} = 150 + 600 + 75$ $= 825 \text{ m}$

Area under v-t graph from $t = 0$ to $t = 20$: $\text{Distance}_{\text{car}} = \text{Triangle} + \text{Rectangle}$ $= \frac{1}{2}(10)(30) + (10)(30)$ $= 150 + 300$ $= 450 \text{ m}$

Motorcycle starts at $t = 5$ with constant acceleration $A$. Time of travel: $20 - 5 = 15$ seconds $s = ut + \frac{1}{2}at^2$ $\text{Distance}_{\text{motorcycle}} = 0 + \frac{1}{2}A(15)^2$ $= \frac{225A}{2}$ $= 112.5A \text{ m}$

Motorcycle catches up with car when $t = 20$: $112.5A = 450$ $A = \frac{450}{112.5}$ $A = 4 \text{ m s}^{-2}$

Take moments about A. The rope at C is $2.1$ m from A and the weight of the uniform rod acts at its midpoint, $1.5$ m from A. Hence $350(2.1)=W(1.5)$.

$735=1.5W$, so $W=490\ \mathrm{N}$.

When P is attached, the tension at C is $600$ N. Taking moments about A: $600(2.1)=490(1.5)+210d$.

$1260=735+210d$, so $525=210d$ and $d=2.5$.

When Q is attached at B and the rod is on the point of tilting, the rope at A is just slack. The rod is about to tilt about C.

Take moments about C. The rod weight acts $2.1-1.5=0.6$ m to the left of C, and Q acts $3-2.1=0.9$ m to the right. Thus $490(0.6)=X(0.9)$, so $X=\dfrac{980}{3}\ \mathrm{N}$.

With constant acceleration, distance equals average speed times time. So $400=\dfrac{u+28}{2}\cdot20$.

$400=10(u+28)$, so $u+28=40$ and $u=12$.

Use $v=u+at$: $28=12+20a$, hence $a=0.8\ \mathrm{m\,s^{-2}}$.

The midpoint of AB is $200$ m from A. Use $s=ut+\dfrac12at^2$: $200=12t+\dfrac12(0.8)t^2$.

$200=12t+0.4t^2$. The positive solution is $t=5\sqrt{29}-15=11.9258\ldots\ \mathrm{s}$.

Let the driving force be $D$. The resultant forward force is $D-260$. Since $ma=1200(0.8)=960$, $D-260=960$.

$D=1220\ \mathrm{N}$.

From $0$ to $4$ s the acceleration is $-0.5\ \mathrm{m\,s^{-2}}$. Therefore $w=8+(-0.5)(4)=6\ \mathrm{m\,s^{-1}}$.

From $10$ to $20$ s the acceleration is $1.2\ \mathrm{m\,s^{-2}}$, while the speed at $t=10$ is still $6$. Hence $v=6+1.2(10)=18\ \mathrm{m\,s^{-1}}$.

The graph starts at $(0,8)$, falls linearly to $(4,6)$, stays horizontal to $(10,6)$, then rises linearly to $(20,18)$.

The distance is the area under the graph. From $0$ to $4$: $\dfrac{8+6}{2}\cdot4=28$. From $4$ to $10$: $6\cdot6=36$. From $10$ to $20$: $\dfrac{6+18}{2}\cdot10=120$.

$AB=28+36+120=184\ \mathrm{m}$.

Take downward as positive during the impact. Before impact, the hammer has momentum $1.8(10)$ and the peg has momentum $0$. After impact, the combined mass is $1.8+0.2=2.0$ kg.

$1.8(10)=2.0v$, so $18=2v$ and $v=9\ \mathrm{m\,s^{-1}}$.

The peg changes from rest to speed $9\ \mathrm{m\,s^{-1}}$ downward. The impulse magnitude is $0.2(9-0)=1.8\ \mathrm{N\,s}$.

The hammer and peg move together from speed $9$ to rest over $0.12$ m. With downward positive, $0=9^2+2a(0.12)$, so $a=-337.5\ \mathrm{m\,s^{-2}}$.

For the combined mass $2.0$ kg, forces are weight $2g$ downward and resistance $R$ upward. With downward positive, $2g-R=2a$.

$19.6-R=2(-337.5)=-675$, so $R=694.6\ \mathrm{N}$, which is about $695\ \mathrm{N}$.

The train reaches $20\ \mathrm{m\,s^{-1}}$ in $T$ seconds. Since the acceleration magnitude is twice the deceleration magnitude, the time needed to decelerate from $20$ to $0$ is $2T$.

The speed-time graph rises linearly from $(0,0)$ to $(T,20)$, stays horizontal at speed $20$ from $T$ to $T+180$, then falls linearly to zero at $3T+180$.

The distance from A to B is $4.8\ \mathrm{km}=4800\ \mathrm{m}$. The area is $\dfrac12(T)(20)+20(180)+\dfrac12(2T)(20)$.

$4800=10T+3600+20T$, so $1200=30T$ and $T=40\ \mathrm{s}$.

During the first $T$ seconds, $20=aT$. With $T=40$, $a=\dfrac{20}{40}=0.5\ \mathrm{m\,s^{-2}}$.

Take P's original direction as positive. Before collision, P has velocity $3u$ and Q is at rest. After collision, Q has velocity $2u$. Let P's velocity be $v$.

$2m(3u)=2mv+4m(2u)$. Thus $6mu=2mv+8mu$, so $2mv=-2mu$ and $v=-u$.

The speed of P is $u$. The negative sign shows that P moves in the opposite direction after the collision.

After collision, Q starts with speed $2u$ and stops after distance $\dfrac{6u^2}{g}$. Use $v^2=u^2+2as$: $0=(2u)^2+2a\left(\dfrac{6u^2}{g}\right)$.

$0=4u^2+\dfrac{12au^2}{g}$, so $a=-\dfrac{g}{3}$. The deceleration magnitude is $\dfrac{g}{3}$.

For Q, $R=4mg$, so friction is $\mu R=4\mu mg$. This friction produces deceleration: $4\mu mg=4m\cdot\dfrac{g}{3}$.

Cancel $4mg$: $\mu=\dfrac13$.

From A to the sign, $u=25$, $a=-6$ and $s=48$. Use $v^2=u^2+2as$.

$v^2=25^2+2(-6)(48)=625-576=49$, so $v=7\ \mathrm{m\,s^{-1}}$.

Use $v=u+at$: $13=25-6t$, so $t=2$ s to slow to $13\ \mathrm{m\,s^{-1}}$.

Distance while slowing is $\dfrac{25+13}{2}\cdot2=38$ m.

The remaining distance is $48-38=10$ m at speed $13$. Total time $=2+\dfrac{10}{13}=\dfrac{36}{13}$ s.

The car first travels at $25\ \mathrm{m\,s^{-1}}$ for $0.2$ s, covering $5$ m.

It then takes another $2$ s to slow from $25$ to $13$, covering $38$ m as before. Total covered before constant-speed part is $5+38=43$ m.

The remaining distance is $48-43=5$ m at speed $13$. Total time $=0.2+2+\dfrac{5}{13}=\dfrac{168}{65}$ s.

Since $\tan\alpha=\dfrac43$, $\sin\alpha=\dfrac45$ and $\cos\alpha=\dfrac35$.

The vertical component of the thrust is $2.5\cos\alpha=2.5\cdot\dfrac35=1.5$ N upward. The weight is $0.2g=1.96$ N downward.

For equilibrium, friction must supply the missing upward force: $F=1.96-1.5=0.46\ \mathrm{N}$. So friction acts upwards.

When $T=6.125$, the vertical component of the thrust is larger than the weight, so friction acts downward at the limiting value.

The normal reaction is the horizontal component of the thrust: $R=6.125\sin\alpha=6.125\cdot\dfrac45=4.9\ \mathrm{N}$.

Vertical equilibrium gives $0.2g+F=6.125\cos\alpha$. Thus $1.96+F=6.125\cdot\dfrac35=3.675$, so $F=1.715$ N.

$F=\mu R$, so $\mu=\dfrac{1.715}{4.9}=0.35$.

The beam length is $8a$. C is at distance $a$ from A, and D is at distance $6a$ from A because $DB=2a$.

Let the reaction at C be $R$. The reaction at D is $3R$.

$R+3R=60g$, so $4R=60g$ and $R=15g$.

The beam's weight acts at distance $x$ from A. Hence $60g x=R(a)+3R(6a)$.

Substitute $R=15g$: $60g x=15ga+45g(6a)=285ga$. Thus $x=\dfrac{285a}{60}=\dfrac{19a}{4}$.

When the box is added, let each support reaction be $S$. Vertical equilibrium gives $2S=(60+M)g$.

The box is at $AE=2a$. Moments about A give $60g\left(\dfrac{19a}{4}\right)+Mg(2a)=S(a)+S(6a)=7aS$.

Substitute $S=\dfrac{(60+M)g}{2}$ and cancel $ag$: $285+2M=\dfrac{7}{2}(60+M)=210+\dfrac72M$. Hence $75=\dfrac32M$ and $M=50$.

The car's graph is a horizontal line at $40\ \mathrm{m\,s^{-1}}$. The motorcyclist's graph starts at zero, rises with gradient $4$ until $t=15$, then rises with smaller gradient $1$ for a further $T$ seconds.

Before $t=15$, the motorcyclist's speed is $4t$. Set this equal to the car's speed: $4t_1=40$, so $t_1=10$ s.

After the first $15$ seconds, the motorcyclist's speed is $4(15)=60\ \mathrm{m\,s^{-1}}$.

During the next $T$ seconds the acceleration is $1\ \mathrm{m\,s^{-2}}$, so the speed at catch-up is $60+T$.

Motorcyclist distance $=\dfrac12(15)(60)+\dfrac12[60+(60+T)]T$. Car distance $=40(15+T)$.

$450+60T+\dfrac12T^2=600+40T$. Hence $\dfrac12T^2+20T-150=0$, so $T^2+40T-300=0$ and $k=40$.

Given: $\tan\theta = \frac{3}{4}$ Using 3-4-5 right triangle: $\sin\theta = \frac{3}{5}, \quad \cos\theta = \frac{4}{5}$

$\perp: \sum F = 0$ $R = mg\cos\theta$ $R = \frac{4mg}{5}$

Package moving UP, so friction acts DOWN the plane: $F = \mu R$ $= \frac{1}{4} \times \frac{4mg}{5}$ $= \frac{mg}{5}$

Taking up the plane as positive: $\parallel: \sum F = ma$ $-mg\sin\theta - F = ma$ $-mg \times \frac{3}{5} - \frac{mg}{5} = ma$ $-\frac{3mg}{5} - \frac{mg}{5} = ma$ $-\frac{4mg}{5} = ma$ $a = -\frac{4g}{5}$ Deceleration = $\left|-\frac{4g}{5}\right| = \frac{4g}{5}$

Package projected up plane, comes to rest at B: ; $u = U$ m s$^{-1}$ (unknown) ; $v = 0$ m s$^{-1}$ (comes to rest) ; $a = -\frac{4g}{5}$ m s$^{-2}$ (from part a) ; $s = 1.5$ m

$v^2 = u^2 + 2as$ $0^2 = U^2 + 2\left(-\frac{4g}{5}\right)(1.5)$ $0 = U^2 - 2 \times \frac{4 \times 9.8}{5} \times 1.5$ $0 = U^2 - \frac{8 \times 9.8 \times 1.5}{5}$ $0 = U^2 - 23.52$ $U^2 = 23.52$ $U = 4.85 \text{ m s}^{-1}$ Or using exact form: $U = \sqrt{2 \times \frac{4g}{5} \times 1.5} = \sqrt{\frac{12g}{5}}$

Package held at rest on slope by force X up the plane. Without X, package would slide down (component down > friction up). For MINIMUM X, friction acts UP the plane (at limiting friction).

Mass = 2 kg, so weight = $2g$ N $R = 2g\cos\theta$ $= 2g \times \frac{4}{5}$ $= \frac{8g}{5} \text{ N}$

Friction acts UP the plane (limiting): $F = \mu R$ $= \frac{1}{4} \times \frac{8g}{5}$ $= \frac{2g}{5} \text{ N}$

For equilibrium (package at rest): $\parallel: \sum F = 0$ Taking up plane as positive: $X + F - 2g\sin\theta = 0$ $X + \frac{2g}{5} - 2g \times \frac{3}{5} = 0$ $X + \frac{2g}{5} - \frac{6g}{5} = 0$ $X = \frac{6g}{5} - \frac{2g}{5}$ $X = \frac{4g}{5}$ $X = \frac{4 \times 9.8}{5}$ $X = 7.84 \text{ N}$

The graph starts at speed $5$ when $t=0$, rises linearly to speed $V$ at time $T_1$, then falls linearly to speed $0$ at time $T_1+T_2$.

The area under the first section is the distance $AB=132$. Thus $\dfrac12(5+V)T_1=132$, so $T_1=\dfrac{264}{V+5}$.

The second section is a triangle of height $V$ and base $T_2$. Since $BC=136$, $\dfrac12VT_2=136$, so $T_2=\dfrac{272}{V}$.

The whole journey takes $28$ s, so $\dfrac{264}{V+5}+\dfrac{272}{V}=28$.

Multiplying by $V(V+5)$ gives $264V+272(V+5)=28V(V+5)$. This simplifies to $7V^2-99V-340=0$, so $(V-17)(7V+20)=0$. Since speed is positive, $V=17$.

For the final section, $T_2=\dfrac{272}{17}=16$ s. The speed falls from $17$ to $0$, so the deceleration is $\dfrac{17}{16}=1.0625\ \mathrm{m\,s^{-2}}$.

For ball A, take upward as positive. Returning to the ground gives $0=8T_1-4.9T_1^2$. The non-zero solution is $T_1=\dfrac{8}{4.9}=1.63\ \mathrm{s}$ to 3 significant figures.

After impact, A rebounds to a maximum height of $2$ m. At the top, $v=0$, so $0=u^2-2g(2)$. Hence $u=\sqrt{4g}=6.261\ \mathrm{m\,s^{-1}}$.

Just before the first impact, A has velocity $-8\ \mathrm{m\,s^{-1}}$. Just after impact, it has velocity $+6.261\ \mathrm{m\,s^{-1}}$. With mass $0.1$ kg, $I=0.1[6.261-(-8)]$.

$I=1.426\ldots\ \mathrm{N\,s}$, so the impulse is $1.43\ \mathrm{N\,s}$ to 3 significant figures.

For the first meeting with B, A is still in its first flight. Its height at time $T_2$ is $h_A=8T_2-4.9T_2^2$.

B starts at time $t=1$, so at time $T_2$ it has been moving for $T_2-1$ seconds. Its height is $h_B=5(T_2-1)-4.9(T_2-1)^2$.

$8T_2-4.9T_2^2=5(T_2-1)-4.9(T_2-1)^2$. Expanding and simplifying gives $6.8T_2=9.9$.

$T_2=\dfrac{9.9}{6.8}=1.456\ldots$, so $T_2=1.46\ \mathrm{s}$ to 3 significant figures.

The car graph is a triangle from $0$ to $30$ plus a rectangle from $30$ to $T$. Distance $=\dfrac12(30)V+V(T-30)$.

$\dfrac12(30)V+V(T-30)=15V+VT-30V=V(T-15)$, as required.

The motorbike starts at $(10,0)$, accelerates with the same gradient as the car until $t=50$, then travels at constant speed until $t=60$, then decelerates linearly to $(T,V)$.

The car acceleration is $\dfrac{V}{30}$. The motorbike accelerates for $40$ s, so its speed is $40\cdot\dfrac{V}{30}=\dfrac{4V}{3}$.

Its distance is $\dfrac12(40)\left(\dfrac{4V}{3}\right)+10\left(\dfrac{4V}{3}\right)+\dfrac12\left(\dfrac{4V}{3}+V\right)(T-60)$.

This becomes $\dfrac{80V}{3}+\dfrac{40V}{3}+\dfrac{7V}{6}(T-60)=40V+\dfrac{7V}{6}(T-60)=\dfrac{7VT}{6}-30V$.

At time $T$, the distances are equal: $V(T-15)=\dfrac{7VT}{6}-30V$. Divide by $V$ to get $T-15=\dfrac{7T}{6}-30$.

Multiplying by $6$ gives $6T-90=7T-180$, so $T=90$.

For car plus trailer, the total mass is $1800$ kg and the total resistance is $2R+R=3R$. Taking up the slope as positive: $3000-1800g\sin\alpha-3R=1800(0.75)$.

Since $\sin\alpha=\dfrac1{12}$, $1800g\sin\alpha=1800\cdot9.8\cdot\dfrac1{12}=1470$. So $3000-1470-3R=1350$, giving $180=3R$ and $R=60\ \mathrm{N}$.

For the trailer while it is being towed, $T-600g\sin\alpha-R=600(0.75)$.

$T-600\cdot9.8\cdot\dfrac1{12}-60=450$, so $T-490-60=450$ and $T=1000\ \mathrm{N}$.

After the tow bar breaks, the trailer is still moving up the slope, but the component of weight and resistance act down the slope. Taking up the slope as positive: $-600g\sin\alpha-60=600a$.

$-490-60=600a$, so $a=-\dfrac{550}{600}=-\dfrac{11}{12}\ \mathrm{m\,s^{-2}}$.

With initial speed $12\ \mathrm{m\,s^{-1}}$ and final speed $0$, use $v^2=u^2+2as$: $0=12^2+2\left(-\dfrac{11}{12}\right)d$. Hence $d=\dfrac{864}{11}=78.5\ldots\ \mathrm{m}$.

For P, at greatest height the speed is zero. Use $v^2=u^2+2as$ with $u=U$, $v=0$, $a=-g$ and $s=H$: $0=U^2-2gH$. Hence $H=\dfrac{U^2}{2g}$.

At this instant P is at height $H$ and has zero velocity. After $t$ seconds, P has fallen $\dfrac12gt^2$, so its height above O is $H-\dfrac12gt^2$.

Q is projected upwards from O with speed $\dfrac12U$, so its height after $t$ seconds is $\dfrac12Ut-\dfrac12gt^2$.

At collision, $H-\dfrac12gt^2=\dfrac12Ut-\dfrac12gt^2$. The gravity terms cancel, so $H=\dfrac12Ut$.

Substitute $H=\dfrac{U^2}{2g}$: $\dfrac{U^2}{2g}=\dfrac12Ut$. Therefore $t=\dfrac{U}{g}$.

Substitute $t=\dfrac{U}{g}$ into P's height: $H-\dfrac12g\left(\dfrac{U}{g}\right)^2=\dfrac{U^2}{2g}-\dfrac{U^2}{2g}=0$. Thus they collide at O.

On a horizontal surface, $R=mg$ and the friction is $F=\mu R=\dfrac17mg$. The friction is the only horizontal force, so the acceleration is opposite the motion with magnitude $\dfrac{F}{m}=\dfrac{g}{7}=1.4\ \mathrm{m\,s^{-2}}$.

From A to B, $u=U$, $v=\dfrac{U}{2}$, $a=-\dfrac{g}{7}$ and $t=0.75$. Using $v=u+at$: $\dfrac{U}{2}=U-\dfrac{g}{7}(0.75)$.

With $g=9.8$, $\dfrac{g}{7}(0.75)=1.05$, so $\dfrac{U}{2}=U-1.05$. Hence $U=2.10\ \mathrm{m\,s^{-1}}$, which is $2.1\ \mathrm{m\,s^{-1}}$ to 2 significant figures.

From B to C, the speed falls from $\dfrac{U}{2}$ to $\dfrac{U}{3}$ with the same acceleration $-1.4$. Thus $\dfrac{U}{3}=\dfrac{U}{2}-1.4t$. With $U=2.1$, this gives $0.7=1.05-1.4t$, so $t=0.25\ \mathrm{s}$.

Use average speed on each section. $AB=\dfrac{U+U/2}{2}(0.75)=1.18125\ \mathrm{m}$ and $BC=\dfrac{U/2+U/3}{2}(0.25)=0.21875\ \mathrm{m}$. Therefore $AC=1.18125+0.21875=1.4\ \mathrm{m}$.

P starts from rest and immediately after the impulse has speed $5\ \mathrm{m\,s^{-1}}$. Since the mass of P is $3m$, $\lambda=3m(5-0)=15m$.

Between the impulse and the collision, P slows from $5$ to $2.5\ \mathrm{m\,s^{-1}}$ over $1.4$ m. Using $v^2=u^2+2as$: $(2.5)^2=5^2+2a(1.4)$.

$6.25=25+2.8a$, so $a=\dfrac{-18.75}{2.8}=-6.696\ldots\ \mathrm{m\,s^{-2}}$.

On the horizontal surface, $R=3mg$ and friction is $F=\mu R=3\mu mg$. This friction is opposite the motion, so $-3\mu mg=3ma$ and therefore $a=-\mu g$.

$\mu=\dfrac{-a}{g}=\dfrac{6.696\ldots}{9.8}=0.683\ldots$, so $\mu=0.68$ to 2 significant figures.

Immediately before collision, P has speed $2.5$ and Q is at rest. Immediately after, Q has speed $2.1$ and P has speed $v$. Conservation of momentum gives $3m(2.5)+m(0)=3mv+m(2.1)$.

$7.5=3v+2.1$, so $3v=5.4$ and $v=1.8\ \mathrm{m\,s^{-1}}$.

From $\tan\alpha=\dfrac34$, use $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$.

The horizontal force has a component into the plane, so $R=5g\cos\alpha+H\sin\alpha=4g+\dfrac35H$.

For the smallest $H$, the particle is on the point of sliding down, so friction acts up the plane. Parallel to the plane, $H\cos\alpha+F=5g\sin\alpha$.

With $F=\dfrac14R$, the parallel equation becomes $\dfrac45H+\dfrac14\left(4g+\dfrac35H\right)=3g$. Hence $0.95H+g=3g$, so $H=\dfrac{40g}{19}=20.6\ \mathrm{N}$.

When $H$ is removed, $R=5g\cos\alpha=4g$ and friction is $\dfrac14R=g$. The component of weight down the plane is $5g\sin\alpha=3g$, so the resultant down the plane is $2g$.

$2g=5a$, so $a=\dfrac{2g}{5}=3.92\ \mathrm{m\,s^{-2}}$ down the plane.

From rest, the particle moves $1.5$ m with acceleration $2g/5$. Using $s=\dfrac12at^2$: $1.5=\dfrac12\left(\dfrac{2g}{5}\right)T^2=\dfrac{g}{5}T^2$. Thus $T=\sqrt{\dfrac{7.5}{9.8}}=0.875\ \mathrm{s}$.

Given $\tan\theta=\dfrac34$, use $\sin\theta=\dfrac35$ and $\cos\theta=\dfrac45$.

The box moves at constant speed, so it is in equilibrium. Vertically, $R+T\sin\theta=mg$, hence $R=mg-T\sin\theta$.

Horizontally, the tension component balances friction: $T\cos\theta=F$.

With $\mu=\dfrac13$, $F=\dfrac13R$. Therefore $T\cos\theta=\dfrac13(mg-T\sin\theta)$.

Substitute $\sin\theta=\dfrac35$ and $\cos\theta=\dfrac45$: $\dfrac45T=\dfrac13\left(mg-\dfrac35T\right)$. This gives $\dfrac45T=\dfrac{mg}{3}-\dfrac15T$, so $T=\dfrac{mg}{3}$.

On the horizontal plane without the rope, $R=mg$ and friction is $\dfrac13mg$. Hence the acceleration is $-\dfrac{g}{3}$.

When the speed has fallen from $u$ to $\dfrac12u$, use $v^2=u^2+2as$: $\left(\dfrac{u}{2}\right)^2=u^2+2\left(-\dfrac{g}{3}\right)d$.

$\dfrac{u^2}{4}=u^2-\dfrac{2gd}{3}$, so $\dfrac{2gd}{3}=\dfrac{3u^2}{4}$ and $d=\dfrac{9u^2}{8g}$.

Take upward as positive from A. At B, the displacement is $-2.5$ m, with $u=24$ and $a=-9.8$. Use $v^2=u^2+2as$: $V^2=24^2+2(-9.8)(-2.5)=625$.

$V=25\ \mathrm{m\,s^{-1}}$.

The velocity at B is $-25\ \mathrm{m\,s^{-1}}$. Using $v=u+at$, $-25=24-9.8t$, so $t=5\ \mathrm{s}$.

C is $10$ m above A. The particle is above C between the two times when $s=10$: $10=24t-4.9t^2$.

$4.9t^2-24t+10=0$, giving $t=0.459\ldots$ and $t=4.438\ldots$.

The length of time above C is $4.438\ldots-0.459\ldots=3.98\ldots\ \mathrm{s}$, about $4.0\ \mathrm{s}$.

The graph starts at speed $24$, decreases linearly to $0$ at the highest point, then increases linearly to speed $25$ at $t=5$. It is a V-shaped speed-time graph.

Given $\tan\alpha=\dfrac{5}{12}$, use $\sin\alpha=\dfrac{5}{13}$ and $\cos\alpha=\dfrac{12}{13}$.

Q has mass $0.5m$ and the system is in equilibrium, so the tension is $T=0.5mg$.

P is on the point of slipping up the plane, so friction acts down the plane. Along the plane, $T=mg\sin\alpha+F$.

$F=T-mg\sin\alpha=\dfrac12mg-\dfrac{5}{13}mg=\dfrac{3}{26}mg$.

$R=mg\cos\alpha=\dfrac{12}{13}mg$. Since $F=\mu R$, $\mu=\dfrac{3mg/26}{12mg/13}=\dfrac18$.

P moves down the plane, so friction acts up the plane. The resultant down the plane is $mg\sin\alpha-F=\dfrac{5}{13}mg-\dfrac18\cdot\dfrac{12}{13}mg=\dfrac{7}{26}mg$.

As P has mass $m$, $a=\dfrac{7g}{26}$ down the plane.

From rest over $0.8$ m, $V^2=2\left(\dfrac{7g}{26}\right)(0.8)$. With $g=9.8$, $V=2.05\ \mathrm{m\,s^{-1}}$ to 3 significant figures.

The beam is $2$ m long. Since $AC=0.6$ m and $DB=0.2$ m, point D is $1.8$ m from A. The uniform beam has weight $mg$ acting $1.0$ m from A.

Let the tension at D be $T$. The tension at C is four times this, so it is $4T$.

$4T+T=pmg+mg$, so $5T=(p+1)mg$.

The particle at A has no moment about A. Hence $4T(0.6)+T(1.8)=mg(1.0)$.

$2.4T+1.8T=mg$, so $4.2T=mg$ and $T=\dfrac{5mg}{21}$.

Substitute into $5T=(p+1)mg$: $5\cdot\dfrac{5mg}{21}=(p+1)mg$. Thus $p+1=\dfrac{25}{21}$ and $p=\dfrac{4}{21}$.

When the particle of mass $qm$ is at A and the beam is on the point of tilting, the rope at D is just slack, so its tension is zero.

About C, the particle at A has moment $qmg(0.6)$ and the beam weight has moment $mg(0.4)$ in the opposite direction. Thus $qmg(0.6)=mg(0.4)$, giving $q=\dfrac{2}{3}$.

The beam is uniform, so its centre of mass is at the midpoint of the beam and its weight acts there.

The parachutist falls freely from rest for $2.5$ s, so $v=gt=9.8(2.5)=24.5\ \mathrm{m\,s^{-1}}$.

The speed-time graph starts at $0$, rises linearly to $24.5$ at $t=2.5$, then decreases linearly with smaller gradient for $T$ seconds, then stays horizontal at speed $V$ until $t=20$.

During the deceleration phase, initial speed is $24.5$, acceleration is $-3.9$, time is $T$, and distance is $73.75$.

$73.75=24.5T+\dfrac12(-3.9)T^2=24.5T-1.95T^2$.

Solving gives $T=5$ or $T=7.56\ldots$. Since $T<6$, take $T=5$.

$V=24.5-3.9(5)=5\ \mathrm{m\,s^{-1}}$.

Height is the total area under the speed-time graph: first triangle $\dfrac12(2.5)(24.5)=30.625$, deceleration distance $73.75$, and final rectangle $(20-2.5-5)(5)=62.5$.

Total height $=30.625+73.75+62.5=166.875$ m, which rounds to $167$ m.

$\uparrow: \sum F_{\text{vertical}} = 0$ $R + P\sin 30° = mg$ $R + 0.5P = 4 \times 9.8$ $R = 39.2 - 0.5P \quad \text{...(1)}$

At limiting equilibrium (on point of sliding): $F = \mu R$ $F = \frac{2}{3} \times R$ $F = \frac{2}{3}(39.2 - 0.5P)$ $F = 26.133... - \frac{P}{3} \quad \text{...(2)}$

Box on point of sliding means $a = 0$: $\rightarrow: \sum F_{\text{horizontal}} = 0$ $P\cos 30° = F$ $P \times \frac{\sqrt{3}}{2} = 26.133... - \frac{P}{3}$ $0.866P = 26.133 - 0.333P$ $0.866P + 0.333P = 26.133$ $1.199P = 26.133$ $P = 21.8 \text{ N (3 s.f.)}$

From equation (1) in part (a): $R = 39.2 - 0.5P$ $R = 39.2 - 0.5(25)$ $R = 39.2 - 12.5$ $R = 26.7 \text{ N}$

Box is now moving, so: $F = \mu R$ $F = \frac{2}{3} \times 26.7$ $F = 17.8 \text{ N}$

$\rightarrow: \sum F = ma$ $P\cos 30° - F = ma$ $25 \times 0.866 - 17.8 = 4a$ $21.65 - 17.8 = 4a$ $3.85 = 4a$ $a = 0.9625$ $a = 0.96 \text{ m s}^{-2} \text{ (2 s.f.)}$

$u = 0$ m s$^{-1}$ (starts from rest) ; $a = 0.9625$ m s$^{-2}$ (from part b) ; $s = 1.5$ m ; $v = ?$

$v^2 = u^2 + 2as$ $v^2 = 0^2 + 2(0.9625)(1.5)$ $v^2 = 2.8875$ $v = \sqrt{2.8875}$ $v = 1.699...$ $v = 1.7 \text{ m s}^{-1} \text{ (2 s.f.)}$

For cyclist P, the distance from A to B is the area under the speed-time graph from $0$ to $42$. This is $\dfrac12(9)V+33V=120$.

$4.5V+33V=37.5V=120$, so $V=3.2\ \mathrm{m\,s^{-1}}$.

Between $t=0$ and $t=9$, P accelerates from $0$ to $3.2$. Hence $a=\dfrac{3.2}{9}=\dfrac{16}{45}=0.356\ldots\ \mathrm{m\,s^{-2}}$.

Cyclist Q starts at $t=6$ from rest, accelerates in a straight line for $T$ seconds to speed $3.6\ \mathrm{m\,s^{-1}}$, then continues horizontally at $3.6\ \mathrm{m\,s^{-1}}$ until $t=54$.

By $t=54$, P has travelled $\dfrac12(9)(3.2)+45(3.2)=14.4+144=158.4\ \mathrm{m}$.

Q travels for $48$ seconds after starting at $t=6$. Her distance is $\dfrac12T(3.6)+(48-T)(3.6)=172.8-1.8T$.

Q catches P at $t=54$, so $172.8-1.8T=158.4$. Thus $1.8T=14.4$ and $T=8$ s.

P has mass $3$ kg and moves upwards, so taking upwards as positive gives $T-3g=3a$.

Q has mass $5$ kg and moves downwards, so taking downwards as positive for Q gives $5g-T=5a$.

Adding the two equations gives $2g=8a$, so $a=\dfrac{g}{4}$. Substitute into $T-3g=3a$: $T=3g+3\left(\dfrac{g}{4}\right)=\dfrac{15g}{4}=36.8\ \mathrm{N}$.

The pulley is pulled by two vertical tensions, each of magnitude $T$. Hence $F=2T=\dfrac{15g}{2}=73.5\ \mathrm{N}$.

Before the string breaks, Q descends $2$ m from rest with acceleration $g/4$. Thus $v^2=2\left(\dfrac{g}{4}\right)(2)=g$, so $v=\sqrt{g}$.

After the string breaks, Q is still $8$ m above the ground and accelerates freely under gravity. Let the impact speed be $w$. Then $w^2=v^2+2g(8)=g+16g=17g$.

$w=\sqrt{17g}=\sqrt{166.6}=12.9\ \mathrm{m\,s^{-1}}$ to 3 significant figures.

Ajaz starts from rest, accelerates linearly to speed $4\ \mathrm{m\,s^{-1}}$, then continues with a horizontal speed-time line at speed $4$.

Beth starts from rest, has a steeper straight acceleration line for $T$ seconds, then a horizontal line. Both graphs finish at the same final time.

Ajaz reaches speed $4$ with acceleration $0.8$, so the time is $\dfrac{4}{0.8}=5$ s.

Distance during acceleration is $\dfrac12(5)(4)=10$ m. The remaining $90$ m at speed $4$ takes $22.5$ s, so Ajaz's total time is $27.5$ s.

Beth accelerates for $T$ seconds to speed $T$, then runs at speed $T$ until time $27.5$. Her distance is $\dfrac12T^2+T(27.5-T)$.

$100=\dfrac12T^2+27.5T-T^2=27.5T-\dfrac12T^2$. Thus $T^2-55T+200=0$. The valid root is $T=3.915\ldots$ s.

Ajaz finishes at $4\ \mathrm{m\,s^{-1}}$. Beth's constant speed is $T=3.915\ldots\ \mathrm{m\,s^{-1}}$.

Difference $=4-3.915\ldots=0.085\ \mathrm{m\,s^{-1}}$ approximately.

Since $\tan\theta=\dfrac5{12}$, $\sin\theta=\dfrac5{13}$ and $\cos\theta=\dfrac{12}{13}$.

$R=6g\cos\theta=6g\cdot\dfrac{12}{13}=\dfrac{72g}{13}$.

With $\mu=\dfrac14$, $F=\dfrac14R=\dfrac{18g}{13}=13.6\ \mathrm{N}$ approximately.

While moving up, both friction and the component of weight act down the plane. Taking up the plane as positive: $6a=-6g\sin\theta-F$.

$6a=-6g\cdot\dfrac5{13}-\dfrac{18g}{13}=-\dfrac{48g}{13}$, so $a=-\dfrac{8g}{13}$.

Use $v^2=u^2+2as$ with $u=5$ and $v=0$: $0=25+2\left(-\dfrac{8g}{13}\right)AB$. Hence $AB=2.07\ldots\ \mathrm{m}$.

When sliding down, friction acts up the plane. Taking down the plane as positive: $6a=6g\sin\theta-F=\dfrac{30g}{13}-\dfrac{18g}{13}=\dfrac{12g}{13}$. Thus $a=\dfrac{2g}{13}$.

Starting from rest at B and reaching speed $5$ at C: $25=2\left(\dfrac{2g}{13}\right)BC$. Hence $BC=8.29\ldots\ \mathrm{m}$.

Tram A starts at $t=0$, accelerates at $2\ \mathrm{m\,s^{-2}}$ for $5$ s, reaching speed $10\ \mathrm{m\,s^{-1}}$, then continues at this constant speed.

Tram B starts at $t=20$, accelerates at $3\ \mathrm{m\,s^{-2}}$ for $4$ s, reaching speed $12\ \mathrm{m\,s^{-1}}$, then continues at this constant speed. Its graph crosses A's distance later.

Tram A's constant speed is $10$. Tram B reaches speed $10$ after $\dfrac{10}{3}$ s of acceleration, so the time from $t=0$ is $20+\dfrac{10}{3}=\dfrac{70}{3}$ s.

A travels $\dfrac12(5)(10)=25$ m in its acceleration phase, then $10\left(\dfrac{70}{3}-5\right)=\dfrac{550}{3}$ m. Total $=\dfrac{625}{3}$ m.

B is still accelerating from rest, so its distance is $\dfrac12\left(\dfrac{10}{3}\right)(10)=\dfrac{50}{3}$ m.

$d=\dfrac{625}{3}-\dfrac{50}{3}=\dfrac{575}{3}$ m.

For $t>24$, A's distance is $25+10(t-5)=10t-25$. B's distance is $\dfrac12(4)(12)+12(t-24)=12t-264$.

Set $10t-25=12t-264$. Then $2t=239$ and $t=119.5$ s.

Using A's distance, $10(119.5)-25=1170$ m.

For the first 5 seconds, P falls freely from rest. Using $s=ut+\dfrac12at^2$, $s=0+\dfrac12(9.8)(5^2)=122.5\ \mathrm{m}$.

After 5 seconds of free fall, $v=0+9.8(5)=49\ \mathrm{m\,s^{-1}}$ downward.

After the parachute opens, weight is $250g=2450$ N downward and resistance is $3200$ N upward. Taking downward as positive, the resultant is $2450-3200=-750$ N, so $a=\dfrac{-750}{250}=-3\ \mathrm{m\,s^{-2}}$.

The remaining distance is $520-122.5=397.5$ m. Using $v^2=u^2+2as$: $v^2=49^2+2(-3)(397.5)=16$.

$v=4\ \mathrm{m\,s^{-1}}$, so P lands with speed $4\ \mathrm{m\,s^{-1}}$.

Using $v=u+at$, $4=49-3t$, so $t=15$ s after the parachute opens.

The total time is $5+15=20$ s. The speed-time graph rises linearly from $0$ to $49$ over the first 5 s, then falls linearly to $4$ at $t=20$; it does not meet the time-axis.

B moves down and A moves up with the same acceleration $a$. For A: $T-2mg=2ma$. For B: $3mg-T=3ma$.

Adding the two equations gives $mg=5ma$, so $a=\dfrac{g}{5}$.

Use $T-2mg=2ma$: $T=2mg+2m\left(\dfrac{g}{5}\right)=\dfrac{12mg}{5}$.

B starts from rest and moves $0.5$ m with acceleration $g/5$. Using $v^2=u^2+2as$, $v^2=2\left(\dfrac{g}{5}\right)(0.5)=\dfrac{g}{5}=1.96$.

$v=\sqrt{1.96}=1.4\ \mathrm{m\,s^{-1}}$.

The time until B hits the surface is found from $0.5=\dfrac12\left(\dfrac{g}{5}\right)t_1^2$. With $g=9.8$, $t_1=\dfrac{5}{7}=0.714285\ldots$ s.

After B hits the surface, A continues upwards freely under gravity. It is then at height $1.00$ m and must rise another $0.06$ m with initial speed $1.4\ \mathrm{m\,s^{-1}}$. So $0.06=1.4t_2-4.9t_2^2$.

Solving $0.06=1.4t_2-4.9t_2^2$ gives the first positive time $t_2=0.0525\ldots$ s. Therefore the total time is $0.714285\ldots+0.0525\ldots=0.767\ \mathrm{s}$.

The force of $18$ N is angled away from the plane, so $R+18\sin40^\circ=2g\cos30^\circ$. Hence $R=2g\cos30^\circ-18\sin40^\circ$.

The parcel moves up the plane, so friction acts down the plane. With $\mu=0.3$, $F=0.3R$.

Taking up the plane as positive: $18\cos40^\circ-F-2g\sin30^\circ=2a$.

$18\cos40^\circ-0.3(2g\cos30^\circ-18\sin40^\circ)-2g\sin30^\circ=2a$.

Using $g=9.8$, this gives $a=1.18\ldots\ \mathrm{m\,s^{-2}}$ up the plane.

From A to B, $u=2$, $s=5$ and $a=1.18\ldots$. Use $v^2=u^2+2as$: $v^2=2^2+2(1.18\ldots)(5)$.

$v=3.98\ldots\ \mathrm{m\,s^{-1}}$, about $4.0\ \mathrm{m\,s^{-1}}$.

At C, with the pulling force removed, $R=2g\cos30^\circ$. The maximum friction is $0.3R=0.3(2g\cos30^\circ)=5.09\ldots\ \mathrm{N}$.

The component of weight down the plane is $2g\sin30^\circ=9.8\ \mathrm{N}$. Since $9.8>5.09\ldots$, friction is not large enough to hold P at rest, so P will not remain at rest at C.

Since $\tan\alpha=\dfrac34$, $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$.

For A, $R=2g\cos\alpha=\dfrac{8g}{5}$. With coefficient $\dfrac14$, friction is $F=\dfrac14R=\dfrac{2g}{5}$.

A moves up the plane. Taking up the plane as positive: $T-2g\sin\alpha-F=2a$.

$T-2g\cdot\dfrac35-\dfrac{2g}{5}=2a$, so $T-\dfrac{8g}{5}=2a$.

B descends. Taking downward as positive: $4g-T=4a$.

From A, $T=2a+\dfrac{8g}{5}$. From B, $T=4g-4a$. Equating gives $6a=\dfrac{12g}{5}$, so $a=\dfrac{2g}{5}$. Then $T=4g-4\left(\dfrac{2g}{5}\right)=\dfrac{12g}{5}$ N.

Before B hits the ground, both particles move distance $h$ from rest with acceleration $\dfrac{2g}{5}$. Thus $v^2=2\left(\dfrac{2g}{5}\right)h=\dfrac{4gh}{5}$.

After B hits the ground, A continues up the plane. The forces down the plane on A are $2g\sin\alpha+F=\dfrac{6g}{5}+\dfrac{2g}{5}=\dfrac{8g}{5}$, so its deceleration is $\dfrac{4g}{5}$.

Let the extra distance be $s$. Then $0=\dfrac{4gh}{5}-2\left(\dfrac{4g}{5}\right)s$, so $s=\dfrac{h}{2}$.

A travels total distance $h+\dfrac{h}{2}=\dfrac{3h}{2}$ before stopping. Since it comes to rest before reaching the pulley, $d>\dfrac{3h}{2}$.

A more realistic model could include friction at the pulley.

Before B hits the floor, it descends $1.4$ m from rest in $2$ s. Use $s=ut+\dfrac12at^2$: $1.4=\dfrac12a(2)^2$. Hence $a=0.7\ \mathrm{m\,s^{-2}}$.

The string is inextensible, so A and B have the same magnitude of acceleration while the string is taut.

For B, taking downward as positive: $3g-T=3(0.7)$. Thus $T=3g-2.1=27.3$ N.

The pulley is pulled by two perpendicular tensions of magnitude $T$. Therefore the resultant is $\sqrt{T^2+T^2}=T\sqrt2=27.3\sqrt2=38.6$ N approximately.

For A before B hits the floor, $T-F=4(0.7)$. Hence $F=27.3-2.8=24.5$ N.

On the horizontal table, $R=4g=39.2$ N and $F=\mu R$. So $\mu=\dfrac{24.5}{39.2}=0.625$.

At this instant, $v=at=0.7(2)=1.4\ \mathrm{m\,s^{-1}}$. A has also moved $1.4$ m towards the pulley.

After B hits the floor, A is slowed only by friction. Its deceleration is $\mu g=0.625(9.8)=6.125\ \mathrm{m\,s^{-2}}$. Use $0=(1.4)^2-2(6.125)s$, giving $s=0.16$ m.

A's total distance is $1.4+0.16=1.56$ m, which is less than $2$ m. Therefore A does not reach the pulley.