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=(mv_0)/(m+M)=((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=(1)/(2)mv_0^2 = (1)/(2)(0.05 kg)v_0^2 = 0.0250 v_0^2`

while the final kinetic energy of the system is

`K_f=(1)/(2)(m+M)v^2 = (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,

`(\Delta K)/(K_i)=(0.0238 v_0^2)/(0.0250 v_0^2)=0.952`

which means 95.2%.