Inelastic Collisions in One Dimension — Question 5

Question 5

A projectile passes through instead of sticking

A $0.010\,\mathrm{kg}$ projectile enters a stationary $0.990\,\mathrm{kg}$ block at $200\,\mathrm{m/s}$ and emerges forward at $50\,\mathrm{m/s}$. The block is free on a frictionless track; external impulse is negligible.

1. Find the final block velocity.

2. Find the kinetic-energy reduction and explain why using the total mass in a common-speed formula would be wrong.

Question 5 - Solution

A projectile passes through instead of sticking

(a) Derivation and calculation

With negligible external impulse during penetration, $$0.010(200)=0.010(50)+0.990V.$$ $$\boxed{V=\frac{1.50}{0.990}=\frac{50}{33}=1.515\ \mathrm{m/s}}.$$

(b) Result and interpretation

The projectile and block retain different speeds: $$K_i=\tfrac12(0.010)(200)^2=200\ \mathrm J,$$ $$K_f=\tfrac12(0.010)(50)^2+\tfrac12(0.990)(50/33)^2=13.636\ \mathrm J.$$ $$\boxed{K_i-K_f=186.364\ \mathrm J}.$$

Physical check

The projectile transfers 1.50 kg m/s of momentum without remaining embedded.