A bullet ofmass m hits a wooden block of mass M, suspended by a string of length l, and gets embedded in it. If the velocity of the bullet is v, find theangular displacement of the block.

A bullet ofmass m hits a wooden block of mass M, suspended by a string

1.	A bullet ofmass m hits a wooden block of mass M, suspended by a string of length l, and gets embedded in it. If the velocity of the bullet is v, find theangular displacement of the block.
Answer» Solution :Let the VELOCITY of the block -bullet system after impact be V and the angular displacement be `theta` [Fig.1.48]. From the lawof conservation of momentum, `mv=(M+m)V or, V=(mv)/(M+m)` Kineticenergy of the block -bullet system at position A `=1/2 (M+m)V^2` `=1/2 (M+m)xx(m^2v^2)/((M+m)^2)` =`(m^2v^2)/(2(M+m))` Potential energy of the system at B `=(M+m)g*AC` `=(M+m)g[OA-OC]` `=(M+m)g(l-lcos theta)=(M+m)gl(1-COS theta)` From the law of conservation of energy, `1/2 (m^2v^2)/(M+m) =(M+m)gl(1- cos theta)` or, `1- cos theta =1/2 (m^2v^2)/((M+m)^2gl)` `or, 2 sin ^2"" theta/2=1/2 ((mv)/(M+m))^2xx1/(gl)` `or, sin "" theta/2 =(mv)/(2(M+m)SQRT(gl)) or, theta/2=sin ^(-1) [(mv)/(2(M+m)sqrt(gl))]` ` therefore theta =2sin ^(-1) [(mv)/(2(M+m)sqrt(gl))]`.