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Momentum conservation in every isolated collision, elastic and perfectly inelastic collisions, restitution, two-dimensional collisions, and the ballistic pendulum.
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By the end of this lesson you will be able to solve elastic and inelastic collisions in one and two dimensions, and combine collisions with energy methods.
The last two lessons showed that the impulse of a force changes momentum and that internal forces cannot move a system's center of mass. From Physics 1 you know conservation of momentum and the difference between elastic and inelastic collisions. This lesson solves collisions systematically, in one and two dimensions, and uses them with energy methods to measure speeds that are hard to measure directly. The skill it builds is deciding, stage by stage, which quantity is conserved, because a collision problem is usually solved by two different laws applied one after the other.
| Term | What it means |
|---|---|
| Isolated system | A system on which the net external impulse is negligible during the process of interest. |
| Elastic collision | A collision that conserves kinetic energy as well as momentum. |
| Inelastic collision | A collision in which some kinetic energy becomes heat, sound or deformation. |
| Perfectly inelastic collision | The bodies stick together; the kinetic energy lost is as large as momentum allows. |
| Coefficient of restitution | $e$, the speed of separation over the speed of approach, from $0$ to $1$. |
| Impulse approximation | Neglecting outside forces during a brief collision because their impulse is tiny. |
| Ballistic pendulum | A hanging block that catches a projectile, whose swing reveals the projectile's speed. |
During a collision the bodies push on each other with large internal forces for a short time. Any external force, such as gravity or friction, delivers an impulse $F\Delta t$ that is tiny by comparison, so the total momentum just after equals the total just before:
$$m_1\vec{v}_1 + m_2\vec{v}_2 = m_1\vec{v}_1' + m_2\vec{v}_2'.$$
This holds for every collision, as a vector equation, one equation per component. Kinetic energy is different. In an elastic collision it is also conserved, which supplies a second equation; for a head-on collision with the second body at rest,
$$v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1, \qquad v_2' = \frac{2m_1}{m_1 + m_2}v_1.$$
In a perfectly inelastic collision the bodies stick, $v' = (m_1v_1 + m_2v_2)/(m_1 + m_2)$, and the kinetic energy lost is the most momentum conservation allows. Real collisions lie between, measured by the coefficient of restitution $e$.
Another way: picture
Picture two carts on a track with a spring between them. They squeeze the spring, stop squeezing, and push apart: whatever the spring does, it pushes on each cart equally and oppositely, so the total momentum never changes. If the spring gives back all its energy, the collision is elastic; replace it with a lump of clay that keeps the energy, and the carts stick.
Another way: steps
By Newton's third law, the force each body exerts on the other is equal and opposite at every instant, so their impulses cancel and the total momentum cannot change. Nothing in that argument says the forces do equal work. If the bodies deform and stay deformed, heat up or make noise, the work done compressing them is not all returned, and kinetic energy is lost.
Momentum and energy are therefore separate constraints, and a collision problem usually needs both. Momentum conservation is exact for any isolated collision. Energy conservation is an extra assumption that holds only for elastic collisions: steel balls, billiard balls and colliding atoms come close; cars, clay and people do not.
For a head-on elastic collision, momentum and energy together give $v_1'$ and $v_2'$. A useful consequence of combining them is that the relative velocity reverses: $v_2' - v_1' = -(v_2 - v_1)$. The bodies separate as fast as they approached, which is often the quickest second equation.
The special cases are worth remembering. Equal masses exchange velocities: the moving ball stops dead, which is how a Newton's cradle works. A light ball hitting a heavy one at rest bounces back at nearly its own speed. A heavy ball hitting a light one keeps going almost unchanged and sends the light one off at nearly twice its speed, which is how a golf club launches a ball.
When the bodies stick, momentum alone fixes the result: $v' = (m_1v_1 + m_2v_2)/(m_1 + m_2)$. The fraction of kinetic energy kept, for a target at rest, is $m_1/(m_1 + m_2)$. A car rear-ending an identical stopped car and locking bumpers keeps half its kinetic energy; the other half goes into crumpling metal.
This is the largest possible loss consistent with momentum conservation. Any loss beyond it would require the center of mass to slow down, which internal forces cannot do. In the frame moving with the center of mass, the total momentum is zero, and a perfectly inelastic collision brings everything to rest: all the kinetic energy in that frame is lost.
Momentum is a vector, so in two dimensions there are two conservation equations, one for each component. For a glancing collision of pool balls, or cars at an intersection, write the $x$ and $y$ momenta before and after separately.
Two equations are not enough for four unknown velocity components, so something else must be known: that the bodies stick, that the collision is elastic, or one of the final directions. A striking result follows for equal masses in an elastic collision with one at rest: the two leave at right angles to each other. Pool players rely on it to predict where the cue ball goes after a cut shot.
Many problems have a collision followed or preceded by something else, and each stage needs its own law. A ballistic pendulum is the classic: a bullet embeds in a hanging block, and the block swings up. Momentum is conserved in the collision, where the energy loss is large and unknown; energy is conserved in the swing, where the only work is done by gravity.
Mixing them is the common error. Energy is not conserved across the collision, and momentum is not conserved during the swing, because the string and gravity then act for a long time. Name the stages, state what each conserves, and apply them in order.
Checking an answer. Total momentum must be the same before and after. Kinetic energy after must not exceed kinetic energy before, unless something like an explosion supplies energy. The bodies must not pass through each other: after the collision the one behind cannot be moving faster than the one in front. And the center-of-mass velocity must be unchanged.
Real collisions lose some but not all kinetic energy. The coefficient of restitution, $e = |v_2' - v_1'|/|v_1 - v_2|$, compares the speed of separation to the speed of approach. It is $1$ for an elastic collision and $0$ for a sticking one. For a ball bouncing off a rigid floor, $e$ is the ratio of rebound speed to impact speed, so a ball dropped from height $h$ bounces to $e^2h$.
Sports governing bodies regulate it. A regulation basketball must bounce to between about $1.2$ and $1.4$ m when dropped from $1.8$ m, which fixes $e$ near $0.85$. Major League Baseball specifies the coefficient of restitution of its balls against a wooden wall at $26$ m/s, near $0.55$. A small change in $e$ changes how far a batted ball flies by meters.
Viewed from a frame moving with the center of mass, every collision looks symmetric: the total momentum is zero before and after, so the two bodies approach each other and recede with momenta equal and opposite. An elastic collision in that frame simply reverses each velocity, or rotates it in two dimensions; an inelastic one shrinks them.
Particle physicists work in this frame because it shows what energy is really available. When two protons collide head-on at equal speeds, as in a collider, all their kinetic energy is available to create new particles. When a proton hits one at rest, most of the energy goes into moving the center of mass. That is why facilities such as Brookhaven's Relativistic Heavy Ion Collider in New York collide beams head-on.
When cars collide at an intersection, police accident reconstructionists work backward from the scene. The skid marks after impact, with the road's coefficient of friction, give the wreck's speed just after the crash through energy: $\tfrac{1}{2}v^2 = \mu_kgd$. The direction of the skid gives its angle.
Then momentum conservation, applied separately north and east, recovers each car's speed before the impact: the northward momentum after must equal the northbound car's momentum before, and likewise east. Reconstructionists trained through programs such as those of the Texas A&M Transportation Institute use exactly these equations, and their results decide who was speeding. The method works because the crash is brief: friction's impulse during the impact is negligible.
A spacecraft swinging past a planet is an elastic collision with no contact: gravity is conservative, so kinetic energy is conserved in the planet's frame. In that frame the craft leaves at the speed it arrived, only turned. But the planet is moving, and in the Sun's frame the craft can gain up to twice the planet's orbital speed, just as a light ball bouncing off a heavy one moving toward it leaves faster.
The planet pays with an imperceptible slowing, conserving momentum. NASA's Voyager 2 used Jupiter, Saturn and Uranus in turn to reach Neptune, and New Horizons used Jupiter to shave three years off its trip to Pluto. The same one-dimensional equations for an elastic collision of very unequal masses, applied in the right frame, give the speed gained.
Because both momentum and energy are conservation laws, it is tempting to use both in every collision. Momentum is conserved in any collision of an isolated system; kinetic energy only in an elastic one. Using energy conservation for a bullet embedding in a block gives a bullet speed that is far too low, because it ignores the energy that went into splintering wood.
A second error is to apply momentum conservation over a long interval. During the swing of a ballistic pendulum, the string and gravity act for a long time and the block's momentum changes; conservation holds only across the brief collision, the few milliseconds in which the collision forces dwarf every other force.
A $2.0$ kg cart at $6.0$ m/s hits a $4.0$ kg cart at rest, elastically. Find the first cart's velocity after.
$v_1' = \dfrac{2.0 - 4.0}{6.0} \times 6.0 = -2.0\ \text{m/s}$
It bounces back, being lighter.
Find the second cart's velocity after.
$v_2' = \dfrac{2 \times 2.0}{6.0} \times 6.0 = 4.0\ \text{m/s}$
Forward.
Check the total momentum.
$2.0 \times (-2.0) + 4.0 \times 4.0 = 12 = 2.0 \times 6.0$
Unchanged.
Check kinetic energy.
$\tfrac{1}{2}(2.0)(4.0) + \tfrac{1}{2}(4.0)(16) = 4.0 + 32 = 36\ \text{J} = \tfrac{1}{2}(2.0)(36)$
Unchanged, as it must be.
Check the separation speed.
$v_2' - v_1' = 4.0 - (-2.0) = 6.0\ \text{m/s}$
Equal to the approach speed.
A $5.0$ g bullet embeds in a $1.995$ kg block, which then rises $0.20$ m. Find the block's speed after impact from the swing.
$V = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.20} = 1.98\ \text{m/s}$
Energy conservation during the swing.
Write momentum conservation for the collision.
$mv = (m + M)V$
The impulse approximation.
Solve for the bullet's speed.
$v = \dfrac{m + M}{m}V = \dfrac{2.000}{0.005} \times 1.98$
Rearrange.
Evaluate the speed.
$v = 792\ \text{m/s}$
A typical rifle bullet.
Find the fraction of kinetic energy kept.
$\dfrac{m}{m + M} = \dfrac{0.005}{2.000} = 0.25\%$
Almost all of it becomes heat and deformation in the wood.
A $1500$ kg car going north at $20$ m/s and a $1200$ kg car going east at $15$ m/s lock together. Write the eastward momentum.
$p_x = 1200 \times 15 = 18{,}000\ \text{kg m/s}$
Only the eastbound car.
Write the northward momentum.
$p_y = 1500 \times 20 = 30{,}000\ \text{kg m/s}$
Only the northbound car.
Find the wreck's velocity components.
$v_x = \dfrac{18{,}000}{2700} = 6.67, \quad v_y = \dfrac{30{,}000}{2700} = 11.1\ \text{m/s}$
Divide by the combined mass.
Find its speed.
$v = \sqrt{6.67^2 + 11.1^2} = 13.0\ \text{m/s}$
Pythagorean theorem.
Find its direction.
$\theta = \arctan\dfrac{11.1}{6.67} = 59°\ \text{north of east}$
From the components.
Find the kinetic energy lost.
$K = 3.0 \times 10^5 + 1.35 \times 10^5 - \tfrac{1}{2}(2700)(13.0)^2 = 2.07 \times 10^5\ \text{J}$
Nearly half, into crumpled metal.
Write the momentum before.
$p = 3.0 \times 4.0 = 12\ \text{kg m/s}$
Only the first cart moves.
Divide by the combined mass.
$v' = \dfrac{12}{4.0}$
They move as one.
Evaluate the speed.
A pool ball moving at $9$ m/s hits an identical ball at rest head-on, in an elastic collision. What happens?
Complete the worked solution: a $1$ kg cart moving at $6$ m/s couples to a $9$ kg cart at rest. Find their common velocity in m/s, and the kinetic energy before and after, in joules.
Divide the momentum by the combined mass.
$v' = \dfrac{m_1v_1}{m_1 + m_2} =$ u
Momentum is conserved and they move together.
Find the kinetic energy before.
$K = \tfrac{1}{2}m_1v_1^2 =$ k
Only the first cart moves.
Find the kinetic energy after.
$K' = \tfrac{1}{2}(m_1 + m_2)v'^2 =$ q
Both carts at the common speed.
Compare the two energies.
$\text{the energy after is smaller}$
The rest went into deforming the coupling, sound and heat.
Match each kind of collision or quantity to its statement.
| momentum conserved | kinetic energy conserved too | a shared final velocity | separation speed over approach speed | |
|---|---|---|---|---|
| any isolated collision | ||||
| elastic collision | ||||
| perfectly inelastic collision | ||||
| coefficient of restitution |
A $7$ kg cart moving at $3$ m/s hits a $3$ kg cart at rest in a head-on elastic collision. Fill in each cart's velocity afterward in m/s, and the speed at which they separate.
| value | |
|---|---|
| first cart after (m/s) | |
| second cart after (m/s) | |
| separation speed (m/s) |
A $7$ kg cart moving at $4$ m/s couples to a cart of mass $m$ at rest, and they roll on together. Write their common speed, in m/s, as a formula in $m$ (kg).
Answer:
A $10$ g bullet moving at $300$ m/s embeds itself in a $990$ g wooden block hanging on a long string. With $g = 10$ m/s², how high does the block swing, in meters?
Answer: m the block rises
A police accident reconstructionist in Texas studies a crash at an intersection. A $2000$ kg car going north at $18$ m/s and a $1500$ kg car going east at $22$ m/s locked together. How fast was the wreck moving just after the impact, in m/s?
Answer: m/s just after impact
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A $4$ kg cart moving at $9$ m/s couples to a cart of mass $m$ at rest, and they roll on together. Write their common speed, in m/s, as a formula in $m$ (kg).
Answer:
You can solve collision problems. Explain to someone why a ballistic pendulum needs momentum for the impact but energy for the swing.
19. Your turn: a $3.0$ kg cart at $4.0$ m/s couples with a $1.0$ kg cart at rest. Find their common speed., step 3
$v' = 3.0\ \text{m/s}$
Three-quarters of the kinetic energy is kept.