Lesson 8 · 40 min
Impact, Momentum and Energy Together
Most real problems happen in stages. A block slides down a ramp, hits another, and the second one skids to a stop. A bullet buries itself in a block, which swings up on a cord. No single principle covers the whole story, but each stage has one that fits. The skill is to cut the motion at the right instants and carry the velocity from one stage to the next.
Learning objectives
- Choose between Newton's second law, work and energy, and impulse and momentum for each stage of a problem.
- Split a problem at the instants just before and just after an impact, and carry velocities across the cuts.
- Solve impact-then-spring, ballistic pendulum and slide-collide-skid problems.
- Explain why energy is never conserved across an impact unless \(e = 1\), even when it is conserved before and after.
Choosing a method
| You need | Use |
|---|---|
| A speed at a position, with forces that depend on position (weights, springs, friction over a distance) | Work and energy, or conservation of energy |
| A velocity after a time, or the response to forces that depend on time | Impulse and momentum |
| An average force over a known contact time | Impulse and momentum |
| Velocities just after a collision | Conservation of momentum and the coefficient of restitution |
| Motion before or after a collision | Usually work and energy, applied separately on each side of the impact |
| An acceleration, or a force at one instant (a normal force on a curve, a cord tension) | Newton's second law |
Problems in stages
- Cut the motion into stages at the instants where the physics changes: just before and just after each impact, where a body leaves a surface, where a cord goes slack.
- Pick a principle for each stage from the table. Across an impact: momentum and \(e\), never energy (unless \(e = 1\)).
- Carry the velocities across each cut: the velocity at the end of one stage is the velocity at the start of the next.
An impact, then a spring
Example 8.1 — Impact, then a spring
Block \(A\) (\(2\ \text{kg}\)) slides at \(6\ \text{m/s}\) on a smooth surface and strikes block \(B\) (\(3\ \text{kg}\)), which is at rest against an unstretched spring with \(k = 600\ \text{N/m}\). The coefficient of restitution is \(0.5\). Find (a) the velocities of \(A\) and \(B\) just after the impact, (b) the kinetic energy lost in the impact, and (c) the maximum compression of the spring.
Show solution
(a) The impact. Positive to the right. The spring force is finite, so it is nonimpulsive during the brief impact, and the momentum of \(A\) and \(B\) is conserved:
\[ 2(6) + 3(0) = 2(v_A)_2 + 3(v_B)_2 \qquad (v_B)_2 - (v_A)_2 = 0.5(6 - 0) = 3 \] \[ (v_A)_2 = 0.6\ \text{m/s},\qquad (v_B)_2 = 3.6\ \text{m/s} \](b) \(T_1 = 36\ \text{J}\) and \(T_2 = \tfrac12(2)(0.6)^2 + \tfrac12(3)(3.6)^2 = 19.8\ \text{J}\): \(16.2\ \text{J}\), or \(45\%\), is lost.
(c) The compression. After the impact only the spring does work on \(B\), and \(A\), at \(0.6\ \text{m/s}\), falls behind:
\[ \tfrac12(3)(3.6)^2 \colVe{{} - \tfrac12(600)\,s^2} = 0 \quad\Rightarrow\quad s = 0.255\ \text{m} \]Two stages, two principles: momentum and restitution across the impact, then work and energy for the compression.
| \(e\) | \((v_A)_2\) | \((v_B)_2\) | Energy lost | Maximum compression |
|---|---|---|---|---|
| \(0\) | \(2.4\ \text{m/s}\) | \(2.4\ \text{m/s}\) | \(21.6\ \text{J}\) (\(60\%\)) | \(0.219\ \text{m}\) |
| \(0.5\) | \(0.6\ \text{m/s}\) | \(3.6\ \text{m/s}\) | \(16.2\ \text{J}\) (\(45\%\)) | \(0.255\ \text{m}\) |
| \(1\) | \(-1.2\ \text{m/s}\) | \(4.8\ \text{m/s}\) | \(0\) | \(0.339\ \text{m}\) |
With \(e = 0\), \(A\) stays pressed against \(B\), so both blocks compress the spring together: \(\tfrac12(5)(2.4)^2 = \tfrac12(600)s^2\). The largest compression comes with \(e = 1\): no energy is lost, and \(B\) leaves with the most momentum.
Watch what happens after the spring throws \(B\) back. With \(e = 0.5\), \(B\) returns at \(3.6\ \text{m/s}\) and meets \(A\), still crawling forward at \(0.6\ \text{m/s}\): a second impact, about \(0.3\ \text{s}\) after the first. Each impact is its own stage.
The ballistic pendulum
Before electronic timers, the speed of a bullet was measured by firing it into a heavy block hanging from cords and measuring how high the block swung. It is the classic two-stage problem.
Example 8.2 — A ballistic pendulum
A \(20\ \text{g}\) bullet traveling horizontally at \(600\ \text{m/s}\) embeds itself in a \(4\ \text{kg}\) block hanging from a light cord \(1.5\ \text{m}\) long. Find (a) the block's speed just after the impact, (b) the fraction of kinetic energy lost, and (c) how high the block swings.
Show solution
(a) The impact: momentum. The bullet stops in the block in a fraction of a millisecond. During that time the cord stays vertical, so it exerts no horizontal impulse, and the weight is nonimpulsive. Horizontal momentum is conserved, and the impact is plastic:
\[ 0.02(600) + 0 = (0.02 + 4)\,v_2 \quad\Rightarrow\quad v_2 = 2.99\ \text{m/s} \](b) \(T_1 = \tfrac12(0.02)(600)^2 = 3600\ \text{J}\), \(T_2 = \tfrac12(4.02)(2.985)^2 = 17.9\ \text{J}\). So \(99.5\%\) of the bullet's kinetic energy goes into deforming the bullet and the block, and into heat.
(c) The swing: energy. The cord's tension is perpendicular to the path and does no work:
\[ \tfrac12(4.02)(2.985)^2 = 4.02(9.81)\,h \quad\Rightarrow\quad h = \frac{v_2^2}{2g} = 0.454\ \text{m} \]The cord turns through \(\cos^{-1}(1 - 0.454/1.5) = 45.8^\circ\). Running the steps backward, a measured \(h\) gives the bullet's speed: \(v = \frac{m + M}{m}\sqrt{2gh}\).
Energy conservation from the bullet in flight to the top of the swing would be badly wrong: it would predict the block rising \(91\ \text{m}\).
Three stages
Example 8.3 — Slide, collide, skid
A \(2\ \text{kg}\) block \(A\) is released from rest on a smooth ramp, \(1.25\ \text{m}\) above the level floor. At the foot of the ramp it strikes a \(4\ \text{kg}\) block \(B\) at rest, with \(e = 0.6\). The floor under \(B\) is rough, with \(\mu_k = 0.4\). How far does \(B\) slide, and what happens to \(A\)?
Show solution
Stage 1, the slide: energy. \(\colVg{2(9.81)(1.25)} = \tfrac12(2)(v_A)_1^2\), so \((v_A)_1 = 4.95\ \text{m/s}\).
Stage 2, the impact: momentum and \(e\). Friction on \(B\) is nonimpulsive. Positive to the right:
\[ 2(4.952) = 2(v_A)_2 + 4(v_B)_2 \qquad (v_B)_2 - (v_A)_2 = 0.6(4.952) \] \[ (v_B)_2 = 2.64\ \text{m/s},\qquad (v_A)_2 = -0.330\ \text{m/s} \]Stage 3, the skid: work and energy. \(\tfrac12(4)(2.641)^2 \colF{{} - 0.4(4)(9.81)\,d} = 0\), so \(d = 0.889\ \text{m}\).
And \(A\)? It rebounds up the smooth ramp at \(0.330\ \text{m/s}\), rises only \(0.330^2/(2g) = 5.6\ \text{mm}\), and comes back down at the same speed. On the rough floor it stops after \(0.330^2/\left(2(0.4)(9.81)\right) = 14\ \text{mm}\), far short of \(B\), so there is no second impact.
Check your understanding
Key takeaways
- Cut a problem into stages at impacts and other changes; carry the velocity across each cut.
- Across an impact: conservation of momentum and \(e\). Before and after: work and energy, or impulse and momentum for time questions.
- Energy is lost in every impact with \(e < 1\), often most of it, as in the ballistic pendulum.
- You have now met the three integrated forms of Newton's second law: at an instant, along the path, and over time. Practise choosing between them in the Practice Lab, then take the Self-Check Quiz.