Question

A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg block sitting at rest on a frictionless surface.

A) What is the speed of the block after the collision?

B) What percentage of the ball's initial energy is "lost"?

Answer 1

A) $0.048 v_0$

The law of conservation of momentum states that the total initial momentum must be equal to the total final momentum, so

$p_i = p_f\\m v_0 = (m+M)v$

where

m = 50 g = 0.05 kg is the mass of the ball

v0 is the initial speed of the ball

M = 1.0 kg is the mass of the block

v is the final speed of the block+ball together

Solving for v, we find

$v=\frac{mv_0}{m+M}=\frac{(0.05 kg)v_0}{0.05 kg+1.0 kg}=0.048 v_0$

B) 95.2 %

The initial kinetic energy of the ball is:

$K_i=\frac{1}{2}mv_0^2 = \frac{1}{2}(0.05 kg)v_0^2 = 0.0250 v_0^2$

while the final kinetic energy of the system is

$K_f=\frac{1}{2}(m+M)v^2 = \frac{1}{2}(0.05 kg+1.0 kg)(0.048 v_0)^2 = 0.0012 v_0^2$

So, the loss in kinetic energy is

$\Delta K=K_i - K_f = 0.0250 v_0^2 - 0.0012 v_0 ^2 =0.0238 v_0^2$

In percentage,

$\frac{\Delta K}{K_i}=\frac{0.0238 v_0^2}{0.0250 v_0^2}=0.952$

which means 95.2%.