Question

A bullet of mass m traveling at velocity v strikes a block of mass 2m that is attached to a rod of length R . The bullet collides with the block at a right angle and gets stuck in the block. The rod is free to rotate. What is the centripetal acceleration of the block after the collision?

• A. $v^2/R$
• B. $\left(1/2\right)v^2/R$
• C. $\left(1/3\right)v^2/R$$
• D. $\left(1/9\right)v^2/R$

Answer 1

Answer: (D)

$\left(1/9\right)v^2/R$

Since no external force acts on the system in the direction of the initial direction of movement of the bullet, momentum in that direction is conserved.

Hence $m_{bullet}v=(m_{bullet}+m_{block})v^{\prime }$

$⟹v^{\prime }=\frac{m}{m+2m}v=\frac{v}{3}$
This is the attained velocity of the bullet stuck in block system.

The centripetal acceleration of the block=$\frac{v^{\prime 2}}{R}$

$=\frac{v^2}{9R}$