Velcro Collision on a Track with a Hidden Rough Patch
In a school laboratory a compressed-air track is used to study collisions. Trolley A (mass 0.60 kg) moves at 1.20 m/s towards trolley B (mass 0.40 kg), which is at rest. A Velcro pad makes them stick together, and light gates record the pair moving at 0.70 m/s immediately after the collision. Beyond that point the track has a rough section 0.35 m long with coefficient of kinetic friction 0.20, and a buffer spring of stiffness 180 N/m is mounted at the far end of the rough section. A student writes that the collision 'conserves energy because the track is horizontal' and predicts that the joined trolleys will compress the spring by about 5 cm. The friction patch is not mentioned in the write-up. g = 9.8 m/s².
1. Use conservation of momentum to calculate the speed of the joined trolleys just after the collision and the fraction of kinetic energy lost. Is the light-gate value of 0.70 m/s consistent with your result?
Read the worked answer
During the short collision the external friction force is negligible, so momentum is conserved:
p = m_A u_A = 0.60 × 1.20 = 0.72 kg m s⁻¹
v = p/(m_A + m_B) = 0.72/1.00 = 0.72 m s⁻¹
KE_before = ½ × 0.60 × 1.20² = 0.432 J
KE_after = ½ × 1.00 × 0.72² = 0.259 J
Loss = 0.173 J, i.e. 0.173/0.432 = 40%. The collision is perfectly inelastic: the lost mechanical energy becomes deformation, sound and thermal energy in the Velcro.
The light gate reads 0.70 m s⁻¹, about 3% low. Over the collision the friction impulse is small but not zero, and the gate measures an average over a finite card width, so a small discrepancy is expected. It does not indicate a momentum-conservation failure.
2. Show whether the joined trolleys reach the spring. What would an observer actually see, and why is the student's 5 cm prediction wrong?
Read the worked answer
Friction on the joined trolleys is f = μ(m_A + m_B)g = 0.20 × 1.00 × 9.8 = 1.96 N, so the deceleration is a = f/m = 1.96 m s⁻².
The stopping distance from 0.72 m s⁻¹ is s = v²/(2a) = 0.5184/3.92 = 0.132 m.
The rough section is 0.35 m long, so the trolleys stop after about 13 cm, well before the spring. The observer sees the trolleys slow down and stop on the rough track; the spring is never compressed.
The student's 5 cm prediction assumes all the post-collision kinetic energy is stored elastically. In fact 1.96 × 0.35 = 0.686 J would be needed just to cross the rough section, which exceeds the 0.259 J available, so the trolleys cannot even reach the spring. Energy is dissipated by friction, not stored.
3. If the rough section were removed, calculate the maximum spring compression. Explain why using the pre-collision kinetic energy would give a wrong answer.
Read the worked answer
With no friction, energy conservation from just after the collision to maximum compression gives:
½ m v² = ½ k x²
0.259 = ½ × 180 × x² = 90x²
x² = 2.88 × 10⁻³ m² → x = 0.054 m ≈ 5.4 cm.
This is close to the student's 5 cm estimate, but the reasoning matters: the energy available is the post-collision kinetic energy 0.259 J, not the pre-collision 0.432 J. Using 0.432 J would give x = 0.069 m (6.9 cm), overestimating the compression by about 30%. The collision removes 40% of the kinetic energy before the spring is involved, so the energy budget must start after the collision.