Quantities, Units and Modelling: 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 trucks join together, so they move with a common speed $v$. Using tonnes consistently, $20(4)+30(0)=(20+30)v$.

$80=50v$, so $v=1.6\ \mathrm{m\,s^{-1}}$.

Truck S has mass $20\,000$ kg. Its speed changes from $4$ to $1.6\ \mathrm{m\,s^{-1}}$.

$I=20000|1.6-4|=20000(2.4)=48000\ \mathrm{N\,s}$.

Take the initial direction of P as positive. Then before impact, $v_P=3u$ and $v_Q=-4u$. The impulse on P acts opposite to P\'s initial motion, so its signed impulse is $-\dfrac{33}{2}mu$.

For P, impulse equals change in momentum: $-\dfrac{33}{2}mu=3m(v_P-3u)$. Dividing by $3m$ gives $v_P-3u=-\dfrac{11u}{2}$, so $v_P=-\dfrac{5u}{2}$.

The negative sign means P has reversed direction. Its speed is therefore $\left|-\dfrac{5u}{2}\right|=\dfrac{5u}{2}$.

Q receives impulse $+\dfrac{33}{2}mu$ in the positive direction. Hence $\dfrac{33}{2}mu=5m(v_Q-(-4u))=5m(v_Q+4u)$.

Dividing by $5m$ gives $v_Q+4u=\dfrac{33u}{10}$, so $v_Q=-\dfrac{7u}{10}$. Q is still moving in its original direction, and its speed is $\dfrac{7u}{10}$.

$\text{Impulse} = \text{Final momentum} - \text{Initial momentum}$ For particle Q: ; Mass = $3m$ ; Initial velocity = $2u$ ; Final velocity = $v$ ; Impulse received = $4.5mu$

$4.5mu = 3m \times v - 3m \times 2u$ $4.5mu = 3mv - 6mu$ $4.5u = 3v - 6u$ $4.5u + 6u = 3v$ $10.5u = 3v$ $v = \frac{10.5u}{3}$ $v = 3.5u$

$\text{Total momentum before} = \text{Total momentum after}$ Before collision: $\text{P momentum} = km \times 5u = 5kmu$ $\text{Q momentum} = 3m \times 2u = 6mu$ $\text{Total} = 5kmu + 6mu$ After collision: $\text{P momentum} = km \times 2u = 2kmu$ $\text{Q momentum} = 3m \times 3.5u = 10.5mu$ $\text{Total} = 2kmu + 10.5mu$

$5kmu + 6mu = 2kmu + 10.5mu$ $5km + 6m = 2km + 10.5m$ $5k + 6 = 2k + 10.5$ $5k - 2k = 10.5 - 6$ $3k = 4.5$ $k = \frac{4.5}{3}$ $k = 1.5$

Take the initial direction of A as positive. Before collision, A has velocity $2x$ and B has velocity $-x$. After collision, both reverse direction, so A has velocity $-y$ and B has velocity $5y$.

$4m(2x)+3m(-x)=4m(-y)+3m(5y)$. The left side is $5mx$ and the right side is $11my$.

$5mx=11my$, so $\dfrac{y}{x}=\dfrac{5}{11}$.

The impulse received by A is the change in momentum of A: $I=4m|-y-2x|=4m(y+2x)$.

Using $y=\dfrac{5x}{11}$, $I=4m\left(\dfrac{5x}{11}+2x\right)=4m\left(\dfrac{27x}{11}\right)=\dfrac{108}{11}mx$.

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 is 6.5 m long and overhangs C by 1.5 m, so C is 5.0 m from A. The rod centre of mass is 3.0 m from A, so it is 2.0 m to the left of C. Point D is 0.5 m from A, so it is 4.5 m to the left of C.

For the smallest $M$, the rod is just prevented from tilting clockwise about C. Taking moments about C: $Mg(4.5)+1.2g(2)=4g(1.5)$.

$4.5M+2.4=6$, so $4.5M=3.6$ and $M=0.8$.

Now the old particles are removed. Since $EB=0.9$ m and $AB=6.5$ m, E is $5.6$ m from A, so E is $0.6$ m to the right of C.

At tilting about C, moments about C give $Xg(0.6)=1.2g(2)$, so $X=4$.

The upward force at C balances the rod and the new particle: $R_C=1.2g+4g=5.2g\ \mathrm{N}$, which is about $51\ \mathrm{N}$.

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 $M=\dfrac{4.0}{9.8}=0.408\ \mathrm{kg}$.

Take the initial direction of A as positive. Before collision, A has velocity $3$ and B has velocity $-x$.

After collision, A still moves in the same direction with velocity $1$. B has speed $1.5$ in the positive direction for the valid solution. Hence $5(3)+x(-x)=5(1)+1.5x$.

$15-x^2=5+1.5x$, so $x^2+1.5x-10=0$. This gives $x=2.5$ or $x=-4$. Since mass is positive, $x=2.5$.

The impulse exerted on A equals the change in momentum of A: $I=5(1-3)=-10\ \mathrm{N\,s}$ using the chosen positive direction.

The magnitude of the impulse is $10\ \mathrm{N\,s}$.

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 $4.32\ \mathrm{s}$.

Take P's original direction as positive. Before collision, P has velocity $3u$ and Q has velocity $-2u$.

The impulse on P is opposite to P's original motion and has magnitude $5mu$. Let P's velocity after collision be $v$. Then $-5mu=2m(v-3u)$.

Divide by $m$: $-5u=2v-6u$, so $2v=u$ and $v=\dfrac{u}{2}$.

The impulse on Q is in P's original direction. Let Q's velocity after collision be $w$. Then $5mu=m(w-(-2u))$.

$5u=w+2u$, so $w=3u$. Therefore the speed of Q is $3u$.

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 $t=1$ 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 $7.2\ \mathrm{m}$.

Take the original direction of A as positive. Before the collision, A has velocity $2u$ and B has velocity $-3u$. After the collision, B has velocity $2u$ because its direction is reversed.

Let the velocity of A after the collision be $v$. Then $4(2u)+2(-3u)=4v+2(2u)$.

$8u-6u=4v+4u$, so $2u=4v+4u$. Hence $4v=-2u$ and $v=-\dfrac{u}{2}$.

The speed of A is the magnitude of its velocity, so it is $\dfrac{u}{2}$. The negative sign shows that A has reversed direction.

Impulse equals change in momentum. For B, $I=2\left(2u-(-3u)\right)=2(5u)=10u$.

The impulse is $10u\ \mathrm{N\,s}$, equivalently $10u\ \mathrm{kg\,m\,s^{-1}}$.

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.

The beam length is $5$ m and $DB=1.3$ m, so D is $3.7$ m from A. The centre of mass of the uniform beam is at its midpoint, so it is $1.2$ m to the left of D. Point C is $0.8$ m to the right of D.

When the boy stands at C, the beam is about to tilt about D. Let the mass of the beam be $M$ kg.

$Mg(1.2)=30g(0.8)$. Hence $1.2M=24$ and $M=20\ \mathrm{kg}$.

With the boy at B, his moment about D is $30g(1.3)$. The block at A is $3.7$ m to the left of D, and the beam's weight acts $1.2$ m to the left of D.

$Xg(3.7)+20g(1.2)=30g(1.3)$.

$3.7X+24=39$, so $3.7X=15$ and $X=\dfrac{150}{37}=4.05\ldots\ \mathrm{kg}$.

The block is modelled as a particle, so its mass is treated as acting at one point, A.

Take P's original direction as positive. Q initially has velocity $-2u$. Since Q reverses direction and then has speed $3u$, its final velocity is $3u$.

Impulse equals change in momentum: $I=5m(3u-(-2u))=25mu$.

Before collision, total momentum is $3m(ku)+5m(-2u)=3mku-10mu$.

If P has final velocity $u$, final momentum is $3m(u)+5m(3u)=18mu$. So $3mku-10mu=18mu$, giving $3k=28$ and $k=\dfrac{28}{3}$.

If P has final velocity $-u$, final momentum is $3m(-u)+5m(3u)=12mu$. So $3mku-10mu=12mu$, giving $3k=22$ and $k=\dfrac{22}{3}$.

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.

$\text{Impulse} = \text{Final momentum} - \text{Initial momentum}$ For particle Q (mass = 3m): ; Initial velocity: $-u$ (moving left, negative) ; Final velocity: $+2w$ (moving right, positive) $\text{Impulse} = 3m(2w) - 3m(-u)$ $\frac{7mu}{2} = 6mw + 3mu$ $\frac{7mu}{2} = 3m(2w + u)$ $\frac{7u}{2} = 6w + 3u$ $6w = \frac{7u}{2} - 3u$ $6w = \frac{7u - 6u}{2}$ $6w = \frac{u}{2}$ $w = \frac{u}{12}$

$\text{Total momentum before} = \text{Total momentum after}$ Before collision: $\text{P momentum} = 7m \times ku = 7mku \text{ (right)}$ $\text{Q momentum} = 3m \times (-u) = -3mu \text{ (left)}$ $\text{Total} = 7mku - 3mu$ After collision: $\text{P momentum} = 7m \times w = 7mw \text{ (right)}$ $\text{Q momentum} = 3m \times 2w = 6mw \text{ (right)}$ $\text{Total} = 7mw + 6mw = 13mw$

$7mku - 3mu = 13mw$ $7ku - 3u = 13w$

$7ku - 3u = 13 \times \frac{u}{12}$ $7ku - 3u = \frac{13u}{12}$ $7ku = \frac{13u}{12} + 3u$ $7ku = \frac{13u + 36u}{12}$ $7ku = \frac{49u}{12}$ $7k = \frac{49}{12}$ $k = \frac{49}{84} = \frac{7}{12}$

P could move either right (velocity = $+w$) or left (velocity = $-w$) Case 1: P moves right with velocity $+w$ Already solved: $k = \frac{7}{12}$ Case 2: P moves left with velocity $-w$ Total momentum after = $-7mw + 6mw = -mw$ $7mku - 3mu = -mw$ $7ku - 3u = -w$ $7ku - 3u = -\frac{u}{12}$ $7ku = 3u - \frac{u}{12}$ $7ku = \frac{36u - u}{12}$ $7ku = \frac{35u}{12}$ $k = \frac{35}{84} = \frac{5}{12}$

Take the original direction of A as positive. Before the collision, A has velocity $1.5$ and B has velocity $-1.5$.

After the collision, A keeps the same direction. Let A's speed be $v$. The valid motion has B moving in A's direction with speed $v+1$.

$3m(1.5)+m(-1.5)=3mv+m(v+1)$.

$4.5m-1.5m=3mv+mv+m$, so $3=4v+1$. Hence $v=0.5$, and B's speed is $v+1=1.5$.

For B, impulse equals change in momentum: $I=m(1.5-(-1.5))=3m$.

The magnitude of the impulse is $3m\ \mathrm{N\,s}$.

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

The boat travels straight, so the sideways components balance: $500\sin40^\circ=P\sin\alpha$.

The speed is constant, so the forward components balance the resistance: $500\cos40^\circ+P\cos\alpha=900$.

From the two equations, $P\sin\alpha=500\sin40^\circ$ and $P\cos\alpha=900-500\cos40^\circ$.

Divide the sine equation by the cosine equation: $\tan\alpha=\dfrac{500\sin40^\circ}{900-500\cos40^\circ}$. This gives $\alpha=31.9^\circ$, so $\alpha=32^\circ$ approximately.

Use $P\sin\alpha=500\sin40^\circ$: $P=\dfrac{500\sin40^\circ}{\sin31.9^\circ}=608.7\ldots\ \mathrm{N}$. Hence $P\approx610\ \mathrm{N}$.

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 $t=\dfrac{95}{10}=9.5\ \mathrm{s}$.

Take P's original direction as positive. Before collision, $v_P=u$ and $v_Q=-u$. After collision, Q has velocity $\dfrac{u}{3}$ because its direction is reversed.

Impulse on Q equals change in momentum: $I=m_2\left(\dfrac{u}{3}-(-u)\right)=m_2\left(\dfrac{4u}{3}\right)=\dfrac{4}{3}m_2u$.

Let the speed of P after collision be $v$. Since P reverses direction, its velocity is $-v$. Hence $m_1u-m_2u=-m_1v+\dfrac13m_2u$.

Rearrange: $m_1v=\dfrac43m_2u-m_1u$, so $v=\dfrac{(4m_2-3m_1)u}{3m_1}$.

Since $u>0$ and $m_1>0$, the numerator must be positive: $4m_2-3m_1>0$. Therefore $m_2>\dfrac34m_1$.

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.

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

Take distances from A. Then C is at $1$ m, D is at $3$ m, the rod's centre of mass is at $2.5$ m, and B is at $5$ m.

Taking moments about D: $R_C(2)+Mg(2)=10g(3)+5g(0.5)$.

$2R_C+2Mg=30g+2.5g=32.5g$, so $R_C=16.25g-Mg=(16.25-M)g$.

Taking moments about C: $10g(1)+R_D(2)=5g(1.5)+Mg(4)$.

$10g+2R_D=7.5g+4Mg$, so $2R_D=(4M-2.5)g$ and $R_D=(2M-1.25)g$.

For the rod to rest on both supports, both reactions must be non-negative.

$R_C\ge0$ gives $16.25-M\ge0$, so $M\le16.25$. Also $R_D\ge0$ gives $2M-1.25\ge0$, so $M\ge0.625$. Therefore $0.625\le M\le16.25$.

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.

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 $X=\dfrac{980}{3}\ \mathrm{N}$.

$AC=0.4$ m and $DB=0.4$ m, so $CD=2.2$ m. The beam is uniform, so its weight $30g$ acts at the midpoint, which is $1.1$ m from D. The person is $2.2-x$ m from D.

Taking moments about D removes $R_D$: $R_C(2.2)=55g(2.2-x)+30g(1.1)$.

Using $g=9.8$, $R_C=\dfrac{539(2.2-x)+294(1.1)}{2.2}=686-245x$.

Given $R_C=4R_D$ and vertical equilibrium gives $R_C+R_D=(55+30)g=833$. Hence $5R_D=833$ and $R_C=666.4$.

$686-245x=666.4$, so $245x=19.6$ and $x=0.08$.

When the package is at A and the beam is on the point of tilting about C, the reaction at D is zero.

Take moments about C: $Mg(0.4)=30g(1.1)$. Thus $M=\dfrac{33}{0.4}=82.5$.

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 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 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 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.