Question

In a snooker game, a player strikes the white ball and imparts a velocity of $4\text{m/s}$. The white ball strikes another red ball which was initially at rest. Assuming that the collision of the balls is perfectly elastic and masses of both balls are the same, what are the velocities of the red and white balls respectively?

• A. $0\text{m/s},4\text{m/s}$
• B. $0\text{m/s},2\text{m/s}$
• C. $2\text{m/s},0\text{m/s}$
• D. $4\text{m/s},0\text{m/s}$

Answer 1

The correct option is D $4\text{m/s},0\text{m/s}$

Given:
Initial velocity of white ball is $4\text{m/s}$ and red ball is $0\text{m/s}$.

From the concept of collision, we know,
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$(m_1\times 4)+(m_2\times 0)=m_1{v}_1+m_2{v}_2$
Divide above equation by $m_1$
$4=v_1+\frac{m_2}{m_1}{v}_2$
As $m_1=m_2$,
$v_1+v_2=4\dots \left(i\right)$

Also from conservation of kinetic energy,
$\frac{1}{2}{m}_1(4{)}^2+0=\frac{1}{2}{m}_1(v_1{)}^2+0+\frac{1}{2}{m}_2(v_2{)}^2+0\dots (ii)$

Substituting (i) in (ii) and solving:
we get $v_1=0\text{m/s}$ and $v_2=4\text{m/s}$

$\therefore$The final velocity of the red ball becomes $4\text{m/s}$ and that of the white ball becomes $0\text{m/s}$.