# Momentum and collisions: what survives a crash

> Momentum, impulse and Newton's third law, and how they fix the outcome of sticky and bouncy collisions when energy alone gives the wrong answer.

Series: Physics (https://www.monocurl.com/monocurl-essays/physics/)
Canonical: https://www.monocurl.com/monocurl-essays/physics/momentum-collisions/
Keywords: conservation of momentum, impulse momentum theorem, elastic collision, inelastic collision, center of mass, Newton's third law, kinetic energy lost in a collision

[Figure 1, animation: A 2 kg cart moving at 3 m/s runs into a 1 kg cart at rest and the two latch together, shown at half speed. The bar splits the total momentum, 6 kg·m/s, between the carts: the split changes at the collision and the total does not.]

A 2 kg cart rolls along a level track at 3 m/s and runs into a 1 kg cart that is standing still. Strips of Velcro on the bumpers make them latch together. How fast does the pair move afterward?

[The last lesson](@physics/energy-conservation) suggests a method. The moving cart has kinetic energy $\tfrac12 mv^2 = \tfrac12(2)(3)^2 = 9$ J. If energy is conserved, the 3 kg pair has 9 J afterward, so $\tfrac12(3)v^2 = 9$ and $v = 2.45$ m/s. Run the experiment and the pair moves off at 2.00 m/s, every time. The energy method is off by more than a fifth, and the kinetic energy afterward is $\tfrac12(3)(2)^2 = 6$ J: a third of it has gone missing in a collision that takes less than a tenth of a second.

So kinetic energy does not survive this crash. But the outcome is the same on every run, so some rule fixes it, and the way to find the rule is to look for a quantity that does survive.

## Four collisions, and a search for what stays the same

Keep the same two carts, A with mass 2 kg and B with mass 1 kg, on a track with so little friction that a cart given a push keeps its speed for the length of the track. Take positions along the track as $x$, increasing to the right, so that a velocity is positive for motion to the right and negative for motion to the left. We can change the bumpers (Velcro, or springs that make the carts bounce apart) and the starting velocities. Here are four runs.

| Run | Bumpers | A before | B before | A after | B after |
|---|---|---|---|---|---|
| 1 | Velcro | $+3$ m/s | $0$ | $+2$ m/s | $+2$ m/s |
| 2 | springs | $+3$ m/s | $0$ | $+1$ m/s | $+4$ m/s |
| 3 | Velcro | $+3$ m/s | $-3$ m/s | $+1$ m/s | $+1$ m/s |
| 4 | springs | $+3$ m/s | $-3$ m/s | $-1$ m/s | $+5$ m/s |

[Figure 2, animation: Run 2 at half speed: the same carts with spring bumpers. Cart A slows from 3 m/s to 1 m/s, cart B leaves at 4 m/s, and the total momentum is 6 kg·m/s before and after.]

**Pause and try.** For each run, compute three candidate totals before and after the collision: the sum of the velocities, the sum of mass times velocity, and the sum of $\tfrac12 \times$ mass $\times$ velocity squared. Which total is the same before and after in all four runs?

**Answer.** *Sum of velocities:* $3 \to 4$, $3 \to 5$, $0 \to 2$, $0 \to 4$. It changes every time. *Sum of $mv$* (in kg·m/s): run 1, $6 + 0 \to 4 + 2$; run 2, $6 + 0 \to 2 + 4$; run 3, $6 - 3 \to 2 + 1$; run 4, $6 - 3 \to -2 + 5$. It is 6, 6, 3 and 3 before, and the same after. *Sum of $\tfrac12 mv^2$* (in J): $9 \to 6$, $9 \to 9$, $13.5 \to 1.5$, $13.5 \to 13.5$. It survives with spring bumpers and not with Velcro.

The sum of mass times velocity is unchanged in every run, whatever the bumpers are made of, provided we keep the signs: in run 4, cart A ends up moving left and contributes $-2$. The reason has to come from Newton's laws, since those govern the carts at every instant, including the instants of contact that the table skips.

## Impulse: what a force does over a time

In the last lesson we asked what a force accomplishes over a *distance*, and the answer was a change in $\tfrac12 mv^2$. Ask the same question about a force acting for a *time*.

Suppose a constant net force $F$ acts on an object of mass $m$ for a time $\Delta t$. By [Newton's second law](@physics/newtons-second-law) the acceleration is $a = F/m$, and with constant acceleration the velocity changes by $\Delta v = a\,\Delta t$. Multiply by $m$:

$$
m\,\Delta v = F\,\Delta t.
$$

The left side is the change in $mv$, the combination that the table singled out.

**Definition.** The **momentum** of an object of mass $m$ moving with velocity $v$ is $p = mv$. In one dimension it carries the sign of the velocity. Its unit is kg·m/s. The **impulse** delivered by a constant force $F$ acting for a time $\Delta t$ is $J = F\,\Delta t$, with unit N·s. The two units agree: $1 \text{ N·s} = 1\ (\text{kg·m/s}^2)\cdot\text{s} = 1$ kg·m/s.

**Theorem.** (Impulse-momentum theorem.) In an inertial frame, if a constant net force $F$ acts on an object of constant mass $m$ for a time $\Delta t$, the object's momentum changes by the impulse of that force:

$$
\Delta p = m v_f - m v_i = F\,\Delta t.
$$

**Proof.** The second law gives constant acceleration $a = F/m$, so $v_f = v_i + a\,\Delta t = v_i + F\,\Delta t/m$. Multiply by $m$ and subtract $mv_i$.

During a collision the force is far from constant. The repair is the one used for work in the last lesson: cut the contact time into slices so short that the force hardly changes during each, apply the theorem to each slice, and add. The momentum changes add up to the total change, and the impulses add up to the area under the graph of force against time. Making "hardly changes" exact is a limit that belongs to [integral calculus](@calculus/integrals); for a force that is constant on each of a finite number of slices the result is fully proved, and that is the form we use.

One consequence is already useful. A given change of momentum can be delivered by a large force briefly or a small force for longer, since only the product $F\,\Delta t$ is fixed. Airbags and crumple zones work by lengthening $\Delta t$.

## The third law makes the total constant

Now look inside the collision of run 2. With spring bumpers of realistic stiffness, the carts are in contact for about 60 thousandths of a second.

Each cart feels three forces. Its weight and the normal force from the track are vertical and cancel, since the cart has no vertical acceleration. The third is the push of the other cart's bumper, which is horizontal. Newton's third law says that these two pushes are equal in size and opposite in direction at every instant: if A pushes B forward with force $F$, then B pushes A backward with force $-F$.

[Figure 3, image: The forces on the two carts while their bumpers touch, each drawn from the cart's center. Each weight is canceled by a normal force. The two contact forces are a third-law pair.]

The two pushes also last exactly as long as each other, because they are two descriptions of one contact. So in every slice of time the impulse on A is minus the impulse on B, and whatever momentum B gains, A loses.

[Figure 4, steps: Run 2 slowed down, with the spring bumpers modeled as an ideal spring. Left: the force on each cart during the 60 ms of contact. Right: the two velocities. The forces are mirror images, so the impulses are equal and opposite.]

  - step: Cart A's bumper pushes cart B forward with a force that rises to 105 N and falls back to zero. The velocities change meanwhile.
  - step: At every instant cart B pushes cart A backward with the same strength, so the blue curve is the mirror image.
  - step: The area under each curve is the impulse: 4.0 N·s forward on B and 4.0 N·s backward on A.
  - step: Each velocity changes by impulse over mass. The 1 kg cart gains 4 m/s and the 2 kg cart loses 2 m/s.
The figure models the bumper as an ideal spring, which gives the force its smooth hump. Rubber, Velcro or crumpling steel would give other curves, and in each case the curve for A would be the mirror image of the curve for B.

**Theorem.** (Conservation of momentum, two bodies.) In an inertial frame, let two objects interact with each other over some interval of time, and suppose the net *external* force on the pair (the sum of all forces exerted on either object by anything other than the two objects) is zero throughout. Then the total momentum $p_A + p_B = m_A v_A + m_B v_B$ has the same value at the end of the interval as at the beginning.

**Proof.** Cut the interval into short slices, on each of which the forces are constant (or nearly so; see the remark on slices above). In one slice of length $\Delta t$, let $F$ be the force that A exerts on B. By Newton's third law the force that B exerts on A is $-F$. Let the external forces on A and B be $E_A$ and $E_B$, with $E_A + E_B = 0$ by hypothesis (for the carts each is zero). By the impulse-momentum theorem, $\Delta p_A = (-F + E_A)\,\Delta t$ and $\Delta p_B = (F + E_B)\,\Delta t$, so $\Delta p_A + \Delta p_B = 0$ in that slice. Adding over all slices, the total momentum does not change.

Nothing was assumed about the nature of the force $F$, which is why Velcro and springs obey the same rule. If the net external force is not zero, the same proof shows that the total momentum changes by the external impulse, $F_{\text{ext}}\,\Delta t$. For more than two objects the internal forces still cancel in third-law pairs, and the theorem holds for the whole collection.

**Key idea.** Forces between the parts of a system come in equal and opposite pairs that act for the same time, so they move momentum from one part to another and never change the total. Only a force from outside can do that.

A real track has some friction, which is an external force. It does little harm, because the collision is so brief: a rolling friction force of 0.05 N acting for 0.06 s delivers 0.003 N·s, against the 4 N·s the carts exchange. Momentum conservation across a collision is an excellent approximation whenever the collision forces are much larger than the external ones.

## Sticky collisions and the missing energy

With the law in hand, the opening question is one line. In a **perfectly inelastic** collision the objects move off together with a common velocity $v$. Let their masses be $m_1, m_2$ and their velocities beforehand $u_1, u_2$. Conservation of momentum gives $m_1 u_1 + m_2 u_2 = (m_1 + m_2)\,v$, so

$$
v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}.
$$

For run 1, $v = (2 \times 3 + 0)/3 = 2$ m/s. For run 3, $v = (6 - 3)/3 = 1$ m/s. Both match the table.

Now the energy. Write $M = m_1 + m_2$. The kinetic energy before minus the kinetic energy after is

$$
K_i - K_f = \tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2 - \frac{(m_1 u_1 + m_2 u_2)^2}{2M}.
$$

Put everything over $2M$. The numerator is $(m_1 u_1^2 + m_2 u_2^2)(m_1 + m_2) - (m_1 u_1 + m_2 u_2)^2$, and when it is multiplied out the terms $m_1^2 u_1^2$ and $m_2^2 u_2^2$ cancel, leaving $m_1 m_2\,(u_1^2 + u_2^2 - 2u_1 u_2)$. So

$$
K_i - K_f = \frac12\,\frac{m_1 m_2}{m_1 + m_2}\,(u_1 - u_2)^2.
$$

This is never negative, and it is zero only if $u_1 = u_2$, in which case the objects never collide. Every sticky collision loses kinetic energy, and the amount depends only on the masses and the velocity of approach $u_1 - u_2$. For run 1 it is $\tfrac12 \cdot \tfrac{2}{3} \cdot 3^2 = 3$ J, the 3 J that went missing at the start. For run 3 it is $\tfrac12 \cdot \tfrac23 \cdot 6^2 = 12$ J, which is $13.5 - 1.5$.

The energy has gone into deforming the bumpers, into sound, and mostly into thermal energy. You might object that this is suspicious: the internal forces cancel well enough to leave momentum alone, so why do they not leave energy alone? The answer is that the two pushes act for the same *time* and not over the same *distance*. While the bumpers are being crushed the gap between the carts is closing, so cart A travels farther than cart B. The force slows A over a longer distance than it speeds B up, and A loses more kinetic energy than B gains. Impulse is force times time, and cancels. Work is force times distance, and need not.

## Bouncy collisions: two conditions, two unknowns

When the carts separate after the collision there are two unknown velocities, $v_1$ and $v_2$, and momentum conservation is only one equation. A second condition has to come from what the bumpers are made of.

The cleanest case is the one where no kinetic energy is lost. A collision is **elastic** if the total kinetic energy is the same after as before. Steel balls and good spring bumpers come close, and runs 2 and 4 are elastic, as their 9 J and 13.5 J totals showed.

For an elastic collision we have

$$
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2, \qquad \tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2 = \tfrac12 m_1 v_1^2 + \tfrac12 m_2 v_2^2.
$$

The second equation is quadratic, and substitution is unpleasant. A better route is to collect the terms for each object on one side:

$$
m_1 (u_1 - v_1) = m_2 (v_2 - u_2), \qquad m_1 (u_1^2 - v_1^2) = m_2 (v_2^2 - u_2^2).
$$

Each side of the second equation is a difference of squares, and it contains the corresponding side of the first as a factor: $u_1^2 - v_1^2 = (u_1 - v_1)(u_1 + v_1)$. Divide the second equation by the first. This is allowed when $u_1 \ne v_1$, that is, when the collision changes the velocities at all. What remains is $u_1 + v_1 = v_2 + u_2$, or

$$
v_1 - v_2 = -(u_1 - u_2).
$$

In an elastic collision the relative velocity reverses: the objects separate at the speed with which they approached. Check run 2: approach at $3 - 0 = 3$ m/s, separation at $4 - 1 = 3$ m/s. (The excluded case, $v_1 = u_1$ and $v_2 = u_2$, also satisfies both conservation laws. It describes the carts passing through each other untouched.)

Now the problem is linear. Substitute $v_2 = v_1 + u_1 - u_2$ into the momentum equation and solve for $v_1$, then for $v_2$.

**Theorem.** (Elastic collision in one dimension.) Two objects of masses $m_1, m_2$ move along a line with velocities $u_1, u_2$, collide, and separate with velocities $v_1, v_2$ different from $u_1, u_2$. If no net external force acts and the total kinetic energy is unchanged, then

$$
v_1 = \frac{(m_1 - m_2)\,u_1 + 2 m_2 u_2}{m_1 + m_2}, \qquad v_2 = \frac{(m_2 - m_1)\,u_2 + 2 m_1 u_1}{m_1 + m_2}.
$$

**Proof.** Momentum conservation and the reversal of relative velocity (derived above from the two conservation laws) give $m_1 v_1 + m_2 v_2 = m_1 u_1 + m_2 u_2$ and $v_2 = v_1 + u_1 - u_2$. Substituting, $(m_1 + m_2)\,v_1 + m_2 (u_1 - u_2) = m_1 u_1 + m_2 u_2$, so $(m_1 + m_2)\,v_1 = (m_1 - m_2)\,u_1 + 2 m_2 u_2$. The formula for $v_2$ follows from $v_2 = v_1 + u_1 - u_2$, or by exchanging the labels 1 and 2, since the hypotheses treat the two objects alike.

For run 4, $v_1 = \bigl((2 - 1)(3) + 2(1)(-3)\bigr)/3 = -1$ m/s and $v_2 = \bigl((1 - 2)(-3) + 2(2)(3)\bigr)/3 = +5$ m/s, as measured.

**Pause and try.** A moving object strikes an object at rest, elastically. Without the formulas, guess the outcome in three cases, then check each guess with $u_2 = 0$: (a) the masses are equal; (b) the moving object is far heavier, like a bowling ball hitting a ping-pong ball; (c) it is far lighter, the ping-pong ball hitting the bowling ball.

**Answer.** With $u_2 = 0$: $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}u_1$ and $v_2 = \dfrac{2m_1}{m_1 + m_2}u_1$. (a) $m_1 = m_2$: $v_1 = 0$ and $v_2 = u_1$. The moving object stops dead and the other leaves with its velocity. (b) $m_1 \gg m_2$: $v_1 \approx u_1$ and $v_2 \approx 2u_1$. The heavy object barely notices, and the light one leaves at twice the heavy one's speed, which most people do not guess. (c) $m_1 \ll m_2$: $v_1 \approx -u_1$ and $v_2 \approx 0$. The light object bounces back at nearly its original speed.

[Figure 5, animation: Three elastic collisions at half speed, each with a cart arriving at 2 m/s. Equal masses exchange velocities. A 10 kg cart barely slows and sends a 1 kg cart off at 3.6 m/s. A 1 kg cart bounces back from a 10 kg cart.]

With masses 10 kg and 1 kg and $u_1 = 2$ m/s, case (b) gives $v_1 = \tfrac{9}{11}(2) = 1.64$ m/s and $v_2 = \tfrac{20}{11}(2) = 3.64$ m/s, and case (c) gives $v_1 = -1.64$ m/s and $v_2 = \tfrac{2}{11}(2) = 0.36$ m/s.

Most real collisions lie between the two extremes: the objects separate, more slowly than they approached. Momentum conservation still holds, and the second condition must be measured.

## The center of mass does not notice the collision

Momentum conservation says $m_1 v_1 + m_2 v_2$ is constant. Divide by the total mass and it says that a certain average velocity is constant. That average is the velocity of a definite point.

**Definition.** The **center of mass** of two objects at positions $x_1, x_2$ with masses $m_1, m_2$ is the point

$$
x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}.
$$

It lies between the objects, closer to the heavier one; for equal masses it is the midpoint.

If the objects move for a time $\Delta t$ with velocities $v_1, v_2$, their positions change by $v_1\Delta t$ and $v_2\Delta t$, so $x_{\text{cm}}$ changes by $(m_1 v_1 + m_2 v_2)\,\Delta t/(m_1 + m_2)$. The velocity of the center of mass is therefore

$$
v_{\text{cm}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{p_{\text{total}}}{M}.
$$

Since $p_{\text{total}}$ is constant when no net external force acts, so is $v_{\text{cm}}$. The carts change velocity abruptly, and the center of mass moves through the collision at a steady velocity. For runs 1 and 2 it is $6/3 = 2$ m/s throughout. The common velocity after a perfectly inelastic collision is exactly $v_{\text{cm}}$, because once the carts are stuck together the center of mass moves with them.

[Figure 6, animation: Run 2 at one third speed, with the position of each cart's center plotted against time. The green marker is the center of mass. Both carts change velocity at the collision; the center of mass keeps a straight line of slope 2 m/s.]

So watch the collision from a camera that rides along with the center of mass. Changing to an observer who moves at a constant velocity $V$ subtracts $V$ from every velocity and leaves every force and every acceleration unchanged, so Newton's laws hold for that observer as well. In the frame moving at $v_{\text{cm}} = 2$ m/s, cart A approaches at $3 - 2 = +1$ m/s and cart B at $0 - 2 = -2$ m/s. Their momenta are $+2$ and $-2$ kg·m/s, total zero.

**Pause and try.** In this frame the total momentum is zero before the collision, so it is zero after. What does that force a sticky collision to look like? And an elastic one, given that the relative velocity reverses?

**Answer.** *Sticky:* the pair has a common velocity $v$ and momentum $3v = 0$, so $v = 0$. Both carts stop dead, and all the kinetic energy in this frame, $\tfrac12(2)(1)^2 + \tfrac12(1)(2)^2 = 3$ J, is lost. *Elastic:* total momentum zero requires $2v_A + v_B = 0$, and reversal requires $v_A - v_B = -3$. So $v_A = -1$ and $v_B = +2$: each cart leaves with its velocity reversed. Adding 2 m/s back gives $+1$ and $+4$ m/s in the track's frame, which is run 2.

[Figure 7, animation: Runs 1 and 2 seen by an observer moving with the center of mass, at half speed. With Velcro both carts stop and the 3 J of kinetic energy in this frame is lost; with springs each reverses and the 3 J is kept. The green marker is the center of mass, which stays where it is.]

This view explains the energy-loss formula. In the track's frame the pair must keep moving at $v_{\text{cm}}$ after any collision, so the kinetic energy of that shared motion, $\tfrac12 M v_{\text{cm}}^2 = \tfrac12(3)(2)^2 = 6$ J, can never be lost. Only the remaining 3 J, the kinetic energy of the motion relative to the center of mass, is available, and a sticky collision loses all of it. That 3 J is $\tfrac12 \frac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2$, the amount computed earlier.

## Worked examples

**Worked example.** **Impulse on a baseball.** A 0.145 kg baseball arrives at the bat at 40 m/s and leaves in the opposite direction at 50 m/s. Contact lasts 0.7 ms. Find the impulse on the ball and the average force.

Momentum has a sign, so choose a direction first: take the direction of the hit ball as positive. Then $v_i = -40$ m/s and $v_f = +50$ m/s. The impulse is $\Delta p = 0.145 \times (50 - (-40)) = 0.145 \times 90 = 13.05$ kg·m/s. The average force is $13.05 / 0.0007 = 18\,600$ N, about 13 000 times the ball's weight of 1.42 N, which is why the weight can be ignored during the contact. Subtracting the speeds, $50 - 40$, would give a force nine times too small: the velocity changed by 90 m/s because it reversed.

**Worked example.** **A rear-end collision.** A 1200 kg car moving at 15 m/s runs into the back of an 1800 kg van moving the same way at 5 m/s, and the bumpers lock. Find the velocity just afterward and the kinetic energy lost.

The collision is perfectly inelastic, and over its short duration the road's friction delivers little impulse, so momentum is conserved: $v = (1200 \times 15 + 1800 \times 5)/3000 = 27\,000/3000 = 9$ m/s. Energy lost: $\tfrac12 \cdot \dfrac{1200 \times 1800}{3000} \cdot (15 - 5)^2 = \tfrac12 (720)(100) = 36\,000$ J.

*Check* directly: before, $\tfrac12(1200)(225) + \tfrac12(1800)(25) = 135\,000 + 22\,500 = 157\,500$ J. After, $\tfrac12(3000)(81) = 121\,500$ J. The difference is 36 000 J.

**Worked example.** **An elastic collision with both objects moving.** A 0.2 kg glider moving at $+4$ m/s on an air track meets a 0.3 kg glider moving at $-1$ m/s, and they collide elastically. Find the velocities afterward.

Use the center-of-mass frame, where an elastic collision reverses each velocity. $v_{\text{cm}} = (0.2 \times 4 + 0.3 \times (-1))/0.5 = 1$ m/s. In that frame the velocities are $4 - 1 = +3$ and $-1 - 1 = -2$ m/s; after the collision they are $-3$ and $+2$ m/s. Back in the track's frame: $v_1 = -3 + 1 = -2$ m/s and $v_2 = 2 + 1 = +3$ m/s.

*Check:* momentum is $0.8 - 0.3 = 0.5$ kg·m/s before and $-0.4 + 0.9 = 0.5$ after; kinetic energy is $1.6 + 0.15 = 1.75$ J before and $0.4 + 1.35 = 1.75$ J after.

## Limits and common mistakes

Conservation of momentum has one hypothesis, zero net external force, and it is a statement about a *system* you choose. A ball bouncing off a wall changes its own momentum from $mv$ to nearly $-mv$, because the wall pushes on it. Include the wall and the Earth it is attached to, and the total is conserved: the Earth recoils with momentum $2mv$, at a speed far too small to detect.

Everything here was along one line. In two or three dimensions momentum is a [vector](@linear-algebra/vectors), the proof runs unchanged for each component, and each component is conserved separately. The elastic formulas are for head-on collisions only.

**Common mistake.** **Conserving kinetic energy in every collision.** The opening calculation did this and got 2.45 m/s in place of 2 m/s. Kinetic energy is conserved only in elastic collisions, which must be stated or justified. Momentum is conserved in every collision with negligible external impulse. When objects stick or crumple, use momentum for the collision itself, and energy only for the motion before and after it.

**Common mistake.** **Dropping the sign.** Momentum is $mv$ with $v$ a velocity. Two 1 kg carts approaching each other at 3 m/s have total momentum $3 + (-3) = 0$, and if they stick they stop. Adding the speeds gives 6 kg·m/s and a pair moving at 3 m/s in no particular direction. Mark the positive direction and give every velocity its sign before writing an equation.

**Check yourself.** A 3 kg cart moving at 2 m/s collides with a 1 kg cart at rest and they stick together. Which statement about the collision is correct?
Answer: The pair moves at 1.5 m/s and 1.5 J of kinetic energy is lost Momentum: $3 \times 2 = 4v$, so $v = 1.5$ m/s. Kinetic energy falls from 6 J to $\tfrac12(4)(1.5)^2 = 4.5$ J.

**Recap.** - Momentum is $p = mv$, with a sign in one dimension. Impulse is $F\,\Delta t$, the area under a force-time graph, and it equals the change in momentum. This follows from $F = ma$.
- By the third law, internal forces deliver equal and opposite impulses, so the total momentum of a system changes only by the impulse of external forces.
- Perfectly inelastic: $v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$, with kinetic energy loss $\tfrac12 \dfrac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2$.
- Elastic: the relative velocity reverses, $v_1 - v_2 = -(u_1 - u_2)$. Equal masses exchange velocities; a heavy object sends a light one off at about twice its speed; a light object bounces back from a heavy one.
- The center of mass moves at $v_{\text{cm}} = p_{\text{total}}/M$, constant through any collision. In its frame the total momentum is zero.

The [next lesson](@physics/simple-harmonic-motion) returns to a single object, on a spring whose force lasts as long as the motion does and pulls the object back and forth.

## Problems

Take the first object's initial direction of motion as positive, and use $g = 9.8$ m/s².

**Problem 1.** A 70 kg driver traveling at 15 m/s is brought to rest in a crash. Find the impulse on the driver, and the average force if the stop takes (a) 0.005 s against a rigid dashboard, (b) 0.1 s against an airbag.

**Solution.** $\Delta p = 70 \times (0 - 15) = -1050$ kg·m/s, the same in both cases. (a) $F = -1050/0.005 = -210\,000$ N. (b) $F = -1050/0.1 = -10\,500$ N. The airbag does not reduce the impulse, which is fixed by the change in momentum. It spreads the impulse over twenty times as long, so the force is twenty times smaller. The minus signs say the force is directed against the initial motion.

**Problem 2.** A 0.4 kg lump of clay moving at 5 m/s hits a 1.6 kg block at rest on a frictionless table and sticks. Find the final velocity and the fraction of the kinetic energy lost.

**Solution.** Momentum: $0.4 \times 5 = (0.4 + 1.6)\,v$, so $v = 1$ m/s. Kinetic energy before: $\tfrac12(0.4)(25) = 5$ J. After: $\tfrac12(2)(1) = 1$ J. Lost: 4 J, which is 80%. Check with the formula: $\tfrac12 \cdot \tfrac{0.4 \times 1.6}{2} \cdot 5^2 = \tfrac12(0.32)(25) = 4$ J.

**Problem 3.** A 0.5 kg glider moving at 4 m/s collides elastically with a 1.5 kg glider at rest. Find both velocities afterward, and verify that momentum and kinetic energy are conserved.

**Solution.** With $u_2 = 0$: $v_1 = \dfrac{0.5 - 1.5}{2}(4) = -2$ m/s and $v_2 = \dfrac{2(0.5)}{2}(4) = +2$ m/s. Momentum: before $0.5 \times 4 = 2$; after $0.5(-2) + 1.5(2) = -1 + 3 = 2$ kg·m/s. Kinetic energy: before $\tfrac12(0.5)(16) = 4$ J; after $\tfrac12(0.5)(4) + \tfrac12(1.5)(4) = 1 + 3 = 4$ J. The relative velocity went from $+4$ to $-4$ m/s.

**Problem 4.** A 60 kg skater at rest on frictionless ice throws a 4 kg ball horizontally at 6 m/s relative to the ice. (a) Find the skater's velocity afterward. (b) Find the total kinetic energy before and after. Where did it come from? (c) Where is the center of mass of the skater and ball 2 s after the throw, relative to where it started?

**Solution.** (a) The total momentum is zero before, so $60v + 4 \times 6 = 0$ and $v = -0.4$ m/s: the skater recoils. (b) Before: 0. After: $\tfrac12(60)(0.16) + \tfrac12(4)(36) = 4.8 + 72 = 76.8$ J. It came from chemical energy in the skater's muscles. Internal forces cannot change the total momentum, and they can change the total kinetic energy; this is a sticky collision run backward. (c) $v_{\text{cm}} = p_{\text{total}}/M = 0$, so the center of mass has not moved. Check: after 2 s the skater is at $-0.8$ m and the ball at $+12$ m, and $(60 \times (-0.8) + 4 \times 12)/64 = 0$.

**Problem 5.** (Ballistic pendulum.) A 10 g bullet is fired horizontally into a 2 kg wooden block hanging at rest from long strings, and stays in it. The block swings up until its center has risen 5 cm. (a) Find the bullet's speed. (b) A student writes $\tfrac12 m v^2 = (m + M)gh$ for the whole process and gets 14 m/s. What is wrong with that?

**Solution.** Split the process into two stages. *Swing:* after the bullet has stopped inside the block, only gravity does work (the string tension is perpendicular to the motion), so mechanical energy is conserved: $\tfrac12 (m + M) V^2 = (m + M) g h$, giving $V = \sqrt{2 \times 9.8 \times 0.05} = 0.990$ m/s just after the impact. *Impact:* it is over before the block has moved appreciably, so the strings are still vertical and exert no horizontal force, and horizontal momentum is conserved: $m v = (m + M)V$, so $v = 2.01 \times 0.990 / 0.010 = 199$ m/s. (b) The student conserved kinetic energy through a perfectly inelastic collision. The fraction of kinetic energy lost in the impact is $M/(m + M) = 2/2.01 = 99.5\%$, so the equation is wrong by a factor of about 200 in energy, and by $\sqrt{201} \approx 14$ in speed.

**Problem 6.** An object of mass $m_1$ moving at $u_1$ collides elastically with an object of mass $m_2$ at rest. (a) Show that the fraction of the first object's kinetic energy that is transferred to the second is $\dfrac{4 m_1 m_2}{(m_1 + m_2)^2}$. (b) Show that this fraction is at most 1, with equality exactly when the masses are equal. (c) Nuclear reactors slow fast neutrons by letting them collide with light nuclei. Compare the fraction transferred per head-on collision with a hydrogen nucleus (mass about equal to the neutron's) and with a lead nucleus (about 207 neutron masses).

**Solution.** (a) $v_2 = \dfrac{2m_1}{m_1 + m_2}u_1$, so the second object's kinetic energy is $\tfrac12 m_2 v_2^2 = \tfrac12 m_2 \dfrac{4m_1^2}{(m_1 + m_2)^2}u_1^2$. Divide by $\tfrac12 m_1 u_1^2$ to get $\dfrac{4m_1 m_2}{(m_1 + m_2)^2}$. (b) $(m_1 + m_2)^2 - 4m_1 m_2 = (m_1 - m_2)^2 \ge 0$, so $4m_1 m_2 \le (m_1 + m_2)^2$, with equality exactly when $m_1 = m_2$. (c) Hydrogen: $4(1)(1)/2^2 = 1$, all of the energy in one head-on collision. Lead: $4(207)/208^2 = 828/43\,264 = 0.019$, under 2%. This is why reactors use water or graphite as moderators, and not lead.