A ball of mass m moving with velocity V makes a head on elastic collision with a ball of the same mass moving with velocity 2V towards it. Taking direction of V as positive, velocities of the two balls after collision area)-V and 2Vb)2V and Vc)V and 2Vd)-2V and VCorrect answer is option 'D'. Can you explain this answer?

Question

Answer

Given:
- A ball of mass m moving with velocity V makes a head-on elastic collision with a ball of the same mass moving with velocity 2V towards it.

To find:
- Velocities of the two balls after collision.

Solution:
1. Elastic Collision:
- An elastic collision is one in which both the momentum and the kinetic energy of the system are conserved.
- In this collision, the total momentum of the system before the collision is equal to the total momentum after the collision.

2. Conservation of Momentum:
- Let the velocities of the two balls after the collision be v1 and v2.
- According to the conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision.
- The initial momentum of the system is mV - 2m(2V) = -3mV.
- The final momentum of the system is mv1 + mv2.
- Therefore, mv1 + mv2 = -3mV.

3. Conservation of Kinetic Energy:
- According to the conservation of kinetic energy, the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision.
- The initial kinetic energy of the system is (1/2)mv^2 + (1/2)m(2V)^2 = 3mV^2.
- The final kinetic energy of the system is (1/2)mv1^2 + (1/2)mv2^2.
- Therefore, (1/2)mv1^2 + (1/2)mv2^2 = 3mV^2.

4. Solving Simultaneous Equations:
- We have two equations and two unknowns (v1 and v2).
- Solving these equations simultaneously, we get v1 = -2V and v2 = V.

5. Final Answer:
- Therefore, the velocities of the two balls after the collision are -2V and V.
- Hence, the correct answer is option 'D'.

Answer

Answer

M1u1+m2u2=m1v1+m2v2m1u1+m2u2=m1v1+m2v2

⇒m(V)+m(−2V)=mv1+mv2⇒m(V)+m(−2V)=mv1+mv2

⇒v1=−V−v2

12m1u12+12m2u22=12m1v12+12m2v2212m1u12+12m2u22=12m1v12+12m2v22

⇒12mu12+12mu22=12mv12+12mv22

V2+(−2V)2=(−V−v2)2+v22
As (−a−b)2=(−1(a+b))2=a2+b2+2ab
V2+4V2=V2+v22+2Vv2+v22

by factorisation
⇒v22+Vv2−2V2=0
v22+2Vv2−Vv2−2V2=0

⇒v2(v2+2V)−V(v2+2V)=0
⇒(v2+2V)(v2−V)=0
Therefore v2=−2V
or v2=V

for v1;
⇒v1=−V−v2
for v2=V
⇒v1=−V−V=−2V

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