Question

A massive ball moving with a speed $v$ collides with a tiny ball having a mass very much smaller than the mass of it. If the collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to:
(A) $v$
(B) $2v$
(C) $\dfrac{v}{2}$
(D) $\infty$

Answer 1

In the elastic collision, both linear momentum and kinetic energy of the system remains conserved before and after the collision. Use the formula for the final velocity of the second object for the elastic collision.

Formula used:
${v_2} = \dfrac{{2{m_1}{u_1} + {u_2}\left( {{m_2} - {m_1}} \right)}}{{{m_1} + {m_2}}}$

Complete step by step answer:
For an inelastic collision, according to the conservation of the linear momentum and kinetic energy, the velocity of the second body is given by the equation,
${v_2} = \dfrac{{2{m_1}{u_1} + {u_2}\left( {{m_2} - {m_1}} \right)}}{{{m_1} + {m_2}}}$
Here, ${m_1}$ is the mass of the massive ball, ${u_1}$ is the initial velocity of the massive ball, ${m_2}$ is the mass of tiny ball, and ${u_2}$ is the initial velocity of the tiny ball.
Since the tiny ball is initially at rest, the initial velocity of the tiny ball is zero. Therefore, the above equation becomes,
${v_2} = \dfrac{{2{m_1}{u_1}}}{{{m_1} + {m_2}}}$
Since the mass of a tiny ball is negligible as compared to the mass of a massive ball, we can substitute ${m_1}$ for ${m_1} + {m_2}$ in the above equation.
${v_2} = \dfrac{{2{m_1}{u_1}}}{{{m_1}}}$
$\Rightarrow {v_2} = 2{u_1}$
Since the initial velocity of the massive ball is v, substitute v for ${u_1}$in the above equation.
${v_2} = 2v$
Therefore, the final velocity of the tiny ball will be twice the initial velocity of the massive ball.
So, the correct answer is option (B).

Note: To get the formula for the final velocity of the second object in elastic collision, use the equation of conservation of linear momentum and conservation of kinetic energy as the kinetic energy and the linear momentum conserved in the elastic collision.