Question

Ball $A$ of mass $10\text{kg}$ moving with a velocity $6\text{m/s}$, collides with ball $B$ of mass $50\text{kg}$ which is at rest. What is the ratio of the kinetic energies of the ball $A$ before and after the elastic collision?

• A. $9:4$
• B. $4:9$
• C. $2:3$
• D. $3:2$

Answer 1

The correct option is A $9:4$

Given:
$m_a=10\text{kg},u_a=6\text{m/s}$
$m_b=50\text{kg},u_b=0\text{m/s}$
Let velocity of $m_a$ is $v_a$ after elastic collision.
$\Rightarrow v_a=\left(\frac{m_a-m_b}{m_a+m_b}\right)u_a+\frac{2m_b{u}_b}{m_a+m_b}$
$=\left(\frac{10-50}{10+50}\right)6+\frac{2\times 50\times 0}{10+50}=-4\text{m/s}$

The ratio of initial and final kinetic energies of block $A$ is given by,
$\frac{K.E_i}{K.E_f}=\frac{\frac{1}{2}m{u}_a^2}{\frac{1}{2}m{v}_a^2}=\frac{\frac{1}{2}m\times 6^2}{\frac{1}{2}m\times (-4{)}^2}=\frac{36}{16}=\frac{9}{4}$
Hence the correct option is (a).