A moving block having mass m, collides with another stationary block having mass 4 m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

A moving block having mass m, collides with another stationary block h

1.	A moving block having mass m, collides with another stationary block having mass 4 m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be
Answer» 0.5 0.25 0.8 0.4 Solution :Let final velocity of the BLOCK of mass `4 m = v.` Intial VELCOITY of block of mass 4 m = 0 Final velocity of block of mass m = 0 . According to law of conservation of linear MOMENTUM, `MV = 4 m xx 0 = 4mv. + 0 " or " , v. = v/4` Coefficient of restitution, `e = ("Relative velocity of separation")/("Relative velocity of APPROACH") = (v//4)/v = 0.25`.