A ball with mass M, moving horizontally at 4.00 m/s, collides elastically with a block with mass 3M that is initially hanging at rest from the ceiling on the end of a 50.0-cm wire. Find the maximum angle through which the block swings after it is hit?

The initial speed of the ball is $u_1=4m/s$

The initial speed of the block is $u_2=0m/s$

The length of wire for the block is $l=50cm=0.5m$

The mass of the block is: 3 M

The mass of the ball is: M

The maximum angle of swing for the block is calculated by using the height raised by the block after the collision.

The coefficient of restitution is given as:

$e=\frac{u_1-u_2}{v_2-v_1}$

Here, e is the coefficient of restitution and its value for perfect elastic collision.

$$\begin{gathered} 1=\frac{u_1-0}{v_2-v_1} \\ {v}_2-v_1=u_1..........\left(1\right) \end{gathered}$$

Apply the momentum conservation for block and ball.

$M{u}_1+3M{u}_2=M{v}_1+3M{v}_2$

Substitute all the values in the above equation.

$$\begin{gathered} M{u}_1+3M\left(0\right)=M{v}_1+3M{v}_2 \\ {v}_1+3v_2=u_1.........\left(2\right) \end{gathered}$$

Solve equation (1) and (2), we get

$$\begin{gathered} v_2=\frac{u_1}{2} \\ {v}_2=\frac{4m/s}{2} \\ {v}_2=2m/s \end{gathered}$$

Apply energy conservation to find the height raised by the block.

$$\begin{gathered} \frac{1}{2}\left(3M\right){\left(v_2\right)}^2=\left(3M\right)gh \\ h=\frac{v_2^2}{2g} \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} h=\frac{{\left(2m/s\right)}^2}{2\left(9.8m/s^2\right)} \\ h=0.204m \end{gathered}$$

The maximum angle of swing for block is given as:

$\cos \theta =\frac{l-h}{l}$

Substitute all the values in the above equation.

$$\begin{gathered} \cos \theta =\frac{0.50m-0.204m}{0.50m} \\ \theta =53.7° \end{gathered}$$

Therefore, the maximum angle of swing for block is $53.7°$.