Question

A steel ball of radius $2\,cm$ is at rest on a horizontal frictionless surface. Another steel ball of radius $4\,cm$ moving at a velocity of 81cm/s collides elastically with the first ball. After collision smaller ball moves with a speed of
A. $81\,cm\,{s^{ - 1}}$
B. $63\,cm\,{s^{ - 1}}$
C. $144\,cm\,{s^{ - 1}}$
D. None of these

Answer 1

An insight into the basics of collision (mainly elastic collision) will help in solving the question. By the help of the radius of bodies we can get a relation between masses of bodies. Applying those values in equations relating velocity and mass of body in both states (i.e. before and after collision) we can achieve our results.

Formula Used:
$\rho = \dfrac{m}{V}$
$\Rightarrow V = \dfrac{4}{3}\pi {r^3}$
Where $\rho =$ density, m = mass and V = volume
${v_1} = \dfrac{{\left( {{m_2} - {m_1}} \right){u_2}}}{{\left( {{m_2} + {m_1}} \right)}}$
Where,${v_1} =$ Final velocity of smaller (radius 2cm) body and ${u_2} = 81\,cm\,{s^{ - 1}}$ Initial velocity of striking (radius 4cm) body.

Complete step by step answer:
To get a relation between mass and radius of balls we need to compare their densities. As the material of the balls are the same and so will be their density.
Let density of balls be $= \rho$
And the mass of balls be ${m_1}$ and ${m_2}$ for 2 cm and 4 cm radii respectively.
According to formula of density,
$\rho = \dfrac{m}{V}$
$\Rightarrow m = \rho \times V$ ….. (i)
Where $\rho =$ density, m = mass and V = volume

Volume of a sphere is,
$V = \dfrac{4}{3}\pi {r^3}$
Putting values of mass and volume and radius in equation (i)
${m_1} = \dfrac{4}{3}\pi {\left( 2 \right)^3}\rho$
$\Rightarrow {m_2} = \dfrac{4}{3}\pi {\left( 4 \right)^3}\rho$
Equating above equations we get,
$8{m_1} = {m_2}$ …… (ii)

For elastic collision velocity of bodies after collision is given by,
${v_1} = \dfrac{{\left( {{m_2} - {m_1}} \right){u_2}}}{{\left( {{m_2} + {m_1}} \right)}}$
Where,${v_1} =$ Final velocity of smaller (radius 2cm) body and ${u_2} = 81\,cm\,{s^{ - 1}}$ Initial velocity of striking (radius 4cm) body.
Putting values of compared mass from equation (ii),
${v_1} = \dfrac{{\left( {8{m_1} - {m_1}} \right)81}}{{\left( {8{m_1} + {m_1}} \right)}}$
Solving the equation we will get,
${v_1} = 63\,cm\,{s^{ - 1}}$
Therefore, the velocity of the smaller body is $63\,cm\,{s^{ - 1}}$.

Hence, the correct answer is option B.

Note: Elastic collision is the state where kinetic energy as well as momentum of the body remains conserved even after collision. After elastic collision by comparing the kinetic energy and momentum of bodies in initial and final states we can achieve a relation between velocity of body and other terms. The momentum of the body itself might get varied but the momentum of the whole system always remains conserved.