Question

Two identical blocks A and B connected by a massless string are placed on a frictionless horizontal plane. A bullet having same mass moving with a speed $u$ strikes block B from behind as shown. If the bullet gets embedded into the block B then find impulse on B due to normal force of the collision?

• A. $\frac{2mu}{3}$
• B. $\frac{mu}{3}$
• C. $\frac{2mu}{5}$
• D. $-\frac{2mu}{3}$

Answer 1

The correct option is A $\frac{2mu}{3}$

velocity of whole system after collision is

by momentum conservation is

$mu=(m+m+m)v$

$v=\frac{u}{3}$

now the change in momentum of A is due to impulse of tension

$Tdt=mv-o=\frac{mu}{3}$ ...(1)

now impulse on B will be due to tension and normal force

So $Ndt-Tdt=mv-o$

from 1 putting the value of $Tdt$

$Ndt=Tdt+mv=\frac{mu}{3}+\frac{mu}{3}=\frac{2mu}{3}$