Question

A bullet of mass m is fired into a block of wood of mass M which hangs on the end of a pendulum and gets embedded into it. When the bullet strikes the wooden block, the pendulum starts swinging with maximum rise R. Then, the velocity of the bullet is given by.

• A. $\frac{(M+m)gR}{m}$
• B. $\frac{M+m}{m}\sqrt{2gR}$
• C. $\frac{(M+m)\sqrt{\left(gR\right)}}{m}$
• D. $\frac{M+m}{mgR}$

Answer 1

The correct option is B $\frac{M+m}{m}\sqrt{2gR}$

since the bullet gets embedded in wooden block,all its momentum is transferred to block$+$bullet system.
$(M+m)V=muu=\frac{(M+m)V}{m}$
and then K.E. is converted into P.E.
$\frac{1}{2}(M+m)V^2=(M+m)gRV=\sqrt{2gR}u=\frac{(M+m)}{m}\sqrt{2gR}$