Question

A led bullet of mass 0.05kg is fired with a velocity of 200m/s into a lead block of mass 0.95kg. Given that the lead block can move freely calculate the kinetic energy after the impact and the loss of energy

Answer 1

From the conservation of energy:

$$mV=(m+M)v$$

$$(0.05)(200)=(0.05+0.95)v$$

$$v=10\frac{m}{s}$$

The kinetic energy after the impact:

$$K=0.5(m+M)v^2=0.5(1)10^2=50\ J$$

The loss of energy:

$$E=0.5mV^2-K=0.5(0.05)(200)^2-50=950\ J$$