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Question: A ball of mass \(m\) moving with a speed \(2{v_0}\) collides head-on with an identical ball at rest....

A ball of mass mm moving with a speed 2v02{v_0} collides head-on with an identical ball at rest. If ee is the coefficient of restitution, then what will be the ratio of the velocity of two balls after collision?
(A) 1−e1+e\dfrac{{1 - e}}{{1 + e}}
(B) 1+e1−e\dfrac{{1 + e}}{{1 - e}}
(C) e−1e+1\dfrac{{e - 1}}{{e + 1}}
(D) e+1e−1\dfrac{{e + 1}}{{e - 1}}

Explanation

Solution

The law of conservation of linear momentum is derived from Newton’s second and third laws of motion. The law of conservation of momentum states that the total momentum of an isolated system will remain the same. The total linear momentum of a system will be constant when the net external force is zero.

Complete step by step solution:
The velocity of the moving ball is 2v02{v_0}. The other ball is at rest. Two balls have the same mass.
According to the law of conservation of linear momentum, the momentum before collision will be the same as the momentum after the collision.
Let the mass of the balls be mm.
m(2v0)+m×0=mv1+mv2m\left( {2{v_0}} \right) + m \times 0 = m{v_1} + m{v_2}
Where v1{v_1} and v2{v_2} be the velocities of the two balls after the collision.
Canceling the common terms, the above equation can be written as
2v0=v1+v22{v_0} = {v_1} + {v_2} ……….(1)(1)
The coefficient of restitution by definition
e=v2−v1u1−u2e = \dfrac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}
Where u1{u_1} and u2{u_2} are the initial velocities of the balls.
Here, u1=2v0{u_1} = 2{v_0} and u2=0{u_2} = 0
Substituting, we get
e=v2−v12v0−0=v2−v12v0e = \dfrac{{{v_2} - {v_1}}}{{2{v_0} - 0}} = \dfrac{{{v_2} - {v_1}}}{{2{v_0}}}
This equation can be written as,
v2−v1=2v0e{v_2} - {v_1} = 2{v_0}e ……….(2)(2)
Adding equation (1)(1) and (2)(2), we get
v1+v2+v2−v1=2v0+2v0e{v_1} + {v_2} + {v_2} - {v_1} = 2{v_0} + 2{v_0}e
That is,
2v2=2v0(1+e)2{v_2} = 2{v_0}\left( {1 + e} \right)
Canceling common terms on both sides,
v2=v0(1+e){v_2} = {v_0}\left( {1 + e} \right)
Subtracting equation (2)(2) from equation (1)(1), we get
v1+v2−v2+v1=2v0−2v0e{v_1} + {v_2} - {v_2} + {v_1} = 2{v_0} - 2{v_0}e
This equation can be written as,
2v1=2v0(1−e)2{v_1} = 2{v_0}\left( {1 - e} \right)
Canceling common terms, we get
v1=v0(1−e){v_1} = {v_0}\left( {1 - e} \right)
Then we can write,
v1v2=v0(1−e)v0(1+e)\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{v_0}(1 - e)}}{{{v_0}(1 + e)}}
Canceling v0{v_0}
v1v2=1−e1+e\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{1 - e}}{{1 + e}}

Note: The coefficient of restitution is the ratio of final velocity to the initial velocity of two objects after they collide with each other. If the total kinetic energy of the system is conserved after a collision then it is an elastic collision. If the total kinetic energy is not conserved, then the collision is inelastic.