Question

A bullet of mass m$=50$ gm strikes $(\Delta t\approx 0)$ a sand bag of mass M$=5$ kg hanging from a fixed point, with a horizontal velocity $\overrightarrow{v_p}$. If bullet sticks to the sand bag then just after collision the ratio of final and initial kinetic energy of the bullet is?

• A. ${10}^{-2}$
• B. ${10}^{-3}$
• C. ${10}^{-6}$
• D. ${10}^{-4}$

Answer 1

The correct option is D ${10}^{-4}$

According to Conservation of Momentum,

mvp = (m+M)V where

m is the mass of the bullet, 

vp is the velocity of the bullet before collision, 

M is the mass of the sandbag and V is the velocity after the collision

0.05* vp  = (0.050+5)V

vp = [5.050/0.05] V = 101 V 

Ratio of final KE to initial KE = [0.5mV2] / [0.5mvp2] = 

10‑4  (approx.)