Question

A ball of mass 0.075 kg is fired horizontally into a ballistic pendulum. The pendulum mass is 0.350 kg. The ball is caught in the pendulum, and the centerof mass of the system rises a vertical distance of 0.145 m in the earth's gravitational field. What was the original speed of the ball? Assume g=9.80 m/s^2. How much kinetic energy was lost in the collision?

Answer 1

1) In the initial situation, the total mechanical energy of the system is given only by the kinetic energy of the ball that is moving with speed v:
$E_i =K= \frac{1}{2}m_b v^2$
where $m_b = 0.075 kg$ is the mass of the ball.

In the final situation, where the system (ball+pendulum) rises a vertical distance of h=0.145 m, the system is stationary (v=0) so the total mechanical energy of the system is the gravitational potential energy:
$E_f = U = (m_b+m_p)gh$
where $m_p = 0.350 kg$ is the mass of the pendulum.

For the law of conservation of energy, $E_i=E_f$ , so we can find the initial speed v of the ball:
$\frac{1}{2}m_bv^2 = (m_b+m_p)gh$
$v= \sqrt{ 2 \frac{m_b+m_p}{m_b}gh } =4.0 m/s$

2) The kinetic energy lost in the collision is the initial kinetic energy of the ball:
$K= \frac{1}{2}m_bv^2= \frac{1}{2}(0.075 kg)(4.0 m/s)^2=0.6 J$