Corrected exercises: momentum, impulse and collisions (203-SN1-RE)

With the axis pointing in the direction of the kick: initial velocity $-12$ m/s, final velocity $+18$ m/s. $\Delta p=m(v_{f}-v_{i})=0.45(18-(-12))=0.45(30)=13.5$ kg·m/s. The classic mistake is to subtract the magnitudes ($18-12=6$): the ball changes DIRECTION, so the velocities have opposite signs.

$F=\frac{\Delta p}{\Delta t}=\frac{13.5}{0.010}=1350$ N.

Weight: $mg=(0.45)(9.8)\approx 4.4$ N. The impact force is about 300 times the weight: during the 10 ms of contact, the impulse of gravity ($4.4\times 0.010=0.044$ kg·m/s) is invisible next to that of the foot (13.5 kg·m/s). During the flight, on the other hand, gravity is the ONLY force and acts for whole seconds: it is the duration of action, not the size of the force alone, that decides what we neglect.

In both cases the change in momentum $\Delta p$ is imposed (going from the impact velocity to rest): it does not change. What changes is the time $\Delta t$ over which it is spread. Bent knees or an airbag lengthen $\Delta t$, and since $F=\frac{\Delta p}{\Delta t}$, the average force drops in proportion. We do not reduce the shock, we dilute it in time.

Because the force is not constant: it rises, levels off, then falls back. The general definition is $\vec{J}=\int_{t_{i}}^{t_{f}}\vec{F}\,dt$, and the impulse-momentum theorem states that $\vec{J}=\Delta\vec{p}$. On the graph, this integral is the AREA between the curve $F(t)$ and the time axis. Since the curve is made of straight segments, a simple geometric decomposition replaces the integral calculus.

Careful with units: the durations are in milliseconds, so they must be converted to seconds. Rising triangle: $\frac{1}{2}\left(4.00\times 10^{-3}\right)(550)=1.10$ N$\cdot$s. Rectangle: $\left(2.00\times 10^{-3}\right)(550)=1.10$ N$\cdot$s. Falling triangle: $\frac{1}{2}\left(4.00\times 10^{-3}\right)(550)=1.10$ N$\cdot$s. Total impulse: $J=3.30$ N$\cdot$s. The total contact time is $10.0$ ms, so the average force is $F_{avg}=\frac{3.30}{10.0\times 10^{-3}}=330$ N. It is indeed below the maximum of $550$ N, like any average of a bell-shaped curve.

Take as positive the direction in which the ball leaves. Before the impact the ball travels the other way: $p_{i}=(0.0580)(-25.0)=-1.45$ kg$\cdot$m/s. Impulse-momentum theorem: $p_{f}=p_{i}+J=-1.45+3.30=1.85$ kg$\cdot$m/s, so $v_{f}=\frac{1.85}{0.0580}\approx 31.9$ m/s, or about $115$ km/h. Forgetting the sign of the incoming velocity is the most common mistake of the chapter: it would give $81.9$ m/s here instead of $31.9$ m/s.

The weight of the ball is $mg=0.0580(9.80)=0.568$ N, almost a thousand times less than the maximum force from the strings. During the $10$ ms of contact, the impulse of the weight reaches $0.568(0.0100)=5.7\times 10^{-3}$ N$\cdot$s, completely negligible next to $3.30$ N$\cdot$s. This quantitative justification is the one to give, rather than simply stating that gravity is weak. With the contact lengthened to $15.0$ ms, same shape and same maximum, every area is multiplied by $1.50$: $J=4.95$ N$\cdot$s, so $v_{f}=\frac{4.95-1.45}{0.0580}\approx 60.3$ m/s. This is the whole logic of the follow-through: lengthening the contact means adding area.

Against the wall: $\Delta p=0.0580\left(20.0+25.0\right)=2.61$ kg$\cdot$m/s, so $F_{avg}=\frac{2.61}{2.00\times 10^{-3}}\approx 1.31\times 10^{3}$ N. That is almost four times the average force of the racquet, even though the change in momentum is smaller. Nearly all the difference comes from the duration: an impulse of the same order ($3.30$ N$\cdot$s against $2.61$ N$\cdot$s) spread over five times as long produces a force about four times smaller. Advice for the elbow: loosen the strings slightly. Softer strings sink in further, lengthen the contact time and lower the maximum force transmitted to the arm, exactly like an airbag or a knee pad.

During the fraction of a second of the coupling, the forces between the cars (internal to the system) are enormous compared with the friction of the rails (an external force): the impulse of friction over that time is negligible. Conservation applies on the scale of the collision, not on the scale of the day.

$v'=\frac{m_{1}v_{1}}{m_{1}+m_{2}}=\frac{(2000)(3)}{5000}=1.2$ m/s.

Before: $K=\frac{1}{2}(2000)(3)^{2}=9000$ J. After: $\frac{1}{2}(5000)(1.2)^{2}=3600$ J. Loss: 5400 J, or 60 % of the initial energy, gone into deformation of the couplers, sound and heat. Momentum, on the other hand, is fully transmitted: 6000 kg·m/s before and after.

With $v'=\frac{m_{1}v_{1}}{m_{1}+m_{2}}$: $\frac{K'}{K}=\frac{\frac{1}{2}(m_{1}+m_{2})v'^{2}}{\frac{1}{2}m_{1}v_{1}^{2}}=\frac{(m_{1}+m_{2})}{m_{1}v_{1}^{2}}\cdot\frac{m_{1}^{2}v_{1}^{2}}{(m_{1}+m_{2})^{2}}=\frac{m_{1}}{m_{1}+m_{2}}$. The fraction lost is therefore $1-\frac{m_{1}}{m_{1}+m_{2}}=\frac{m_{2}}{m_{1}+m_{2}}$: the speed has disappeared from the result. Here: $\frac{3000}{5000}=60$ %, in agreement with c). If $m_{2}\gg m_{1}$ (a ball of putty against a wall), the fraction tends to 100 %: almost all the energy is dissipated, which matches intuition.

Momentum: $m_{1}v_{1}=m_{1}v_{1}'+m_{2}v_{2}'$, that is $4=v_{1}'+3v_{2}'$. Kinetic energy (elastic collision): $\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{1}v_{1}'^{2}+\frac{1}{2}m_{2}v_{2}'^{2}$, that is $16=v_{1}'^{2}+3v_{2}'^{2}$.

For a projectile hitting a stationary target, solving the system gives $v_{1}'=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1}=\frac{1-3}{4}(4)=-2$ m/s and $v_{2}'=\frac{2m_{1}}{m_{1}+m_{2}}v_{1}=\frac{2}{4}(4)=2$ m/s. The light cart BOUNCES BACK ($v_{1}'<0$): it hits something heavier than itself; the heavy cart moves forward at 2 m/s.

Momentum: before $(1)(4)=4$ kg·m/s; after $(1)(-2)+(3)(2)=4$ kg·m/s. Energy: before $\frac{1}{2}(1)(16)=8$ J; after $\frac{1}{2}(1)(4)+\frac{1}{2}(3)(4)=2+6=8$ J. Both laws are satisfied.

Before: $v_{1}-v_{2}=4$ m/s (closing speed). After: $v_{1}'-v_{2}'=-2-2=-4$ m/s (separating at the same rate): the relative velocity has indeed reversed. This relation is LINEAR: combined with conservation of $p$ (also linear), it replaces the energy equation (quadratic) and gives a system of two first-degree equations, quick to solve. This is the standard trick for elastic collisions in one dimension: the quadratic equation was only needed to establish the property once and for all.

In space, no external force acts along the line of the throw on the astronaut-wrench system: no friction, no support. The forces of the throw are internal to the system: the total $P$ stays what it was, that is, zero.

$0=m_{a}v_{a}+m_{w}v_{w}$, so $v_{a}=-\frac{(2)(8)}{80}=-0.2$ m/s: the astronaut recoils at 0.2 m/s toward the station. Time: $t=\frac{30}{0.2}=150$ s, that is 2 min 30 s. The trip is slow but sure: nothing will slow the astronaut down.

$K=\frac{1}{2}(2)(8)^{2}+\frac{1}{2}(80)(0.2)^{2}=64+1.6=65.6$ J. There is no violation: conservation of KINETIC energy is a law only for elastic collisions. Here the energy comes from the chemical energy of the astronaut's muscles: the throw is an anti-collision, a miniature explosion that CREATES kinetic energy from another form.

At a fixed throwing speed $v$, the recoil is $v_{a}=\frac{m_{w}v}{m_{a}}$: it grows with the thrown mass, so throwing heavy is more effective. The energy is shared according to the speeds: the wrench carries off $\frac{64}{65.6}\approx 97.6$ % of the energy produced, and the astronaut recovers only 2.4 %. This is general: when the total $P$ is zero, the energies are in inverse proportion to the masses ($E=\frac{p^{2}}{2m}$ with the same $p$): the light body carries off almost everything. The muscles pay dearly for each m/s of recoil, and this is exactly the problem a real thruster solves, or a rocket, which throws gas very fast and continuously.

$x_{cm}=\frac{\sum m_{i}x_{i}}{\sum m_{i}}=\frac{2.00(0)+3.00(4.00)+5.00(2.00)}{10.0}=\frac{22.0}{10.0}=2.20$ m. $y_{cm}=\frac{2.00(0)+3.00(0)+5.00(3.00)}{10.0}=\frac{15.0}{10.0}=1.50$ m. The centre of mass is at $(2.20, 1.50)$ m. It is pulled toward the $5.00$ kg mass, which is the expected behaviour of a weighted average.

No external horizontal force acts on the angler-boat system: the centre of mass therefore stays perfectly at rest relative to the shore. Let $D$ be the displacement of the boat relative to the shore. The angler moves $4.00$ m relative to the boat, so $D+4.00$ relative to the shore. The condition of a fixed centre of mass is $60.0\left(D+4.00\right)+80.0D=0$, that is $140D=-240$ and $D=-1.71$ m. The boat therefore moves back $1.71$ m, opposite to the direction of the walk. The angler moves $-1.71+4.00=+2.29$ m relative to the shore.

Measure positions from the shore, the boat initially occupying the interval from $0$ to $4.00$ m, the angler at $0$ and the centre of the boat at $2.00$ m. Before: $x_{cm}=\frac{60.0(0)+80.0(2.00)}{140}=\frac{160}{140}\approx 1.14$ m. After: the angler is at $2.29$ m and the centre of the boat at $2.00-1.71=0.286$ m, so $x_{cm}=\frac{60.0(2.29)+80.0(0.286)}{140}=\frac{137.1+22.9}{140}\approx 1.14$ m. The same, as it must be.

No, never. By walking, the angler only exchanges INTERNAL forces with the boat: the centre of mass cannot move, and the bow backs away as the angler advances. At best the angler gains the $2.29$ m calculated in b), always less than the length of the boat. To reach the dock, a force EXTERNAL to the system is needed: pushing on the bottom with a pole, rowing and so pushing on the water, or throwing a heavy object backward, which amounts to removing that object from the system. The same impossibility prevents an isolated astronaut from moving by waving the arms and legs.

The explosion is an INTERNAL event: the forces it releases change nothing about the motion of the centre of mass, which carries on along the original parabola undisturbed. Since the highest point is $60.0$ m from the gun, the total range of the intact shell would have been $120$ m by symmetry: that is where the centre of mass lands. Both fragments leave the highest point with the same zero vertical velocity and fall $40.0$ m, so they land at the same instant, $t=\sqrt{\frac{2(40.0)}{9.80}}\approx 2.86$ s later, and the centre of mass relation applies as is at landing: $\frac{4.00(60.0)+8.00x_{2}}{12.0}=120$, so $x_{2}=150$ m. Check with horizontal momentum: the shell was moving forward at $\frac{60.0}{2.86}=21.0$ m/s, so $p=12.0(21.0)=252$ kg$\cdot$m/s; afterward, $4.00(0)+8.00\left(\frac{90.0}{2.86}\right)=8.00(31.5)=252$ kg$\cdot$m/s. The two methods agree.

$P_{x}=(1200)(20)=24{,}000$ kg·m/s (east); $P_{y}=(1600)(15)=24{,}000$ kg·m/s (north). Each car feeds only one axis: this is what makes the problem readable component by component.

Total mass 2800 kg, and each component is conserved: $v_{x}=\frac{24{,}000}{2800}\approx 8.57$ m/s, $v_{y}\approx 8.57$ m/s. Magnitude: $v'=\sqrt{v_{x}^{2}+v_{y}^{2}}\approx 12.1$ m/s; direction: $\tan\alpha=\frac{P_{y}}{P_{x}}=1$, that is 45° north of east.

Before: $\frac{1}{2}(1200)(20)^{2}+\frac{1}{2}(1600)(15)^{2}=240{,}000+180{,}000=420{,}000$ J. After: $\frac{1}{2}(2800)(12.12)^{2}\approx 205{,}700$ J. Loss: about 214,000 J, or 51 %: half the energy went into crumpled metal, a typical order of magnitude for perpendicular urban collisions.

The direction of the wreck encodes the ratio of the momenta: $\tan 38^{\circ}=\frac{P_{y}}{P_{x}}=\frac{(1600)(15)}{1200\,v_{1}}$, so $v_{1}=\frac{24{,}000}{1200\tan 38^{\circ}}=\frac{20}{0.781}\approx 25.6$ m/s, or about 92 km/h. The limit of 50 km/h corresponds to 13.9 m/s: the measured angle requires almost double. The marks contradict the driver: the CLOSER to the east axis the wreck leaves, the faster the eastbound car was going. This is exactly how momentum serves as an expert witness in accident reconstructions.

Conservation of momentum: $m\vec{v}_{0}=m\vec{v}_{1}+m\vec{v}_{2}$, and since the masses are equal they cancel: $\vec{v}_{0}=\vec{v}_{1}+\vec{v}_{2}$. Conservation of kinetic energy, the collision being elastic: $\frac{1}{2}mv_{0}^{2}=\frac{1}{2}mv_{1}^{2}+\frac{1}{2}mv_{2}^{2}$, that is $v_{0}^{2}=v_{1}^{2}+v_{2}^{2}$. The first result is a vector equation and the second a scalar one: it is precisely the confrontation of the two that gives everything.

Square the vector relation, that is, take the dot product of $\vec{v}_{0}$ with itself: $v_{0}^{2}=\left(\vec{v}_{1}+\vec{v}_{2}\right)\cdot\left(\vec{v}_{1}+\vec{v}_{2}\right)=v_{1}^{2}+v_{2}^{2}+2\,\vec{v}_{1}\cdot\vec{v}_{2}$. But conservation of energy requires $v_{0}^{2}=v_{1}^{2}+v_{2}^{2}$. Comparing, what remains is $\vec{v}_{1}\cdot\vec{v}_{2}=0$. Two nonzero vectors with a zero dot product are perpendicular: the balls leave at $90^{\circ}$ from each other. The proof assumes neither the geometry of the contact, nor the point of impact, nor the initial speed. It assumes only equal masses and an elastic collision, which explains why every billiards player knows the rule without knowing the proof.

Since the angle between $\vec{v}_{1}$ and $\vec{v}_{2}$ is $90^{\circ}$, the three vectors form a right triangle with $\vec{v}_{0}$ as the hypotenuse. The object ball leaving at $35.0^{\circ}$: $v_{2}=v_{0}\cos 35.0^{\circ}=3.00(0.8191)\approx 2.46$ m/s. The cue ball: $v_{1}=v_{0}\sin 35.0^{\circ}=3.00(0.5736)\approx 1.72$ m/s, and it leaves at $55.0^{\circ}$ on the other side of the initial direction, since $35.0+55.0=90.0$.

Along the initial axis: $2.457\cos 35.0^{\circ}+1.721\cos 55.0^{\circ}=2.013+0.987=3.00$ m/s, which is indeed $v_{0}$. Along the perpendicular axis: $2.457\sin 35.0^{\circ}-1.721\sin 55.0^{\circ}=1.410-1.410=0$, as before the collision. Kinetic energy: $v_{1}^{2}+v_{2}^{2}=2.961+6.039=9.00=v_{0}^{2}$. All three checks pass.

If the collision is not elastic, some kinetic energy is lost: $v_{1}^{2}+v_{2}^{2}<v_{0}^{2}$. Substituting into the identity of part b), $v_{0}^{2}=v_{1}^{2}+v_{2}^{2}+2\,\vec{v}_{1}\cdot\vec{v}_{2}$, we get $2\,\vec{v}_{1}\cdot\vec{v}_{2}>0$, so the $\cos$ of the angle is strictly positive: the angle becomes LESS than $90^{\circ}$. The two paths close up toward the front, and more so the softer the collision. This is indeed what is seen in billiards with worn balls or a damp cloth. As for the electron and the proton, the proof collapses on the first line: it required cancelling the masses, which assumes $m_{1}=m_{2}$. The proton is nearly $1{,}840$ times more massive than the electron, and the electron then bounces almost straight back while the proton hardly moves. Historically, it was this $90^{\circ}$ test in bubble chambers that made it possible to identify a collision between two particles of the same mass.

Phase 1, the impact (brief, violent): momentum is conserved (external forces negligible during the millisecond of the collision), but NOT kinetic energy: the projectile lodges in the wood, a perfectly inelastic collision. Phase 2, the rise (slow, no collision): mechanical energy is conserved (the tension in the strings does no work), but momentum is NOT conserved: gravity and tension, external forces, act throughout the rise and eventually stop the block.

Rise: $\frac{1}{2}(M+m)V^{2}=(M+m)gh$, so $V=\sqrt{2(9.8)(0.050)}=\sqrt{0.98}\approx 0.99$ m/s.

Impact: $mv=(M+m)V$, so $v=\frac{2.50}{0.010}(0.99)\approx 247$ m/s. The ballistic pendulum turns a speed that cannot be measured directly (247 m/s) into a height of 5 cm, measurable with a ruler: it is a multiplier.

$\frac{K'}{K}=\frac{\frac{1}{2}(M+m)V^{2}}{\frac{1}{2}mv^{2}}$ with $V=\frac{mv}{M+m}$ gives $\frac{m}{M+m}=\frac{0.010}{2.50}=0.4$ %. So 99.6 % of the projectile's energy disappears in the wood (heating, tearing of fibres): only the fraction $\frac{m}{M+m}$ survives, and the fraction lost is indeed $\frac{M}{M+m}$. If we naively wrote $\frac{1}{2}mv^{2}=(M+m)gh$ (conservation of energy from start to finish), we would find $v=\sqrt{2(9.8)(0.050)\times 250}\approx 16$ m/s, fifteen times too small: we would have ignored the 99.6 % of the energy swallowed by the impact. Hence the rule of the chapter: momentum steps over the collision, energy steps over the rise, never the reverse.

The person-plus-cart system is subject to no external horizontal force, so its momentum is conserved. Before: $p=\left(60.0+40.0\right)(2.00)=200$ kg$\cdot$m/s. After: $60.0(-3.00)+40.0v=200$, so $40.0v=200+180=380$ and $v=9.50$ m/s. The cart is strongly pushed forward, which makes sense: the person carried off backward a negative momentum, and it had to be compensated.

If the $3.00$ m/s is relative to the cart, the velocity of the person relative to the ground is $v-3.00$, where $v$ is the final velocity of the cart. Conservation is written $60.0\left(v-3.00\right)+40.0v=200$, that is $100v=380$ and $v=3.80$ m/s; the person then ends up at $0.800$ m/s forward, not backward. The two answers differ because velocity is relative to the chosen frame of reference: $3.00$ m/s backward relative to a cart that is moving forward is not $3.00$ m/s backward relative to the ground. The second formulation is the more realistic, because the person's legs push on the CART: it is indeed a relative velocity that the jump imposes. Reading carefully the frame of reference of each velocity is half the work in this chapter.

Nothing external. A rocket does not push on the air or on the ground: it violently ejects part of its own mass backward, and by conservation of momentum the rest moves forward. This is exactly the jump of part a), repeated thousands of times per second. The thrust is the rate of transfer of momentum to the gas: $F=v_{ex}\frac{dm}{dt}=2{,}500(150)=3.75\times 10^{5}$ N.

Far from any star or planet, thrust is the only force. At launch: $a_{i}=\frac{3.75\times 10^{5}}{3.00\times 10^{4}}=12.5$ m/s$^{2}$. At burnout only the structure remains, $3.00\times 10^{4}-2.10\times 10^{4}=9.00\times 10^{3}$ kg, so $a_{f}=\frac{3.75\times 10^{5}}{9.00\times 10^{3}}\approx 41.7$ m/s$^{2}$, more than three times as much. Burn time: $\frac{2.10\times 10^{4}}{150}=140$ s. The acceleration increases because the thrust stays constant while the mass to be accelerated keeps decreasing: the rocket is the perfect counterexample to the reflex of treating mass as a constant in $F=ma$.

$\Delta v=2{,}500\ln\frac{3.00\times 10^{4}}{9.00\times 10^{3}}=2{,}500\ln 3.33\approx 3.01\times 10^{3}$ m/s. The naive estimate would give $a_{i}t=12.5(140)=1{,}750$ m/s, or $42\,\%$ less: assuming the acceleration is constant and equal to its initial value badly underestimates the result, since the acceleration only increases. The true average acceleration is $\frac{3{,}010}{140}\approx 21.5$ m/s$^{2}$, comfortably between $12.5$ and $41.7$. Finally, the logarithm explains staging: $\Delta v$ depends only on the RATIO of masses, and that ratio is dragged down by the empty tanks, which stay heavy once useless. Dropping them restarts the mass ratio at a favourable value for the next stage, and the $\Delta v$ values add up. Without staging, no chemical launch vehicle could reach the $7.9$ km/s needed for low Earth orbit.