A bob of mass m is released from rest from a horizontal position. At its lowest point, it makes a head-on elastic collision with a stationary block of mass M kept on a smooth horizontal surface. After the collision, the bob rebounds with a kinetic energy equal to 1 4 of its kinetic energy just before the collision. The ratio M m is

Step-by-step solution

Let the velocity of the bob of mass m just before the collision be u . Kinetic energy just before collision, K_i = 1 2 mu^2 After the collision, the bob rebounds with kinetic energy K_f = 1 4 K_i = 1 8 mu^2 . Let its velocity after collision be v₁ . Since it rebounds, its direction is reversed. 1 2 mv₁^2 = 1 8 mu^2 v₁ = - u 2 Let the velocity of block M after collision be v₂ . By conservation of linear momentum: mu + M(0) = mv₁ + Mv₂ mu = m (- u 2 ) + Mv₂ Mv₂ = 3 2 mu For a perfectly elastic collision, the coeffici