A simple pendulum , made of a string of length l and a bob of mass m, is released from a small angle `theta_(0)` it strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations , elastically , it bounces back and goes up to an angle `theta_(1)` then M is given by :

Options

A. `(m)/(2)((theta_(0)+theta_(1))/(theta_(0)-theta_(1)))`
B. `m((theta_(0)-theta_(1))/(theta_(0)+theta_(1)))`
C. `m((theta_(0)+theta_(1))/(theta_(0)-theta_(1)))`
D. `(m)/(2)((theta_(0)-theta_(1))/(theta_(0)+theta_(1)))`

Answer

C

Explanation

`omega=sqrt((g)/(l))`.
Velocity just before collision `u= (sqrt((g)/(l))theta_(0))l=sqrt(gl)theta_(0)`
v = velocity of ball after collision `v=(sqrt((g)/(l))theta_(1))l=sqrt(gl)theta_(1)`.
Momentum conservation : m u = Mv'-mv ………(i)
`e=1: `
`v' +v=u` ......(ii)
`omega=sqrt((g)/(l))` From (i) and (ii)
`m(u+v)=M(u-v)`
`m(theta_0+ theta_1)=M(theta_0 -theta_1)`
`M=m((theta_0 +theta_1)/(theta_0 -theta_1))`.
.