Question

A simple pendulum of length $1m$ has a wooden bob of mass $1kg$. It is struck by a bullet of mass ${10}^{-2}kg$ moving with a speed of $2\times {10}^{2}m{s}^{-1}$. The height to which the bob rises before swinging back is (Take $g=10m{s}^{-2})$.

• A. $0.2m$
• B. $0.6m$
• C. $8m$
• D. $1m$

Answer 1

The correct option is A $0.2m$

Simple pendulum of length$=1m$

Mass$=1kg$

Bullet of mass$={10}^{-2}kg$

Speed $=2\times {10}^{2}m/s$

Height to which the bob rises before swimming back$(g=10m/s^2)$

Now finding speed of bullet $-$ bob combination after collison $=V$

using the law of conservation of momentum

$mv=(m+M)V\left(0.01\right)\left(200\right)=(0.01+1)VV=1.98m/s$

$h=$ Height to which the bob of pendulum is raised

using conservation of energy

electric potential energy gained$-$

Kinetic energy lost

$(m+M)gh=\left(0.5\right)(m+M)V^{2}V=\sqrt{2gh}1.98=\sqrt{2\times 10\times h}h=(1.98{)}^2\times 2\times 10=0.2m$