Newton's Second Law: past paper questions & worked solutions

When the string becomes taut, A and B move with the same speed $S$. Before this, only A is moving. Hence $mU=(m+3m)S=4mS$.

From $mU=4mS$, cancel $m$ to get $S=\dfrac{U}{4}$.

The impulse on A equals the magnitude of its change in momentum: $I=m\left|\dfrac{U}{4}-U\right|$.

$I=m\left(\dfrac{3U}{4}\right)=\dfrac{3mU}{4}$.

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.

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: $2mg+3mg-T=2ma$.

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

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

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 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 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.

Take P's initial direction as positive. Then P initially has velocity $4u$ and Q initially has velocity $-3u$.

When the string becomes taut, both particles move with common speed $v$. Hence $3m(4u)+2m(-3u)=(3m+2m)v$.

$12mu-6mu=5mv$, so $v=\dfrac{6u}{5}$.

Impulse on Q equals change in momentum: $I=2m\left(\dfrac{6u}{5}-(-3u)\right)$.

$I=2m\left(\dfrac{6u}{5}+\dfrac{15u}{5}\right)=2m\cdot\dfrac{21u}{5}=\dfrac{42}{5}mu$.

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: $912 + 16W + 6W = 1440$ $22W = 528$ $W = 24$

$\sum F_{\perp} = 0 \text{ (no sideways motion)}$ Perpendicular component of 800 N rope: $800 \sin 25° = 800 \times 0.4226 = 338.1 \text{ N}$ Perpendicular component of 500 N rope: $500 \sin \alpha$

$800 \sin 25° = 500 \sin \alpha$ $338.1 = 500 \sin \alpha$ $\sin \alpha = \frac{338.1}{500}$ $\sin \alpha = 0.6762$ $\alpha = \sin^{-1}(0.6762)$ $\alpha = 42.5°$

$\sum F_{\parallel} = ma$ Parallel component of 800 N rope: $800 \cos 25° = 800 \times 0.9063 = 725.0 \text{ N}$ Parallel component of 500 N rope: $500 \cos 42.5° = 500 \times 0.7373 = 368.7 \text{ N}$ Resistance force: $750$ N (opposes motion) Mass: $15$ tonnes $= 15000$ kg

$725.0 + 368.7 - 750 = 15000 \times a$ $343.7 = 15000a$ $a = \frac{343.7}{15000}$ $a = 0.02291...$ $a = 0.023 \text{ m s}^{-2} \text{ (2 sf)}$

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

The applied force increases the normal reaction in this model, so $R=W+P\sin\alpha$.

For the particle not to move, friction must be able to balance the horizontal component of the applied force, so $P\cos\alpha\le F_{\max}$.

Since $\mu=\dfrac14$, $F_{\max}=\dfrac14R=\dfrac14(W+P\sin\alpha)$.

Thus $P\cdot\dfrac35\le\dfrac14\left(W+P\cdot\dfrac45\right)$.

Multiplying by $20$ gives $12P\le5W+4P$. Hence $8P\le5W$, so $P\le\dfrac58W$.

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 $p=2m$.

The tractor has mass $6000$ kg. Its forward driving force is $7400$ N, and the backward forces are $200$ N resistance and $6000$ N tension.

$7400-200-6000=6000a$, so $1200=6000a$ and $a=0.2\ \mathrm{m\,s^{-2}}$.

The block has mass $2000$ kg. The forward force is the tension $6000$ N and the resistance is $R$ N.

$6000-R=2000(0.2)=400$. Therefore $R=5600\ \mathrm{N}$.

The cable is inextensible, so the tractor and block move together with the same acceleration.

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.

For the trailer, take up the road as positive. The forces along the road are tension $2060$ N up the road, resistance $300$ N down the road, and weight component $400g\sin\alpha$ down the road.

$2060-300-400g\sin\alpha=400a$. Since $\sin\alpha=\dfrac17$, this becomes $2060-300-400g\cdot\dfrac17=400a$.

Using $g=9.8$, $400g\cdot\dfrac17=560$. Hence $1200=400a$ and $a=3\ \mathrm{m\,s^{-2}}$.

For the car: $D-420-800g\sin\alpha-2060=800a$.

$D-420-800g\cdot\dfrac17-2060=800(3)$. Since $800g\cdot\dfrac17=1120$, $D=420+1120+2060+2400=6000$ N.

The resultant force is $(2p\mathbf{i}-3p\mathbf{j})+(q\mathbf{i}+\mathbf{j})+(3\mathbf{i}-q\mathbf{j})=(2p+q+3)\mathbf{i}+(-3p+1-q)\mathbf{j}$.

The resultant acts in the direction $4\mathbf{i}-5\mathbf{j}$, so the ratio of the components is the same: $\dfrac{-3p+1-q}{2p+q+3}=\dfrac{-5}{4}$.

Cross-multiplying gives $4(-3p+1-q)=-5(2p+q+3)$. Therefore $-12p+4-4q=-10p-5q-15$, so $2p-q=19$.

With $p=7$, the relation gives $14-q=19$, hence $q=-5$.

Substitute $p=7$ and $q=-5$ into the resultant: $\mathbf{F}=12\mathbf{i}-15\mathbf{j}$. Since the mass is $0.5$ kg, $\mathbf{F}=m\mathbf{a}$ gives $12\mathbf{i}-15\mathbf{j}=0.5\mathbf{a}$.

Dividing by $0.5$ gives $\mathbf{a}=24\mathbf{i}-30\mathbf{j}\ \mathrm{m\,s^{-2}}$.

$\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=(6\mathbf{i}+8\mathbf{j})+(-16\mathbf{i}+2\mathbf{j})+(-2\mathbf{i}+8\mathbf{j})=-12\mathbf{i}+18\mathbf{j}$.

The mass is $2.5$ kg, so $-12\mathbf{i}+18\mathbf{j}=2.5\mathbf{a}$. Hence $\mathbf{a}=-4.8\mathbf{i}+7.2\mathbf{j}\ \mathrm{m\,s^{-2}}$.

$|\mathbf{a}|=\sqrt{(-4.8)^2+7.2^2}=8.653\ldots$, so the magnitude is $8.7\ \mathrm{m\,s^{-2}}$ to 2 significant figures. Exactly, $|\mathbf{a}|=\dfrac{12\sqrt{13}}{5}$.

With $\mathbf{F}_4=p\mathbf{i}+p\mathbf{j}$, the new resultant is $(p-12)\mathbf{i}+(p+18)\mathbf{j}$.

The resultant acts in the direction $7\mathbf{i}+2\mathbf{j}$, so $\dfrac{p-12}{p+18}=\dfrac{7}{2}$.

$2(p-12)=7(p+18)$, so $2p-24=7p+126$. Therefore $-150=5p$ and $p=-30$.

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 $x=0.63\ \mathrm{m}$.

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 $M=\dfrac{78}{0.4}=195$.

Vertically, $R_C+R_D=50g+25g=75g$. Given $2R_D=3R_C$, we have $R_D=1.5R_C$. Therefore $2.5R_C=75g$, so $R_C=30g$ and $R_D=45g$.

$D$ is $6-1.8=4.2$ m from A. The support C is $4.2-0.9=3.3$ m to the left of D, the child at E is $4.2-3=1.2$ m to the left of D, and the centre of mass is $x$ m to the left of D.

Balancing moments about D gives $50gx+25g(1.2)=R_C(3.3)$.

Substitute $R_C=30g$: $50x+30=99$, so $50x=69$ and $x=\dfrac{69}{50}=1.38$.

When the block at B is on the point of making the beam tilt, the beam is about to pivot about D and the reaction at C is zero.

The child and the beam act to the left of D, and the block at B acts $1.8$ m to the right of D. Hence $25g(1.2)+50g(1.38)=Mg(1.8)$.

$30+69=1.8M$, so $M=\dfrac{99}{1.8}=55$.

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 $162+39=201\ \mathrm{m}$.

$\mathbf{F}=(-3\mathbf{i}+2\mathbf{j})+(p\mathbf{i}+q\mathbf{j})=(p-3)\mathbf{i}+(q+2)\mathbf{j}$.

Since the particle starts from rest, its motion is in the direction of the resultant force. This direction is $\mathbf{i}-2\mathbf{j}$, so $\dfrac{q+2}{p-3}=-2$.

$q+2=-2(p-3)=-2p+6$, hence $2p+q-4=0$.

Substitute $p=5$ into $2p+q-4=0$: $10+q-4=0$, so $q=-6$.

The resultant force is $(5-3)\mathbf{i}+(-6+2)\mathbf{j}=2\mathbf{i}-4\mathbf{j}$ N.

The mass is $0.5$ kg, so $\mathbf{a}=\dfrac{2\mathbf{i}-4\mathbf{j}}{0.5}=4\mathbf{i}-8\mathbf{j}$.

Starting from rest, $\mathbf{v}=\mathbf{a}t=(4\mathbf{i}-8\mathbf{j})4=16\mathbf{i}-32\mathbf{j}$.

Speed $=\sqrt{16^2+(-32)^2}=\sqrt{1280}=16\sqrt5\ \mathrm{m\,s^{-1}}$, about $36\ \mathrm{m\,s^{-1}}$.

Equilibrium gives $\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}$. Now $\mathbf{F}_1+\mathbf{F}_2=(5\mathbf{i}+2\mathbf{j})+(-3\mathbf{i}+\mathbf{j})=2\mathbf{i}+3\mathbf{j}$.

So $\mathbf{F}_3=-2\mathbf{i}-3\mathbf{j}$.

The vector $-\mathbf{j}$ points vertically downward. For $\mathbf{F}_3=-2\mathbf{i}-3\mathbf{j}$, the angle from $-\mathbf{j}$ satisfies $\tan\theta=\dfrac{2}{3}$.

$\theta=33.7^\circ$, so the angle is about $34^\circ$.

With $\mathbf{F}_4=\lambda(\mathbf{i}+3\mathbf{j})$, the resultant is $(2+\lambda)\mathbf{i}+(3+3\lambda)\mathbf{j}$.

The mass is $4$ kg and acceleration magnitude is $3.25$, so the resultant force magnitude is $4(3.25)=13$ N.

$(2+\lambda)^2+(3+3\lambda)^2=13^2$.

Expanding gives $10\lambda^2+22\lambda-156=0$. The roots are $3$ and $-5.2$, and since $\lambda$ is positive, $\lambda=3$.

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 $ma=1200(0.8)=960$, $D-260=960$.

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

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 $695\ \mathrm{N}$.

For equilibrium, $\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}$. Hence $\mathbf{F}_3=-(\mathbf{F}_1+\mathbf{F}_2)$.

$\mathbf{F}_1+\mathbf{F}_2=(3c-14)\mathbf{i}+(4c+7)\mathbf{j}$, so $\mathbf{F}_3=(14-3c)\mathbf{i}+(-7-4c)\mathbf{j}$.

When $c=2$, $\mathbf{F}_1=6\mathbf{i}+8\mathbf{j}$ and $\mathbf{F}_2=-14\mathbf{i}+7\mathbf{j}$. Therefore the resultant is $-8\mathbf{i}+15\mathbf{j}$.

The resultant is in the second quadrant. Its angle from the positive $\mathbf{i}$ direction is $180^\circ-\tan^{-1}\left(\dfrac{15}{8}\right)=118.1^\circ$, so the angle is about $118^\circ$.

$|\mathbf{R}|=\sqrt{(-8)^2+15^2}=17\ \mathrm{N}$.

The acceleration magnitude is $8.5\ \mathrm{m\,s^{-2}}$, so $17=8.5m$. Hence $m=2$.

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

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 two tensions are equal and symmetric. Their vertical components support the weight of C, so $2T\sin\beta=3mg$. Hence $T=\dfrac{3mg}{2\sin\beta}$.

For ring A, the normal reaction balances its own weight and the downward vertical component of the string tension: $R=mg+T\sin\beta$.

From $2T\sin\beta=3mg$, we have $T\sin\beta=\dfrac32mg$. Therefore $R=mg+\dfrac32mg=\dfrac52mg$.

The rings are just about to slide towards each other, so friction is limiting. With $\mu=\dfrac45$, $F=\mu R=\dfrac45\cdot\dfrac52mg=2mg$.

The horizontal component of the tension is balanced by friction: $T\cos\beta=F=2mg$.

Substitute $T=\dfrac{3mg}{2\sin\beta}$: $\dfrac{3mg}{2\sin\beta}\cos\beta=2mg$. Hence $\dfrac32\cot\beta=2$, so $\cot\beta=\dfrac43$ and $\tan\beta=\dfrac34$.

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 $d=\dfrac{864}{11}=78.5\ldots\ \mathrm{m}$.

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.

Resolve vertically. The upward component of the applied force is $14\sin30^\circ$, so $R+14\sin30^\circ=5g$.

$R+7=49$, so $R=42\ \mathrm{N}$.

$F_{\max}=\mu R=\dfrac37(42)=18\ \mathrm{N}$.

The horizontal component of the applied force is $14\cos30^\circ=12.1\ldots\ \mathrm{N}$. Since this is less than the maximum possible friction, the block does not move.

For equilibrium, the actual friction equals the horizontal pull, so $F=12.1\ \mathrm{N}$ approximately.

When the block is on the point of slipping, friction is limiting. Resolve vertically: $R+P\sin30^\circ=5g$, so $R=5g-P\sin30^\circ$.

At the limiting case, $P\cos30^\circ=\dfrac37R=\dfrac37(5g-P\sin30^\circ)$.

Substitute $g=9.8$, $\sin30^\circ=\dfrac12$: $P\cos30^\circ=\dfrac37(49-\dfrac12P)$. Solving gives $P=19.4\ldots\ \mathrm{N}$.

The total force is $(p\mathbf{i}+q\mathbf{j})+(2q\mathbf{i}+p\mathbf{j})=(p+2q)\mathbf{i}+(p+q)\mathbf{j}$. Since the mass is $2$ kg and acceleration is $\mathbf{i}-\mathbf{j}$, the total force is $2\mathbf{i}-2\mathbf{j}$.

$p+2q=2$ and $p+q=-2$.

Subtract the equations to get $q=4$. Then $p+4=-2$, so $p=-6$.

The acceleration is $\mathbf{i}-\mathbf{j}$. Its dot product with $\mathbf{j}$ is $-1$, so the angle with the positive $\mathbf{j}$ direction is obtuse. Since the components are equal in magnitude, the angle is $135^\circ$.

$\mathbf{v}=(3\mathbf{i}-4\mathbf{j})+T(\mathbf{i}-\mathbf{j})=(3+T)\mathbf{i}+(-4-T)\mathbf{j}$.

The velocity is parallel to $11\mathbf{i}-13\mathbf{j}$, so $\dfrac{3+T}{11}=\dfrac{-4-T}{-13}$.

$13(3+T)=11(4+T)$, so $39+13T=44+11T$. Hence $2T=5$ and $T=2.5$.

The graph starts at $(0,0)$, rises in a straight line with gradient $0.8$ until time $T$ and speed $V$, then stays horizontal at speed $V$ until $t=30$.

Uniform acceleration gives $V=0.8T$.

The distance is the area under the speed-time graph: $\dfrac12TV+V(30-T)=200$.

Using $V=0.8T$: $\dfrac12T(0.8T)+(0.8T)(30-T)=200$.

$0.4T^2+24T-0.8T^2=200$, so $-0.4T^2+24T-200=0$. Multiplying by $-2.5$ gives $T^2-60T+500=0$, so $k=60$.

$T^2-60T+500=(T-10)(T-50)=0$, so $T=10$ or $50$. Since the whole race lasts $30$ s, $T=10$.

A more realistic model could include reaction time at the start, and could replace the instantaneous change from acceleration to constant speed with a smoother transition.

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

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 ; $v = ?$

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

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.

Given $\tan\alpha=\dfrac{7}{24}$, use $\sin\alpha=\dfrac{7}{25}$ and $\cos\alpha=\dfrac{24}{25}$.

For block A, $R=mg\cos\alpha=\dfrac{24mg}{25}$. Since $\mu=\dfrac13$, friction is $F=\dfrac13R=\dfrac{8mg}{25}$.

A moves up the plane, so friction and the component of weight act down the plane. Taking up the plane as positive: $T-mg\sin\alpha-F=ma$.

Substitute $T=\dfrac43mg$, $mg\sin\alpha=\dfrac{7mg}{25}$ and $F=\dfrac{8mg}{25}$: $ma=\dfrac43mg-\dfrac{15mg}{25}=\left(\dfrac43-\dfrac35\right)mg=\dfrac{11mg}{15}$. Hence $a=\dfrac{11g}{15}$.

Block B moves downwards with the same acceleration. Taking downward as positive: $kmg-T=kma$.

Substitute $T=\dfrac43mg$ and $a=\dfrac{11g}{15}$: $kmg-\dfrac43mg=k m\dfrac{11g}{15}$. Divide by $mg$ to get $k-\dfrac43=\dfrac{11k}{15}$, so $15k-20=11k$ and $k=5$.

The pulley is pulled by two tensions of magnitude $T$. One is vertical and the other is along the plane. The horizontal component of the resultant is $T\cos\alpha$ and the vertical component is $T+T\sin\alpha$.

The magnitude is $T\sqrt{\cos^2\alpha+(1+\sin\alpha)^2}$. With $\sin\alpha=\dfrac{7}{25}$ and $\cos\alpha=\dfrac{24}{25}$, this is $T\sqrt{\dfrac{576}{625}+\dfrac{1024}{625}}=\dfrac{8T}{5}$. Since $T=\dfrac43mg$, the resultant is $\dfrac{32}{15}mg$.

For Q on the horizontal table, $R=mg$, so the friction is $F=\mu mg$.

P moves downward and Q moves towards the pulley. For P, taking downward as positive: $mg-T=ma$.

For Q, taking motion towards the pulley as positive: $T-\mu mg=ma$.

From P, $a=g(1-k)$. From Q, $a=g(k-\mu)$.

Equate the two accelerations: $1-k=k-\mu$. Hence $2k=1+\mu$ and $k=\dfrac{1+\mu}{2}$.

$a=g(1-k)=g\left(1-\dfrac{1+\mu}{2}\right)=\dfrac{g(1-\mu)}{2}$.

Q starts from rest and travels distance $d$, so $d=\dfrac12at^2=\dfrac12\cdot\dfrac{g(1-\mu)}{2}t^2$.

$d=\dfrac{g(1-\mu)t^2}{4}$, so $t=\sqrt{\dfrac{4d}{g(1-\mu)}}$.

If $\mu\ge1$, the maximum available friction on Q is at least $mg$, so it can balance the pull caused by the hanging particle. The system would not start moving.

Q lies on a smooth horizontal plane. The only horizontal force on Q is the tension, so $T=ma$.

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

For P, $R=4mg\cos\alpha=4mg\cdot\dfrac45=\dfrac{16mg}{5}$. With coefficient $\dfrac14$, friction is $F=\dfrac14R=\dfrac{4mg}{5}$.

P moves down the plane. Taking down the plane as positive: $4mg\sin\alpha-T-F=4ma$.

$4mg\cdot\dfrac35-T-\dfrac{4mg}{5}=4ma$, so $\dfrac{8mg}{5}-T=4ma$.

From Q, $a=\dfrac{T}{m}$. Substitute into the equation for P: $\dfrac{8mg}{5}-T=4T$.

$\dfrac{8mg}{5}=5T$, so $T=\dfrac{8}{25}mg$.

The pulley is pulled by two tensions of magnitude $T$: one horizontal and one along the inclined string. Their resultant has components $T-T\cos\alpha$ horizontally and $T\sin\alpha$ vertically.

Magnitude $=\sqrt{(T-T\cos\alpha)^2+(T\sin\alpha)^2}=T\sqrt{(1-\tfrac45)^2+(\tfrac35)^2}=\dfrac{T}{5}\sqrt{10}$. With $T=\dfrac{8}{25}mg$, the resultant is $\dfrac{8\sqrt{10}}{125}mg$.

The string is light, so the tension can be treated as the same on the P side and the Q side of the pulley.

Since $\tan\theta=\dfrac43$, use a $3$-$4$-$5$ triangle: $\sin\theta=\dfrac45$ and $\cos\theta=\dfrac35$.

With one $0.2$ kg block in the bucket, the bucket side has mass $0.45+0.2=0.65$ kg. Since the package is on the point of slipping down the plane, friction on P acts up the plane.

The tension equals the bucket weight, so $T=0.65g$. Along the plane for P in equilibrium, $T+F=g\sin\theta$. Hence $0.65g+F=\dfrac45g$, so $F=\dfrac{3g}{20}$.

Perpendicular to the plane, $R=g\cos\theta=\dfrac35g$. Since $F=\mu R$, $\mu=\dfrac{F}{R}=\dfrac{3g/20}{3g/5}=\dfrac14$.

When P is about to move up the plane, friction acts down the plane. With $n$ blocks in the bucket, the bucket-side mass is $0.45+0.2n$. The bucket weight must exceed the downslope resistance: $(0.45+0.2n)g>g\sin\theta+\mu R$.

Using $g\sin\theta=\dfrac45g$ and $\mu R=\dfrac14\cdot\dfrac35g=\dfrac{3g}{20}$, the condition is $0.45+0.2n>\dfrac45+\dfrac{3}{20}=0.95$. Hence $0.2n>0.50$, so $n>2.5$ and the least integer is $n=3$.

For $n=3$, the bucket-side mass is $0.45+0.6=1.05$ kg and the total moving mass is $1+1.05=2.05$ kg. The driving force is $1.05g-\dfrac45g-\dfrac{3g}{20}=0.10g$.

Using $F=ma$, $0.10g=2.05a$, so $a=\dfrac{0.10g}{2.05}=\dfrac{2g}{41}=0.478\ \mathrm{m\,s^{-2}}$ to 3 significant figures.

$\mathbf{F}_{\text{resultant}} = \mathbf{F}_1 + \mathbf{F}_2$ $\mathbf{F}_{\text{resultant}} = (-2\mathbf{i} + 3\mathbf{j}) + (4\mathbf{i} + 2\mathbf{j})$ $= (-2 + 4)\mathbf{i} + (3 + 2)\mathbf{j}$ $= 2\mathbf{i} + 5\mathbf{j} \text{ N}$

$\mathbf{F} = m\mathbf{a}$ $2\mathbf{i} + 5\mathbf{j} = 2 \times \mathbf{a}$ $\mathbf{a} = \frac{2\mathbf{i} + 5\mathbf{j}}{2}$ $\mathbf{a} = \mathbf{i} + 2.5\mathbf{j} \text{ m s}^{-2}$ Or: $\mathbf{a} = \mathbf{i} + \frac{5}{2}\mathbf{j}$ m s$^{-2}$

$\mathbf{v} = \mathbf{u} + \mathbf{a}t$ Given: ; $\mathbf{u} = 3\mathbf{i} - 4\mathbf{j}$ m s$^{-1}$ ; $\mathbf{a} = \mathbf{i} + 2.5\mathbf{j}$ m s$^{-2}$ ; $t = 3$ seconds $\mathbf{v} = (3\mathbf{i} - 4\mathbf{j}) + (\mathbf{i} + 2.5\mathbf{j}) \times 3$ $= 3\mathbf{i} - 4\mathbf{j} + 3\mathbf{i} + 7.5\mathbf{j}$ $= 6\mathbf{i} + 3.5\mathbf{j} \text{ m s}^{-1}$

$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$ $\text{Speed} = \sqrt{6^2 + 3.5^2}$ $= \sqrt{36 + 12.25}$ $= \sqrt{48.25}$ $= 6.946...$ $= 6.95 \text{ m s}^{-1} \text{ (3 s.f.)}$

$\mathbf{F}_{\text{resultant}} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3$ $= (-2\mathbf{i} + 3\mathbf{j}) + (4\mathbf{i} + 2\mathbf{j}) + (b\mathbf{i} + c\mathbf{j})$ $= (-2 + 4 + b)\mathbf{i} + (3 + 2 + c)\mathbf{j}$ $= (2 + b)\mathbf{i} + (5 + c)\mathbf{j}$

Given that resultant = $\lambda(\mathbf{i} + \mathbf{j})$, we can write: $(2 + b)\mathbf{i} + (5 + c)\mathbf{j} = \lambda\mathbf{i} + \lambda\mathbf{j}$

For this to be in the form $\lambda(\mathbf{i} + \mathbf{j})$, the coefficients must be equal: $2 + b = \lambda$ $5 + c = \lambda$ Therefore: $2 + b = 5 + c$ $b = 5 + c - 2$ $b = 3 + c$ $b - c = 3$

Resultant = $(2 + b)\mathbf{i} + (5 + c)\mathbf{j}$ Since resultant = $\lambda(\mathbf{i} + \mathbf{j})$, we have $2 + b = 5 + c = \lambda$ $|\mathbf{F}_{\text{resultant}}| = \lambda\sqrt{1^2 + 1^2} = \lambda\sqrt{2}$ Given: $|\mathbf{F}_{\text{resultant}}| = 10\sqrt{2}$ $\lambda\sqrt{2} = 10\sqrt{2}$ $\lambda = 10 \quad \text{or} \quad \lambda = -10$ (The negative value comes from considering the opposite direction)

Case 1: $\lambda = 10$ $2 + b = 10 \quad \Rightarrow \quad b = 8$ $5 + c = 10 \quad \Rightarrow \quad c = 5$ Check: $b - c = 8 - 5 = 3$ ✓ Therefore: $\mathbf{F}_3 = 8\mathbf{i} + 5\mathbf{j}$ N Case 2: $\lambda = -10$ $2 + b = -10 \quad \Rightarrow \quad b = -12$ $5 + c = -10 \quad \Rightarrow \quad c = -15$ Check: $b - c = -12 - (-15) = 3$ ✓ Therefore: $\mathbf{F}_3 = -12\mathbf{i} - 15\mathbf{j}$ N

Forces on Q (mass = 5m): Taking downward as positive for Q: $\text{For Q: } \sum F = ma$ $5mg - T = 5ma$

$\perp \text{ to plane: } \sum F = 0$ $R = 2mg\cos\alpha$ $R = 2mg \times \frac{12}{13}$ $R = \frac{24mg}{13}$

P is moving UP the plane, so friction acts DOWN the plane: $F = \mu R$ $= \frac{1}{8} \times \frac{24mg}{13}$ $= \frac{3mg}{13}$

Taking up the plane as positive: $\parallel \text{ to plane: } \sum F = ma$ $T - 2mg\sin\alpha - F = 2ma$ $T - 2mg \times \frac{5}{13} - \frac{3mg}{13} = 2ma$ $T - \frac{10mg}{13} - \frac{3mg}{13} = 2ma$ $T - \frac{13mg}{13} = 2ma$ $T - mg = 2ma \quad \text{...(1)}$

From part (a): $5mg - T = 5ma$ ...(2) Add equations (1) and (2): $(T - mg) + (5mg - T) = 2ma + 5ma$ $4mg = 7ma$ $a = \frac{4g}{7}$

From equation (1): $T = mg + 2ma$ $= mg + 2m \times \frac{4g}{7}$ $= mg + \frac{8mg}{7}$ $= \frac{7mg + 8mg}{7}$ $= \frac{15mg}{7}$

An inextensible string means the string cannot stretch. This has the following consequence: For an inextensible string over a pulley: The magnitude of acceleration of both connected particles must be equal. In this problem: $|a_P| = |a_Q| = a$ How this was used in part (b): When we wrote the equations of motion: ; For P: $T - mg = 2ma$ ; For Q: $5mg - T = 5ma$ We used the same acceleration $a$ for both particles because the string is inextensible. If the string could stretch, the two particles could have different accelerations.

The string from Q is vertical (downward). The string from P is along the plane at angle $\alpha$ to horizontal. Angle between the two strings = $90^\circ + \alpha$

Let's take horizontal (right) and vertical (up) as positive directions. Tension from Q: ; Horizontal component: $0$ ; Vertical component: $-T = -\frac{15mg}{7}$ (downward) Tension from P along plane: The plane makes angle $\alpha$ with horizontal, so: ; Horizontal component: $-T\cos\alpha = -\frac{15mg}{7} \times \frac{12}{13} = -\frac{180mg}{91}$ ; Vertical component: $T\sin\alpha = \frac{15mg}{7} \times \frac{5}{13} = \frac{75mg}{91}$

$R_x = 0 - \frac{180mg}{91} = -\frac{180mg}{91}$ $R_y = -\frac{15mg}{7} + \frac{75mg}{91}$ Convert $\frac{15mg}{7}$ to denominator 91: $\frac{15mg}{7} = \frac{195mg}{91}$ $R_y = -\frac{195mg}{91} + \frac{75mg}{91} = -\frac{120mg}{91}$

$|R| = \sqrt{R_x^2 + R_y^2}$ $= \sqrt{\left(\frac{180mg}{91}\right)^2 + \left(\frac{120mg}{91}\right)^2}$ $= \frac{mg}{91}\sqrt{180^2 + 120^2}$ $= \frac{mg}{91}\sqrt{32400 + 14400}$ $= \frac{mg}{91}\sqrt{46800}$ $= \frac{mg}{91} \times 216.333...$ $= 2.377... mg$ $= 2.38mg \text{ (3 s.f.)}$ Therefore: $k = 2.38$

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.

A moves up the plane, so friction and the component of weight act down the plane. Taking up the plane as positive: $T-2mg\sin\alpha-F=2ma$.

B moves downward. Taking downward as positive: $3mg-T=3ma$.

Since $\tan\alpha=\dfrac34$, $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$. The normal reaction on A is $R=2mg\cos\alpha=\dfrac{8mg}{5}$, so $F=\dfrac12R=\dfrac{4mg}{5}$.

$T-2mg\cdot\dfrac35-\dfrac{4mg}{5}=2ma$, so $T-2mg=2ma$.

From B, $T=3mg-3ma$. Substitute into $T-2mg=2ma$: $3mg-3ma-2mg=2ma$. Thus $mg=5ma$ and $a=\dfrac{g}{5}$.

Using $3mg-T=3m\left(\dfrac{g}{5}\right)$ gives $T=3mg-\dfrac{3mg}{5}=\dfrac{12mg}{5}$.

The two tensions are one vertical and one along the plane. Components of the resultant are $T\cos\alpha$ horizontally and $T+T\sin\alpha$ vertically.

Resultant $=T\sqrt{\cos^2\alpha+(1+\sin\alpha)^2}=T\sqrt{(\tfrac45)^2+(\tfrac85)^2}=\dfrac{4\sqrt5}{5}T$. With $T=\dfrac{12mg}{5}$, this is $\dfrac{48\sqrt5}{25}mg$.

The pulley is smooth, so the tension is the same on both sides of the pulley.

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 $0.714285\ldots+0.0525\ldots=0.767\ \mathrm{s}$.

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

For A, $R=2mg\cos\alpha=2mg\cdot\dfrac{12}{13}=\dfrac{24mg}{13}$. Since $\mu=\dfrac{11}{36}$, $F=\mu R=\dfrac{11}{36}\cdot\dfrac{24mg}{13}=\dfrac{22mg}{39}$.

B moves down as A moves up. Taking downward as positive for B: $3mg-T=3ma$.

Taking up the plane as positive for A: $T-2mg\sin\alpha-F=2ma$. Substituting $\sin\alpha=\dfrac{5}{13}$ and $F=\dfrac{22mg}{39}$ gives $T-\dfrac{10mg}{13}-\dfrac{22mg}{39}=2ma$.

From B, $T=3mg-3ma$. Substitute into A equation: $3mg-3ma-\dfrac{10mg}{13}-\dfrac{22mg}{39}=2ma$. The constant part is $\dfrac{5mg}{3}$, so $\dfrac{5mg}{3}=5ma$ and $a=\dfrac{g}{3}$.

While B descends distance $h$, A moves distance $h$ up the plane from rest with acceleration $g/3$. Thus $v^2=2\left(\dfrac{g}{3}\right)h=\dfrac{2gh}{3}$.

After B reaches the ground, the string no longer pulls A. The forces down the plane on A are $2mg\sin\alpha+F=\dfrac{10mg}{13}+\dfrac{22mg}{39}=\dfrac{4mg}{3}$. Since A has mass $2m$, its deceleration is $\dfrac{2g}{3}$.

Let the extra distance be $d$. Using $0=v^2-2\left(\dfrac{2g}{3}\right)d$ with $v^2=\dfrac{2gh}{3}$ gives $d=\dfrac{h}{2}$. Total distance travelled by A is $h+\dfrac{h}{2}=\dfrac{3h}{2}$.

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.

Particle P (mass $kM$) moves upward with acceleration $\frac{g}{5}$: ; Tension T (upward) ; Weight $kMg$ (downward)

Taking upward as positive: $\sum F = ma$ $T - kMg = kM \times \frac{g}{5}$

Particle Q (mass $3M$) moves downward with acceleration $\frac{g}{5}$: Taking downward as positive for Q: $3Mg - T = 3M \times \frac{g}{5}$ $3Mg - T = \frac{3Mg}{5}$ $T = 3Mg - \frac{3Mg}{5}$ $T = \frac{12Mg}{5}$

From part (a): $T - kMg = \frac{kMg}{5}$ $\frac{12Mg}{5} - kMg = \frac{kMg}{5}$ $\frac{12Mg}{5} = kMg + \frac{kMg}{5}$ $\frac{12Mg}{5} = \frac{5kMg + kMg}{5}$ $\frac{12Mg}{5} = \frac{6kMg}{5}$ $12 = 6k$ $k = 2$

From part (b): $T = \frac{12Mg}{5}$ Given: $M = 0.5$ kg $T = \frac{12 \times 0.5 \times g}{5}$ $= \frac{6g}{5}$ $= \frac{6 \times 9.8}{5}$ $= 11.76 \text{ N}$

The pulley experiences two tension forces: ; Horizontal tension from P: $T = 11.76$ N ; Vertical tension from Q: $T = 11.76$ N These are perpendicular, so use Pythagoras: $F = \sqrt{T^2 + T^2}$ $= \sqrt{2T^2}$ $= T\sqrt{2}$ $= 11.76\sqrt{2}$ $= 16.63...$ $= 16.6 \text{ N (3 sf)}$ Or: $F = \frac{6g\sqrt{2}}{5} = \frac{12g}{5} = 2.4g = 23.5$ N Wait, let me recalculate... Actually: $F = 2T$ (both tensions pull on pulley) $F = 2 \times \frac{12Mg}{5}$ $= \frac{24Mg}{5}$ $= \frac{24 \times 0.5 \times 9.8}{5}$ $= \frac{117.6}{5}$ $= 23.5 \text{ N (3 sf)}$ Or: $\frac{12g}{5} = 23.5$ N

Using SUVAT with: ; $u = 0$ (starts from rest) ; $a = \frac{g}{5} = \frac{9.8}{5} = 1.96$ m s$^{-2}$ ; $s = 2.5$ m $v^2 = u^2 + 2as$ $v^2 = 0 + 2 \times \frac{g}{5} \times 2.5$ $v^2 = \frac{5g}{5}$ $v^2 = g$ $v = \sqrt{g} = \sqrt{9.8} = 3.130... \text{ m s}^{-1}$ Or exact: $v = \sqrt{g} = \sqrt{\frac{7g}{5}}$ Actually: $v^2 = 2 \times 1.96 \times 2.5 = 9.8$, so $v = \sqrt{9.8} = 3.13$ m s$^{-1}$

$\text{Impulse} = m(v_{\text{after}} - v_{\text{before}})$ Taking downward as positive before impact: ; Velocity before: $v_{\text{before}} = +3.13$ m s$^{-1}$ (downward) ; Velocity after: $v_{\text{after}} = -0.4$ m s$^{-1}$ (upward, rebounds) ; Mass of Q: $3M = 3 \times 0.5 = 1.5$ kg $I = 1.5 \times (-0.4 - 3.13)$ $= 1.5 \times (-3.53)$ $= -5.295...$ $|I| = 5.3 \text{ N s (2 sf)}$ Or using exact values: $|I| = 1.5(\sqrt{g} + 0.4) = 1.5(3.13 + 0.4) = 5.3$ N s

Using $s = ut + \frac{1}{2}at^2$ with $u = 0$, $s = 2.5$ m, $a = \frac{g}{5}$: $2.5 = 0 + \frac{1}{2} \times \frac{g}{5} \times t_1^2$ $2.5 = \frac{g t_1^2}{10}$ $25 = g t_1^2$ $t_1^2 = \frac{25}{9.8} = 2.551...$ $t_1 = \sqrt{\frac{25}{9.8}} = 1.597... \text{ s}$ Or exact: $t_1 = \sqrt{\frac{5 \times 5}{g}} = \frac{5}{\sqrt{g}}$ Better: $t_1 = \sqrt{\frac{5}{g/5}} = \sqrt{\frac{25}{g}}$ Actually: $2.5 = \frac{1}{2} \times 1.96 \times t_1^2$ $t_1^2 = \frac{5}{1.96} = 2.551$, so $t_1 = 1.597$ s

At time $t_1$, P has velocity: $v_P = 0 + \frac{g}{5} \times t_1$ $= \frac{9.8}{5} \times 1.597$ $= 1.96 \times 1.597$ $= 3.13 \text{ m s}^{-1}$ Or: $v_P = \sqrt{g}$ (same as Q)

After Q hits ground, string goes slack. P continues upward under gravity alone. Using $v = u + at$ with $v = 0$, $u = 3.13$ m s$^{-1}$, $a = -g$: $0 = 3.13 - 9.8 t_2$ $t_2 = \frac{3.13}{9.8}$ $= 0.3194... \text{ s}$ Or: $t_2 = \frac{\sqrt{g}}{g} = \frac{1}{\sqrt{g}}$

$T = t_1 + t_2$ $= 1.597 + 0.319$ $= 1.916...$ $= 1.9 \text{ s (2 sf)} \text{ or } 1.92 \text{ s (3 sf)}$

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.