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Question: A ball of mass m moving with velocity \[V\] makes a head on elastic collision with a ball of the sam...

A ball of mass m moving with velocity VV makes a head on elastic collision with a ball of the same mass moving with velocity 2V2V towards it.Taking direction of as positive velocities of the two balls after collision are
A. . - V and 2V{\text{. - V and 2V}}
B. . 2V and - V{\text{. 2V and - V}}
C. . V and - 2V{\text{. V and - 2V}}
D. . - 2V and V{\text{. - 2V and V}}

Explanation

Solution

To solve the given question we use concepts of elastic collision and conservation of linear momentum. Moreover, the elastic collision definition states that the overall kinetic energy and the momentum are preserved in case of this category of collision.Therefore as masses and velocities are given we use linear momentum conservation and kinetic energy conservation to solve the question.

Complete step by step answer:
The velocity of a moving ball is VV. The other ball is moving at a velocity 2V2V towards the first ball. Two balls have the same mass mm.According to the law of conservation of linear momentum, the momentum before collision will be the same as the momentum after the collision.
m1u1+m2u2=m1v1+m2v2{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}

Where masses are given as the same we write m1=m2=m{m_1} = {m_2} = m. Then initial velocity of first ball whose direction is said to be taken positive we write,
u1=V{u_1} = V
The initial velocity of second body will be u2=−2V{u_2} = - 2V(as it was moving towards the first ball i.e. it was moving in opposite direction of first ball)

Now we substitute these values in the conservation of momentum equation
m1u1+m2u2=m1v1+m2v2{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}
⇒m(V)+m(−2V)=mv1+mv2\Rightarrow m\left( V \right) + m\left( { - 2V} \right) = m{v_1} + m{v_2}
⇒v1=−V−v2\Rightarrow {v_1} = - V - {v_2}
Now let us consider kinetic energy conservation
12m1u12+12m2u22=12m1v12+12m2v22\dfrac{1}{2}{m_1}{u_1}^2 + \dfrac{1}{2}{m_2}{u_2}^2 = \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}{v_2}^2
⇒12mu12+12mu22=12mv12+12mv22\Rightarrow \dfrac{1}{2}m{u_1}^2 + \dfrac{1}{2}m{u_2}^2 = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2

Taking values of velocities and cancelling common terms we get
V2+(−2V)2=(−V−v2)2+v22{V^2} + {\left( { - 2V} \right)^2} = {\left( { - V - {v_2}} \right)^2} + {v_2}^2
As (−a−b)2=(−1(a+b))2=a2+b2+2ab{\left( { - a - b} \right)^2} = {\left( { - 1\left( {a + b} \right)} \right)^2} = {a^2} + {b^2} + 2ab
V2+4V2=V2+v22+2Vv2+v22{V^2} + 4{V^2} = {V^2} + {v_2}^2 + 2V{v_2} + {v_2}^2
⇒v22+Vv2−2V2=0\Rightarrow {v_2}^2 + V{v_2} - 2{V^2} = 0
Using factorization ⇒v22+2Vv2−Vv2−2V2=0 \Rightarrow {v_2}^2 + 2V{v_2} - V{v_2} - 2{V^2} = 0
⇒v2(v2+2V)−V(v2+2V)=0\Rightarrow {v_2}\left( {{v_2} + 2V} \right) - V\left( {{v_2} + 2V} \right) = 0
⇒(v2+2V)(v2−V)=0\Rightarrow \left( {{v_2} + 2V} \right)\left( {{v_2} - V} \right) = 0
Therefore v2=−2V{v_2} = - 2V or v2=V{v_2} = V

As the body cannot have the same velocity v2≠−2V{v_2} \ne - 2V
Now let us find v1{v_1}
⇒v1=−V−v2\Rightarrow {v_1} = - V - {v_2} for v2=V{v_2} = V
⇒v1=−V−V=−2V\Rightarrow {v_1} = - V - V = - 2V
Therefore we can conclude that due to elastic collision of bodies having equal masses,their velocities get interchanged.
∴v1=−2V,v2=V\therefore {v_1} = - 2V,{v_2} = V

Therefore,the correct answer is option (D).

Note: 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. 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.