Question

A bullet of mass $m$ travelling at velocity $v$ strikes a block of mass $2m$ that is attached to a rod of length $R$ and gets stuck to the block. The rod is free to rotate. Calculate 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. $\left(1/4\right)v^2/R$
• E. $\left(1/9\right)v^2/R$

Answer 1

Answer: (E)

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

By law of conservation of linear momentum ,

momentum before collision=momentum after collision

$mv+0=(m+2m)v^{\prime }$ as block is at rest before collision so its momentum is taken zero,

or $v^{\prime }=v/3$

therefore centripetal acceleration $a={\left(v^{\prime }\right)}^2/R$

or $a=\frac{{\left(v/3\right)}^2}{R}$

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