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

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

1.	A moving blockhaving mass m , collides with another stationary block having mass 4m , The lighter block comes to rest after collision . When the intial velocity of the lighter block is v, then the value of coefficient of restitution (e ) will be
Answer» `0.4` `0.5` `0.8` `0.25` Solution :From the law conservation of VELOCITY , ` " P=p" mv =4mv_(2)"" :. V_(2) = (v_(1))/4` ` :. ` Restitutional VALUE , `(e)= (v_(2)-v_(1))/(v_(1)-v_(2))=((v_(1))/4-0)/(v_(1)-0)=1/4 =0.25`.