Question

A simple pendulum of length $2m$ has a wooden bob of mass $1kg$ . It is struck by a bullet of mass ${10}^{-3}kg$ moving with a speed of $3\times {10}^2$ . The bullet gets embedded into the bob. What is the height to which the bob rises before swinging?

• A. $0.3m$
• B. $0.2m$
• C. $0.01m$
• D. $0.004m$

Answer 1

The correct option is D $0.004m$

The bullets of mass $m$ gets embedded into the bob of mass $M$ so it follows two conservation laws

According to the conservation of momentum

$mv=\left(M+m\right)V$

We get the velocity of the pendulum

$V=\frac{mv}{\left(m+M\right)}$

$10^{-3}\times 3\times 10^2=\left(10^{-3}+1\right)V$

$V=0.299m/s$

By the conservation of energy we get

$\left(m+M\right)gh=\frac{1}{2}\left(m+M\right)V^2$

$V=\sqrt{2gh}$

By substituting the given values we get

$h=\frac{0.0894}{2\times 10}$

Hence the height is $0.004m$