Which Cart Gets Hit Harder?
On a low-friction track, a 1.0 kg cart (A) rolls at 2.0 m/s and collides with a 3.0 kg cart (B) that is at rest. After the collision, A rolls backward at about 0.5 m/s and B rolls forward at about 0.8 m/s. A student says, 'B got the bigger force - it is heavier, so it must have been pushed harder. And A is the one that reversed, which proves A was hit hardest of all.' The class repeats the collision with a 6.0 kg cart in place of B, keeping A's speed the same, and finds the contact lasts about 0.05 s in both trials.
1. Evaluate the claim that B received a bigger force than A during the collision.
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The claim confuses force with the effect of force. During contact, A pushes B and B pushes A with forces that are equal in size and opposite in direction, and both forces act for the same 0.05 s. So neither cart 'got the bigger force.'
Why do they behave so differently? Because acceleration depends on mass:
- a = F / m, so the 1.0 kg cart gets three times the acceleration of the 3.0 kg cart from the same force.
- That is why A's velocity changes so much more dramatically (it even reverses) while B barely gets going.
The reversal of A is evidence of a large acceleration, not a large force.
2. Using the speed changes, show that the 3-to-1 mass ratio fits the data.
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Speed change of A: from +2.0 m/s to -0.5 m/s, so |change| = 2.5 m/s. Speed change of B: from 0 to 0.8 m/s, so |change| = 0.8 m/s.
Ratio of speed changes: 2.5 / 0.8 = about 3.1.
If the forces are equal and the contact time is the same, then change in speed = acceleration x time, and acceleration is inversely proportional to mass. The mass ratio is 3.0 kg / 1.0 kg = 3, so the speed-change ratio should be about 3. The measured ratio of about 3.1 matches within the rounding of the data (0.5 and 0.8 are approximate). This is a consistency check, not proof - but it fits the equal-force explanation much better than 'B was pushed harder.'
3. Predict how the 6.0 kg trial changes B's motion and what stays the same.
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What stays the same: the pair of forces during contact is still equal in size and opposite in direction, and the contact time is again about 0.05 s.
What changes: with twice the mass, B's acceleration is halved (a = F/m), so B's forward speed after the collision should be smaller than 0.8 m/s. Cart A should also rebound more slowly, because less of its motion is transferred in the shorter, softer push against a heavier object.
One caution: the actual size of the force in the 6 kg trial is not guaranteed to be identical to the 3 kg trial - it depends on how much the carts deform during contact. The reliable conclusion is the pairing: whatever force A feels, B feels the same size at the same moment.