Corrected exercises: momentum, impulse and collisions, PHYS 101

Momentum is $p = mv$, and $v$ carries the sign of the chosen axis. With east positive, $p_{i} = (0.150)(+20.0) = +3.00$ kg m/s. The unit kg m/s is the same thing as the N s of an impulse, which is exactly why the two quantities can be set equal later.

The ball comes back, so its velocity is now negative: $v_{f} = -15.0$ m/s and $p_{f} = (0.150)(-15.0) = -2.25$ kg m/s. Nothing about the ball has changed except the sign, and that sign is the whole content of the chapter.

$\Delta p = p_{f} - p_{i} = (-2.25) - (+3.00) = -5.25$ kg m/s. In words: the wall delivered $5.25$ N s of impulse directed west. The classic loss of marks here is to subtract the two magnitudes, $3.00 - 2.25 = 0.75$, which is the answer to no question at all. Writing the signed values in a two line table before subtracting costs five seconds and saves the whole part.

The putty ends at rest: $\Delta p = 0 - (+3.00) = -3.00$ kg m/s. The bouncing ball therefore receives $5.25/3.00 = 1.75$ times the momentum change of the putty, although it arrives with exactly the same speed and the same mass. Reversing a velocity is more violent than cancelling it, and this single ratio explains why a rebounding hailstone dents a roof more than a snowball of the same mass, and why crumple zones are designed to absorb rather than to bounce.

$p = (6.00)(0.500) = +3.00$ kg m/s, identical to a). The kinetic energies are not: $K_{\text{ball}} = \frac{1}{2}(0.150)(20.0)^{2} = 30.0$ J against $K_{\text{bowling}} = \frac{1}{2}(6.00)(0.500)^{2} = 0.750$ J, a factor $40$. Equal momentum never means equal energy, because $p$ grows like $v$ while $K$ grows like $v^{2}$. Two objects can be matched on one quantity and separated by a factor of forty on the other, and any argument that slides from one to the other is worth zero marks.

The impulse momentum theorem reads $J = \Delta p = m v_{f} - m v_{i}$. Here $J = (70.0)(0) - (70.0)(15.0) = -1050$ N s, so $1050$ N s in magnitude, directed backwards. Note the unit: N s and kg m/s are the same unit written two ways, and marking the equality of the two on the copy is what makes the next line legitimate.

$J = F_{\text{av}} \Delta t$ gives $F_{\text{av}} = \dfrac{1050}{0.080} = 13125$ N, that is $1.31 \times 10^{4}$ N. This is an AVERAGE over the contact; the instantaneous peak is higher still, typically one and a half times the average for a real crash pulse.

$F_{\text{av}} = \dfrac{1050}{0.40} = 2625$ N. The ratio is $\dfrac{13125}{2625} = 5.0$, exactly the ratio of the two contact times, because the numerator $1050$ N s has not moved by a single unit.

The weight is $mg = (70.0)(9.80) = 686$ N. Without the airbag, $\dfrac{13125}{686} = 19.1$ times the body weight; with it, $\dfrac{2625}{686} = 3.83$ times. The human thorax tolerates a few times its own weight, not twenty, and that gap is the whole design brief of the device.

The airbag changes NOTHING to the impulse. The driver arrives with $1050$ kg m/s of momentum and must end at zero, so $1050$ N s must be delivered whatever the padding: that number is fixed by the mass and the speed alone. What the bag changes is the time over which the same area is spread, and since $F_{\text{av}} = J/\Delta t$, five times the time gives one fifth of the force. Bending the knees on landing, a helmet liner and a boxer riding a punch are the same sentence with different words. Saying that the airbag reduces the impulse is the single most expensive sentence of this chapter, because it means the theorem has not been understood at all.

Work in seconds: $4.0$ ms $= 0.0040$ s. Rising triangle: $\frac{1}{2}(0.0040)(400) = 0.80$ N s. Plateau rectangle, from $0.0040$ s to $0.0100$ s: $(0.0060)(400) = 2.40$ N s. Falling triangle, $4.0$ ms wide: $\frac{1}{2}(0.0040)(400) = 0.80$ N s. Total $J = 0.80 + 2.40 + 0.80 = 4.00$ N s. The milliseconds are where the marks go: leaving the times in ms multiplies the answer by one thousand, and an impulse of $4000$ N s on a tennis ball should stop the pen.

$J = \Delta p = m v_{f} - m v_{i}$ with $v_{i} = 0$, so $v_{f} = \dfrac{J}{m} = \dfrac{4.00}{0.200} = 20.0$ m/s. Ordinary for a served ball, which is the check to run before writing the answer down.

$F_{\text{av}} = \dfrac{J}{\Delta t} = \dfrac{4.00}{0.0140} = 286$ N. The peak is $400$ N, so the average is $0.714$ of the peak. Reading the top of the graph as the average is a standard trap: it would give $400 \times 0.0140 = 5.60$ N s, forty per cent too much, because the two triangles are half empty.

Choose the direction of the impulse as positive. The ball arrives against it, so $p_{i} = (0.200)(-8.00) = -1.60$ kg m/s. Then $p_{f} = p_{i} + J = -1.60 + 4.00 = +2.40$ kg m/s and $v_{f} = \dfrac{2.40}{0.200} = 12.0$ m/s. Compare with b): the same impulse produces a smaller outgoing speed, because part of it was spent undoing the incoming motion. Adding the two speeds instead, $8.00 + 20.0$, is the magnitude error again.

A rectangle of width $14.0$ ms and height $286$ N has area $4.00$ N s, the same as the real curve. It is the constant force that would do exactly the same job over the same time: same impulse, same final velocity, and a flat top instead of a peak. That is what makes the average force the right quantity to compare with a tolerance limit, while the peak is what breaks the string.

Momentum of the system: $p = (1.20)(+2.50) + (0.800)(0) = 3.00$ kg m/s. After coupling the two move as one body of mass $2.00$ kg, so $v = \dfrac{3.00}{2.00} = 1.50$ m/s, still to the right. Note that the answer is smaller than $2.50$ m/s and larger than $0$: it must be, since the pair is dragged by one cart and held back by the other.

$K_{i} = \frac{1}{2}(1.20)(2.50)^{2} = 3.75$ J. $K_{f} = \frac{1}{2}(2.00)(1.50)^{2} = 2.25$ J. Lost: $1.50$ J, that is $40.0$ per cent of the initial energy, gone into the deformation of the coupling and into heat. Nothing here is wrong: momentum is conserved and energy is not, and both statements are true at the same time.

Now $v_{B} = -1.00$ m/s. $p = (1.20)(+2.50) + (0.800)(-1.00) = 3.00 - 0.80 = 2.20$ kg m/s, so $v = \dfrac{2.20}{2.00} = 1.10$ m/s, still to the right but slower. Energy: $K_{i} = 3.75 + \frac{1}{2}(0.800)(1.00)^{2} = 3.75 + 0.40 = 4.15$ J, and $K_{f} = \frac{1}{2}(2.00)(1.10)^{2} = 1.21$ J. Lost: $2.94$ J, that is $70.8$ per cent. A head on collision destroys far more energy than a rear end one at comparable speeds, which is the physics behind the design of a divided highway.

Adding as positive numbers gives $p = 3.00 + 0.80 = 3.80$ and $v = \dfrac{3.80}{2.00} = 1.90$ m/s. The range check does not catch it: $1.90$ lies inside $[-1.00 , +2.50]$, the band every perfectly inelastic answer must sit in. What catches it is a comparison with part a): with B at rest the pair already moved at $1.50$ m/s, and a cart B that comes the OTHER way can only slow the pair down, so any answer above $1.50$ m/s is wrong before it is read. The one line fix is to write the signed table before the equation: $v_{A} = +2.50$, $v_{B} = -1.00$.

During the contact, A pushes B and B pushes A with forces equal in magnitude and opposite in direction, so their contributions to the total momentum cancel exactly and the sum cannot change. Energy has no such law: the same internal forces do work on deforming metal and rubber, and that work leaves the system as heat and sound, which is why the kinetic energy after is smaller and never larger.

With $m_{A} = m_{B}$ the first formula gives $v_{A}' = 0$ and the second $v_{B}' = v_{A} = 0.600$ m/s. Momentum: $(0.400)(0.600) = 0.240$ kg m/s before and after. Kinetic energy: $\frac{1}{2}(0.400)(0.600)^{2} = 0.0720$ J before and after. Equal masses in an elastic head on collision simply swap velocities, which is the one special case worth knowing by heart, and it is exactly what a Newton cradle shows.

$v_{A}' = \dfrac{0.600-0.200}{0.800}(4.00) = \dfrac{0.400}{0.800}(4.00) = 2.00$ m/s and $v_{B}' = \dfrac{2(0.600)}{0.800}(4.00) = 6.00$ m/s. Momentum: before $(0.600)(4.00) = 2.40$ kg m/s, after $(0.600)(2.00)+(0.200)(6.00) = 1.20+1.20 = 2.40$ kg m/s. Energy: before $\frac{1}{2}(0.600)(4.00)^{2} = 4.80$ J, after $\frac{1}{2}(0.600)(2.00)^{2}+\frac{1}{2}(0.200)(6.00)^{2} = 1.20+3.60 = 4.80$ J. Both hold, which is the definition of elastic. The light glider leaves faster than A arrived, and that is not a mistake: a light target always outruns the projectile that hit it.

Momentum still holds, whatever the bumpers: $(0.600)(2.50)+(0.200)v_{B}' = 2.40$, so $(0.200)v_{B}' = 2.40-1.50 = 0.90$ and $v_{B}' = 4.50$ m/s. Energy: $K_{f} = \frac{1}{2}(0.600)(2.50)^{2}+\frac{1}{2}(0.200)(4.50)^{2} = 1.875+2.025 = 3.90$ J, so $0.90$ J lost, that is $18.75$ per cent. Notice what could be used and what could not: momentum was an equation, energy was only an audit performed afterwards.

$v = \dfrac{2.40}{0.800} = 3.00$ m/s and $K_{f} = \frac{1}{2}(0.800)(3.00)^{2} = 3.60$ J, so $1.20$ J lost, that is $25.0$ per cent.

The kinetic energy after lies between $3.60$ J and $4.80$ J. The upper end is set by the elastic case, since no collision can create kinetic energy out of nothing when nothing explodes. The lower end is set by the perfectly inelastic case: once the two bodies move together there is no relative motion left to destroy, so no collision can lose more. Every real collision, the $3.90$ J of part c) included, sits somewhere in that band, and the only way to know where is to be told one more piece of information, a final velocity or an energy.

With $v_{1i} = v$ and $v_{2i} = 0$, the second term of each line vanishes:

Dimensions: each fraction is kilograms over kilograms, a pure number, so both answers are in m/s like $v$. A version with $m_{1}m_{2}$ on top and $m_{1}+m_{2}$ underneath would carry kilograms and fail on sight.

Case $m_{1} = m_{2}$: $v_{1f} = 0$ and $v_{2f} = v$. The gliders swap velocities, which is part a) again.

Case $m_{1} \gg m_{2}$: the fractions tend to $\frac{m_{1}}{m_{1}} = 1$ and $\frac{2m_{1}}{m_{1}} = 2$, so $v_{1f} \to v$ and $v_{2f} \to 2v$.

The heavy glider carries on almost unaffected, and the light one leaves at almost TWICE its speed: seen from the heavy glider, the light one comes in at $v$ and bounces back at $v$, and adding back the motion of the heavy glider gives $2v$.

Case $m_{1} \ll m_{2}$: $v_{1f} \to -v$ and $v_{2f} \to 0$. The light glider bounces back at its own speed, like a ball off a wall, and the heavy one barely moves.

Sense of variation: $v_{2f}$ grows with $m_{1}$, from $0$ towards $2v$ and never beyond. A heavier projectile always launches the target faster, but no projectile launches it faster than $2v$.

Pitfall: answering $v_{2f} = v$ for the light target, as if the heavy glider simply carried it along. It contradicts the limit just found, and part b) already sent B off at $6.00$ m/s after A arrived at $4.00$ m/s.

Kinetic energy after, in letters:

Factor $m_{1}$ in the numerator: $m_{1}\left[(m_{1}-m_{2})^{2} + 4m_{1}m_{2}\right] = m_{1}(m_{1}+m_{2})^{2}$, so $K_{f} = \frac{1}{2}m_{1}v^{2}$, the kinetic energy before. The two lines of the sheet ARE the elastic case.

Share of the target: $\dfrac{\frac{1}{2}m_{2}v_{2f}^{2}}{\frac{1}{2}m_{1}v^{2}} = \dfrac{4m_{1}m_{2}}{(m_{1}+m_{2})^{2}}$, a pure number, equal to $1$ when $m_{1} = m_{2}$ and tending to $0$ in both other limits.

Check with b): $\frac{4(0.600)(0.200)}{(0.800)^{2}} = 0.75$, and B did take $3.60$ J of the $4.80$ J.

Numbers: $v_{1f} = \dfrac{0.100-0.900}{1.000}(1.50) = -1.20$ m/s and $v_{2f} = \dfrac{2(0.100)}{1.000}(1.50) = 0.300$ m/s.

The light glider bounces back at $1.20$ m/s, already close to the limit $-v$ of a very heavy target.

Momentum: $(0.100)(1.50) = 0.150$ kg m/s before, $(0.100)(-1.20) + (0.900)(0.300) = -0.120 + 0.270 = 0.150$ kg m/s after.

Energy: $\frac{1}{2}(0.100)(1.50)^{2} = 0.1125$ J before, $\frac{1}{2}(0.100)(1.20)^{2} + \frac{1}{2}(0.900)(0.300)^{2} = 0.0720 + 0.0405 = 0.1125$ J after.

Share of the target: $\frac{4(0.100)(0.900)}{(1.000)^{2}} = 0.36$, which is indeed $\frac{0.0405}{0.1125}$. A projectile nine times HEAVIER would hand over the same $36$ per cent, since the formula does not change when $m_{1}$ and $m_{2}$ swap.

Pitfall: $v_{1f} = +1.20$ m/s. The minus sign is the rebound; without it the momentum after would be $0.390$ kg m/s instead of $0.150$.

Before the shot everything is at rest, so $p_{\text{total}} = 0$, and it stays zero. With the bullet direction positive: $0 = (0.00900)(+750) + (4.50)v$, so $v = -\dfrac{6.75}{4.50} = -1.50$ m/s. The minus sign IS the recoil; dropping it makes the rifle follow the bullet, which is the version of this exercise that scores zero. Convert the grams before anything else: $9.00$ g $= 0.00900$ kg, and a bullet of $9.00$ kg would give a recoil of $1500$ m/s.

$K_{\text{bullet}} = \frac{1}{2}(0.00900)(750)^{2} = 2531$ J and $K_{\text{rifle}} = \frac{1}{2}(4.50)(1.50)^{2} = 5.06$ J. The ratio is $\dfrac{2531}{5.06} = 500$, and $500$ is exactly $\dfrac{4.50}{0.00900}$, the mass ratio. The reason is $K = \dfrac{p^{2}}{2m}$: since the two momenta are equal in magnitude, the energies go like $1/m$, so the light piece takes almost everything. That is why a rifle bruises a shoulder while a bullet goes through a plank, and why an explosion sends the small fragments far and leaves the heavy casing nearby.

$0 = (60.0)(+1.20) + (45.0)v$, so $v = -\dfrac{72.0}{45.0} = -1.60$ m/s: the lighter skater leaves at $1.60$ m/s the other way. The lighter one always leaves faster, in the ratio of the masses, $\dfrac{60.0}{45.0} = 1.33$.

Here the astronaut alone is $80.0$ kg and the tool is separate, so $0 = (5.00)(+8.00) + (80.0)v$ and $v = -\dfrac{40.0}{80.0} = -0.500$ m/s. Half a metre per second is enough to cross ten metres in twenty seconds, which is why a tool thrown away from a spacecraft is a serious matter and not a joke.

The push is an INTERNAL force: skater on skater, hand on tool. Whatever its size and whatever its duration, it acts on both bodies with the same magnitude and opposite directions, so its two impulses cancel in the total. The total momentum therefore stays at the value it had before the push, namely zero, and the only equation left is $m_{1}v_{1} = -m_{2}v_{2}$. A gentler push gives both skaters smaller speeds, but the RATIO of the speeds is fixed by the masses alone. To find the individual speeds you would need the force and the time, that is to say an impulse; to find their ratio you need nothing at all.

Stage one, the embedding, lasts a few milliseconds. Momentum is conserved, kinetic energy is not: $(0.0100)(300) = (1.50)V$, so $V = \dfrac{3.00}{1.50} = 2.00$ m/s. Note that the string is vertical during the impact, so gravity and the tension have no horizontal component and the horizontal momentum really is conserved.

Stage two, the swing, lasts about a second. Now no energy is lost, since the block is a single body sliding along a circle with no friction, and momentum is NOT conserved, since the string pulls on the system. Energy: $\frac{1}{2}(1.50)(2.00)^{2} = (1.50)(9.80)h$, so $h = \dfrac{V^{2}}{2g} = \dfrac{4.00}{19.6} = 0.204$ m. The mass cancels, which is worth noticing: the rise depends only on the speed just after impact.

$K_{i} = \frac{1}{2}(0.0100)(300)^{2} = 450$ J and $K_{\text{after}} = \frac{1}{2}(1.50)(2.00)^{2} = 3.00$ J. Lost: $447$ J, that is $99.3$ per cent, turned into heat and splintered wood. This is a perfectly inelastic collision at its most extreme, and it is precisely why the energy law cannot be carried across the impact.

That line gives $h = \dfrac{450}{(1.50)(9.80)} = 30.6$ m, a block that clears a ten storey building on a string of $1.20$ m. It is wrong because it silently assumes that the $450$ J survive the embedding, while part c) has just shown that $99.3$ per cent of them do not. The rule to carry away is the order of the two laws: momentum ACROSS the collision, energy AFTER it, never one law for the whole story.

The block rises $h = L - L\cos\theta$, so $\cos\theta = 1 - \dfrac{h}{L} = 1 - \dfrac{0.204}{1.20} = 0.830$ and $\theta = 33.9^{\circ}$. The check is immediate: $h$ must be smaller than $L$, and an angle under $90^{\circ}$ means the string never went slack. A value of $\cos\theta$ outside $[-1 , 1]$ would have signalled an arithmetic slip one line earlier, which is the cheapest safety net of the whole problem.

FALSE. Momentum is a vector: in one dimension it carries the sign of the velocity with respect to the chosen axis. For a cart at $+3.0$ m/s and one at $-2.0$ m/s, the total is $m_{1}(+3.0) + m_{2}(-2.0)$, and the two contributions subtract. Correct version: the total momentum is the ALGEBRAIC sum of the signed momenta, once an axis has been chosen. Cost: the whole question, every time, because the final velocity comes out too large and often in the wrong direction; this is the single most expensive error of the chapter.

FALSE, and it confuses the two laws. Momentum is conserved in EVERY collision of an isolated system, whatever happens to the energy, because the internal forces are action reaction pairs whose impulses cancel. Correct version: in a perfectly inelastic collision momentum is conserved and kinetic energy is not. Cost: the student refuses to write the only equation available and hands in a blank page, so four or five marks.

TRUE. The three readings of the same quantity are $J = F_{\text{av}}\Delta t$, $J$ = area under the force time curve, and $J = \Delta p$. In PHYS 101 that area is always cut into triangles and rectangles, which is all that is needed. Nothing to correct.

FALSE. The impulse is fixed by the change of momentum, $\Delta p = m\Delta v$, which depends only on the mass of the driver and on the speed of the car. Correct version: the airbag leaves the impulse unchanged and stretches the contact time, and since $F_{\text{av}} = J/\Delta t$ a time five times longer gives a force five times smaller. Cost: two marks and, worse, a qualitative answer that contradicts the theorem the question is testing.

TRUE. Momentum is a vector, so its conservation is one equation per axis: $\sum p_{x}$ before $= \sum p_{x}$ after, and the same for $y$. Energy conservation, when it holds, is a single scalar equation and never splits by axis. The practical consequence is that a two dimensional collision gives two equations and therefore two unknowns, which is exactly what is needed when two angles are given. Nothing to correct.

The same force for the same time delivers the same impulse $J = F\Delta t$, and both sleds start from rest, so they end with the SAME momentum $p$.

$K = \dfrac{p^{2}}{2m}$, so at equal $p$ the kinetic energy goes like $\dfrac{1}{m}$:

Answer: the $20.0$ kg sled, with four times the kinetic energy of the other.

With a force of, say, $40.0$ N: $J = 120$ N s for each, so the light sled leaves at $6.00$ m/s with $360$ J and the heavy one at $1.50$ m/s with $90.0$ J. Any other force changes both numbers and keeps the ratio $4$.

The $80.0$ kg sled is wrong: that answer reads $K = \frac{1}{2}mv^{2}$ as growing with $m$ and forgets that $v$ is four times smaller and enters squared.

Both the same is wrong: equal force and equal TIME give equal momentum. Equal kinetic energy would need equal force over an equal DISTANCE, and the light sled covers four times more ground during the pull, $9.00$ m against $2.25$ m at $40.0$ N, so the force does four times more work on it.

Not enough information is wrong: the ratio needs only the masses, since the unknown impulse is the same for both sleds and cancels.

Pitfall: same impulse means same momentum, never same kinetic energy. It is the conceptual twin of part e) of exercise 1.

Momentum is conserved as a VECTOR, which means one equation per axis. Along $x$: $(0.200)(3.00) = (0.200)v_{A}'\cos 30.0^{\circ} + (0.200)v_{B}'\cos 60.0^{\circ}$. Along $y$, where the total was zero before: $0 = (0.200)v_{A}'\sin 30.0^{\circ} - (0.200)v_{B}'\sin 60.0^{\circ}$, the minus sign because B goes BELOW the axis. The common factor $0.200$ divides out, which is what makes the equal mass case quick.

The $y$ equation gives $0.500 v_{A}' = 0.866 v_{B}'$, so $v_{A}' = 1.732 v_{B}'$. Substituting into the $x$ equation: $(1.732 v_{B}')(0.866) + 0.500 v_{B}' = 3.00$, that is $1.500 v_{B}' + 0.500 v_{B}' = 3.00$, so $v_{B}' = 1.50$ m/s and $v_{A}' = 2.60$ m/s. Solving the $y$ equation FIRST is what keeps the algebra short: it is the equation with a zero on one side.