Momentum, Impulse & Collisions: past paper questions & worked solutions

Take the original direction of Q as positive. Q has mass $3m$ and initial velocity $2u$. Let its velocity after collision be $v$.

The impulse has magnitude $4mu$ and acts opposite to Q's original motion, so $4mu=3m(2u-v)$.

Cancel $m$: $4u=6u-3v$. Thus $3v=2u$ and $v=\dfrac{2u}{3}$.

Since $v$ is positive in Q's original direction, Q continues in the same direction after the collision.

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: $I=m\left|\dfrac{U}{4}-U\right|$.

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

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

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 $10\ \mathrm{N\,s}$.

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 $3u$.

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: $I=2m\left(\dfrac{6u}{5}-(-3u)\right)$.

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

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 $10u\ \mathrm{kg\,m\,s^{-1}}$.

Take the original direction of P as positive. Before the collision, P has velocity $3u$ and Q has velocity $-u$.

After collision, each direction is reversed and each speed is halved. So P has velocity $-\dfrac32u$ and Q has velocity $\dfrac12u$.

$km(3u)+m(-u)=km\left(-\dfrac32u\right)+m\left(\dfrac12u\right)$.

Divide by $mu$: $3k-1=-\dfrac32k+\dfrac12$. Hence $\dfrac92k=\dfrac32$, so $k=\dfrac13$.

Impulse on Q equals change in momentum of Q: $I=m\left(\dfrac12u-(-u)\right)=\dfrac32mu$.

The magnitude of the impulse is $\dfrac32mu$.

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 $k=\dfrac{22}{3}$.

$\text{Impulse} = \text{Final momentum} - \text{Initial momentum}$ 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 $3m\ \mathrm{N\,s}$.

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 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 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 $695\ \mathrm{N}$.

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

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.

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 $v=1.8\ \mathrm{m\,s^{-1}}$.

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.

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: $t_2 = \frac{\sqrt{g}}{g} = \frac{1}{\sqrt{g}}$

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