
A block having mass m collides with another stationary block having mass 2m. The lighter box comes to rest after a collision. If the velocity of first block is v, then the value of coefficient of restitution must be
(A) $ 0.5 $
(B) $ 0.4 $
(C) $ 0.6 $
(D) $ 0.8 $
Answer
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Hint
As we can see from the question, there is no external force acting upon the two bodies, hence we can apply the law of conservation of momentum that is the total momentum remains constant.
$ {m_1}{v_1} + {m_2}{v_2} = $ constant
Complete step by step answer
In the question two blocks are given of mass m and 2m respectively. On the application of the law of conservation of momentum which states that in the absence of external force on a system of two bodies, the total momentum is constant, we have
Momentum before collision is $ mv + 2m \times 0 $
Momentum after collision is $ m \times 0 + 2mv $
By momentum conservation law,
$ \Rightarrow mv = 2m{v'} $ where v is the velocity of first block and $ v' $ is velocity of second block
$ \Rightarrow v' = \dfrac{v}{2} $
Coefficient of restitution is the ratio of final relative velocity to initial relative velocity on collision
$ \Rightarrow e = \dfrac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}} $ , where $ {v_2} - {v_1} $ is final relative velocity and $ {u_1} - {u_2} $ is the initial relative velocity.
Substituting the values, $ {u_1} = v,{u_2} = 0and{v_1} = 0,{v_2} = \dfrac{v}{2} $
$ \Rightarrow e = \dfrac{{v/2 - 0}}{{v - 0}} = \dfrac{1}{2} = 0.5 $
Hence, the correct option is (A).
Additional Information
There are two types of collision –
Elastic collision: if force between the colliding bodies is conservative, then kinetic energy and momentum are conserved. This type of collision is known as ‘elastic’. Example- atomic and nuclear collisions.
Inelastic collision: on collision if there is change in kinetic energy, it is said to be ‘inelastic’ collision. In this case also the momentum and total energy is conserved. Example – a bullet getting embedded in a body after hitting it is a perfectly inelastic collision.
Note
The value of coefficient (e) of restitution generally lies between 0 and 1. If e is 1, then it would be a perfect elastic collision. If e is 0 then it is a perfectly inelastic collision.
As we can see from the question, there is no external force acting upon the two bodies, hence we can apply the law of conservation of momentum that is the total momentum remains constant.
$ {m_1}{v_1} + {m_2}{v_2} = $ constant
Complete step by step answer
In the question two blocks are given of mass m and 2m respectively. On the application of the law of conservation of momentum which states that in the absence of external force on a system of two bodies, the total momentum is constant, we have
Momentum before collision is $ mv + 2m \times 0 $
Momentum after collision is $ m \times 0 + 2mv $
By momentum conservation law,
$ \Rightarrow mv = 2m{v'} $ where v is the velocity of first block and $ v' $ is velocity of second block
$ \Rightarrow v' = \dfrac{v}{2} $
Coefficient of restitution is the ratio of final relative velocity to initial relative velocity on collision
$ \Rightarrow e = \dfrac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}} $ , where $ {v_2} - {v_1} $ is final relative velocity and $ {u_1} - {u_2} $ is the initial relative velocity.
Substituting the values, $ {u_1} = v,{u_2} = 0and{v_1} = 0,{v_2} = \dfrac{v}{2} $
$ \Rightarrow e = \dfrac{{v/2 - 0}}{{v - 0}} = \dfrac{1}{2} = 0.5 $
Hence, the correct option is (A).
Additional Information
There are two types of collision –
Elastic collision: if force between the colliding bodies is conservative, then kinetic energy and momentum are conserved. This type of collision is known as ‘elastic’. Example- atomic and nuclear collisions.
Inelastic collision: on collision if there is change in kinetic energy, it is said to be ‘inelastic’ collision. In this case also the momentum and total energy is conserved. Example – a bullet getting embedded in a body after hitting it is a perfectly inelastic collision.
Note
The value of coefficient (e) of restitution generally lies between 0 and 1. If e is 1, then it would be a perfect elastic collision. If e is 0 then it is a perfectly inelastic collision.
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