Head-on Collision of Two Trolleys
Trolley A, of mass 2.0 kg, moves to the right at 3.0 m/s on a smooth horizontal track. It collides head-on with trolley B, of mass 3.0 kg, which is moving to the left at 2.0 m/s. The trolleys stay on the same straight line during and after the collision. The coefficient of restitution between them is 0.60. The track is smooth, so external horizontal forces are negligible during the short collision. A student observes that the total momentum before the collision is zero and concludes that both trolleys must be at rest afterwards. You are asked to find the actual velocities after the collision, the impulse exerted on B, and the kinetic energy lost, and then to evaluate the student’s conclusion using the coefficient of restitution.
1. Find the velocities of A and B immediately after the collision.
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Take right as positive. Before the collision, v_A = +3.0 m/s and v_B = −2.0 m/s. Total momentum: p = 2.0 × 3.0 + 3.0 × (−2.0) = 6.0 − 6.0 = 0. Let the velocities after the collision be v_A and v_B. Momentum conservation gives 2v_A + 3v_B = 0, so v_A = −1.5v_B.
The approach speed before collision is 3.0 − (−2.0) = 5.0 m/s. Newton’s restitution law gives the separation speed after collision: v_B − v_A = e × 5.0 = 0.60 × 5.0 = 3.0 m/s. Substitute v_A = −1.5v_B: v_B − (−1.5v_B) = 2.5v_B = 3.0, so v_B = +1.2 m/s and v_A = −1.8 m/s.
Thus A moves left at 1.8 m/s and B moves right at 1.2 m/s. The negative sign for A means its direction has reversed.
2. Calculate the impulse exerted on trolley B and the kinetic energy lost in the collision.
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The impulse on B equals its change in momentum: J_B = m_B(v_B − u_B) = 3.0 × (1.2 − (−2.0)) = 3.0 × 3.2 = 9.6 N s to the right. By Newton’s third law, the impulse on A is 9.6 N s to the left. Check with A: J_A = 2.0 × (−1.8 − 3.0) = 2.0 × (−4.8) = −9.6 N s, which is consistent.
Kinetic energy before: KE_before = ½ × 2.0 × 3.0² + ½ × 3.0 × 2.0² = 9.0 + 6.0 = 15.0 J. Kinetic energy after: KE_after = ½ × 2.0 × 1.8² + ½ × 3.0 × 1.2² = 3.24 + 2.16 = 5.4 J. Loss = 15.0 − 5.4 = 9.6 J. So 9.6 J of kinetic energy is converted into other forms, such as sound, heat and deformation.
3. Evaluate the student’s conclusion that zero total momentum means both trolleys must stop.
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The student is wrong. Zero total momentum means the vector sum of the momenta is zero, not that each trolley stops. Both stopping would conserve momentum, but it would give a separation speed of zero after the collision. That would imply e = 0 in the restitution equation v_B − v_A = e × 5.0. Since e = 0.60, the trolleys must separate after the collision.
The correct result has equal and opposite momenta after the collision: 2.0 × (−1.8) = −3.6 kg m/s and 3.0 × 1.2 = +3.6 kg m/s, which sum to zero. The case where both stop is the perfectly inelastic case e = 0, not e = 0.60. The coefficient of restitution is the extra piece of information that rules out the student’s conclusion.