Elastic and Inelastic Collisions
Two bumper cars crash head-on at the fair. One bounces back hard, the other barely moves, and you hear a loud thud. Where did the motion go, and can you predict who ends up moving how fast? Momentum always gives you an answer. Kinetic energy tells you what kind of collision it was.
Momentum in a collision
A collision is a short, strong interaction between objects. During it, the two objects push on each other with forces that are equal in magnitude and opposite in direction (Newton's third law). Those are internal forces if you choose both objects as your system.
If the net external force on the system is zero (or tiny compared with the collision forces), the total momentum cannot change:
Why does this work even with friction on a track or gravity on the carts? During the very short collision, the forces between the objects are far larger than any external force, so the external ones barely change the momentum.
Two carts collide on a track and stick together. Cart A has mass $1.0\ \text{kg}$ and moves at $2.0\ \text{m/s}$; cart B has mass $1.0\ \text{kg}$ and is at rest. Find…
Two carts collide on a track and stick together. Cart A has mass $1.0\ \text{kg}$ and moves at $2.0\ \text{m/s}$; cart B has mass $1.0\ \text{kg}$ and is at rest. Find the speed of the combined carts after the collision.
Three kinds of collisions
Momentum tells you about the system, but it does not tell you how much motion energy is left. For that, compare the total kinetic energy $K = \tfrac{1}{2}mv^2$ before and after.
| Type | Momentum | Kinetic energy | Example |
|---|---|---|---|
| Elastic | conserved | conserved | steel balls, air-track gliders with springy bumpers |
| Inelastic | conserved | decreases | bumper cars, a basketball hitting the floor |
| Perfectly inelastic | conserved | decreases the most | clay hitting a wall, football tackle where players move together |
In a perfectly inelastic collision the objects stick and move with the same final velocity. That is the largest loss of kinetic energy that is still allowed by momentum conservation. It is not a loss of all the kinetic energy: if the system keeps moving, it still has some.
Where does the missing kinetic energy go? It is transformed into the internal (thermal) energy of the objects, plus sound and deformation. Total energy is still conserved. It is only the kinetic part that decreases.
Classify each collision as elastic, inelastic, or perfectly inelastic, and justify briefly: (a) two steel balls bounce off each other with no loss of total kinetic…
Classify each collision as elastic, inelastic, or perfectly inelastic, and justify briefly: (a) two steel balls bounce off each other with no loss of total kinetic energy; (b) a lump of clay hits a wall and sticks; (c) two bumper cars collide and bounce apart, with some kinetic energy turned into thermal energy and sound.
A $2.0\ \text{kg}$ cart moving at $3.0\ \text{m/s}$ to the right collides with a $1.0\ \text{kg}$ cart at rest. After the collision, the $2.0\ \text{kg}$ cart moves…
A $2.0\ \text{kg}$ cart moving at $3.0\ \text{m/s}$ to the right collides with a $1.0\ \text{kg}$ cart at rest. After the collision, the $2.0\ \text{kg}$ cart moves right at $1.0\ \text{m/s}$ and the $1.0\ \text{kg}$ cart moves right at $4.0\ \text{m/s}$. (a) Verify that momentum is conserved. (b) Determine whether the collision is elastic, and justify your answer with a calculation.
Perfectly inelastic collisions: how much energy is lost?
When objects stick, you have one unknown, the common final velocity $v'$:
$$m_1 v_1 + m_2 v_2 = (m_1 + m_2)\,v' \quad\Longrightarrow\quad v' = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$Then find the kinetic energy lost:
$$\Delta K = K_{\text{after}} - K_{\text{before}} \qquad(\text{negative when energy is lost})$$Notice the fraction lost, $\dfrac{m_2}{m_1+m_2} = \dfrac{1000}{2500} = 40\%$, when one object starts at rest. The more massive the target, the more kinetic energy is lost.
A $1200\ \text{kg}$ car moving east at $15\ \text{m/s}$ rear-ends a $1800\ \text{kg}$ truck moving east at $10\ \text{m/s}$. After the collision, the two vehicles lock…
A $1200\ \text{kg}$ car moving east at $15\ \text{m/s}$ rear-ends a $1800\ \text{kg}$ truck moving east at $10\ \text{m/s}$. After the collision, the two vehicles lock together. (a) Find their common velocity. (b) Calculate the kinetic energy lost in the collision.
Elastic collisions
In an elastic collision you have two conserved quantities, so you can find two unknown final velocities:
$$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \qquad\text{(momentum)}$$ $$\tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2 = \tfrac{1}{2}m_1 v_1'^2 + \tfrac{1}{2}m_2 v_2'^2 \qquad\text{(kinetic energy)}$$Solving the second equation (it has squares) along with the first is messy, but you can do it. Move all the $m_1$ terms to one side and all the $m_2$ terms to the other:
$m_1(v_1 - v_1') = m_2(v_2' - v_2)$ (from momentum) and $m_1(v_1^2 - v_1'^2) = m_2(v_2'^2 - v_2^2)$ (from kinetic energy).
Factor the differences of squares in the second equation, then divide it by the first. The $m$'s and the common factors cancel, leaving $v_1 + v_1' = v_2 + v_2'$. This gives a shortcut for head-on elastic collisions in one dimension. It is not on the AP equation sheet, so you can derive it as above, or just solve the momentum and kinetic energy equations directly:
Use this equation together with momentum conservation. Both are linear, so you just solve two equations for two unknowns.
Two special cases are worth remembering:
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Equal masses, one at rest: the moving object stops and the other leaves with the same velocity (like a cue ball hitting another ball head-on, or Newton's cradle).
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Light object hits a much heavier one at rest: the light object bounces back with nearly the same speed, and the heavy one hardly moves.
A $0.20\ \text{kg}$ ball moving at $5.0\ \text{m/s}$ collides head-on with an identical ball at rest in an elastic collision. State the final velocities of both balls…
A $0.20\ \text{kg}$ ball moving at $5.0\ \text{m/s}$ collides head-on with an identical ball at rest in an elastic collision. State the final velocities of both balls and verify that the total kinetic energy is unchanged.
Cart A ($m_A = 0.30\ \text{kg}$) moves right at $2.0\ \text{m/s}$ and collides with cart B ($m_B = 0.90\ \text{kg}$) moving left at $1.0\ \text{m/s}$. The collision is…
Cart A ($m_A = 0.30\ \text{kg}$) moves right at $2.0\ \text{m/s}$ and collides with cart B ($m_B = 0.90\ \text{kg}$) moving left at $1.0\ \text{m/s}$. The collision is perfectly elastic. (a) Write the two conservation equations you would use. (b) Find the velocity of each cart after the collision. (c) Verify with the kinetic energy that the collision is elastic. (d) Explain, using your answer to (b), how the relative speed of approach compares with the relative speed of separation.
Two-stage problems: the ballistic pendulum
Some problems chain a collision with another event. The key is to split them into stages and use the right conservation law for each:
| Stage | What happens | Use |
|---|---|---|
| 1. Collision (very short) | objects stick | momentum (not energy: $K$ is lost) |
| 2. Swing up | block and bullet rise, no more thermal loss | mechanical energy ($K \to U_g$) |
A $0.050\ \text{kg}$ bullet moving at $400\ \text{m/s}$ embeds itself in a $2.0\ \text{kg}$ wooden block hanging at rest from a long string. (a) Find the speed of the…
A $0.050\ \text{kg}$ bullet moving at $400\ \text{m/s}$ embeds itself in a $2.0\ \text{kg}$ wooden block hanging at rest from a long string. (a) Find the speed of the block and bullet just after the collision. (b) Find the maximum height the block rises. Use $g = 9.8\ \text{m/s}^2$.
On an air track, a $0.40\ \text{kg}$ glider moving at $1.5\ \text{m/s}$ collides with a $0.60\ \text{kg}$ glider at rest. A student measures the final velocities in two…
On an air track, a $0.40\ \text{kg}$ glider moving at $1.5\ \text{m/s}$ collides with a $0.60\ \text{kg}$ glider at rest. A student measures the final velocities in two trials. Trial 1: the gliders stick together. Trial 2: the $0.40\ \text{kg}$ glider moves at $-0.30\ \text{m/s}$ and the $0.60\ \text{kg}$ glider at $+1.20\ \text{m/s}$. (a) For each trial, find the final velocities (Trial 1) or check momentum conservation (Trial 2). (b) Compute the fraction of the initial kinetic energy lost in each trial. (c) Classify each trial, and argue which trial involves the larger energy loss and where that energy goes.
Back to the bumper cars: use $m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$ with signs to predict the velocities. Then compare the total $K$ before and after to see how much was lost to heat and sound. If $K$ is the same it was elastic, and if it dropped it was inelastic.