Question:

A 0.010-kg bullet is fired toward a 2.0-kg pendulum. The bullet is sucked in the pendulum block and both are raised up to 0.20-m height. What is the initial velocity of the bullet?

Answer:

The masses of the bullet and the block can be denoted as \(m\) and \(M\), respectively. The initial velocity of the bullet and the final velocity of both the bullet and block are \(v\) and \(V\). Consider the states (a) through (b) in the figure. According to the conservation of linear momentum, the total initial momentum must be equal to the total final momentum.

\[  mv = (m+M)V  \]

Now, consider the states (b) to (c). For those states, we use the conservation of mechanical energy.

\[  \frac{1}{2}(m+M)V^2 = (m+M)gh  \]

\(v\) and \(V\) are unknown. In order to obtain the initial velocity, solve for \(V\) first from equation (1).

\[  V = \frac{mv}{m+M}  \]

From equation (2), we also have

\[  V = \sqrt{2gh}  \]

Therefore, we can equate them

\[  \frac{mv}{m+M} = \sqrt{2gh}  \]

The initial velocity is solved as

\[  v = \frac{m+M}{m}\sqrt{2gh} = \frac{0.010+2.00}{0.010}\sqrt{2\cdot 9.8\cdot 0.20} = 398\quad\mathrm{m/s}  \]