Question

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Body ‘P’ having mass M moving with speed ‘u’ has head-on collision elastically with another body ‘Q’ having mass ‘m’ initially at rest. If m < < M, body ‘Q’ will have a maximum speed equal to ‘2u’ after the collision.
Reason R: During an elastic collision, the momentum and kinetic energy are both conserved.
In the light of the above statements, choose the most appropriate answer from the options given below:
A. A is correct but R is not correct.
B. Both A and R are correct but R is NOT the correct explanation of A
C. A is not correct but R is correct.
D. Both A and R are correct and R is the correct explanation of A.

Answer 1

Hint:Collisions are broadly classified into two categories, elastic and inelastic collisions. When there is no net loss in kinetic energy after the collision, such a type of collision is termed an elastic collision. If there is a loss of kinetic energy or the energy is changed into a different type of energy as a result of the collision, then the collision is termed inelastic. Kinetic energy is conserved in an elastic collision only. Momentum is conserved in both elastic and inelastic collisions.

Formula used:
Kinetic energy,
\[T = \dfrac{1}{2}m{v^2}\]
Linear momentum, \[p = mv\]
Where, m = mass of the object
v = velocity of the object.

Complete step by step solution:
Given: Elastic collision between a body P of mass M and velocity u and body Q of mass m at rest such that M > > m.

The conditions for elastic collision are:
- The total kinetic energy before and after the collision remains unchanged, that is the kinetic energy of the system is conserved.
- The total linear momentum before and after the collision remains unchanged, that is the linear momentum of the system is conserved.

Thus, using the condition for elastic collision we get,
From conservation of kinetic energy,
\[\dfrac{1}{2}{m_1}u_1^2 + \dfrac{1}{2}{m_2}u_2^2 = \dfrac{1}{2}{m_1}v_1^2 + \dfrac{1}{2}{m_2}v_2^2\]----- (1)
From the conservation of linear momentum,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]----- (2)
Where \[{m_1},{u_1},{v_1}\] are the mass and initial and final velocities of the first object (P) and \[{m_2},{u_2},{v_2}\] are the mass and initial and final velocities of the second object (Q).

According to the question,
\[{m_1} = M,{u_1} = u = {v_1}\] and \[{m_2} = m,{u_2} = 0\]---- (3)
\[\Rightarrow m < < M\]
Using equations (1) and (2), we get the coefficient of collision, e is given by:
\[e = \dfrac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}\]---- (4)
\[\Rightarrow e = 1\] for elastic collisions ---- (5)
Substituting values from (3) and (5) in (4), we get,
\[1 = \dfrac{{{v_2} - u}}{{u - 0}}\]
\[\therefore {v_2} = 2u\]
Hence object Q has a final velocity of \[{v_2} = 2u\].

Hence option D is the correct answer.

Note: Alternatively, we can put values of masses and velocities of objects P and Q in equations (1) and (2) and solve the two systems of equations generated as a result to obtain this result instead of using the concept of collision coefficient. The value of e is not equal to 1 for the inelastic coefficient. An elastic collision is ideal while an inelastic collision is more realistic.