Question

A ball of mass 'm' moving with a velocity 'v' undergoes an oblique elastic collision with another ball of the same mass 'm' but at rest. After the collision, if the two balls move with different speeds after the collision, the angle between their directions of motion will be

• A. Less than
• B. More than
• C. Exactly
• D. Exactly

Answer 1

The correct option is C Exactly

Let $v_1$ and $v_2$ be the speeds of the balls after the collision. Then Equation (ii) above gives
$v_{1}sin{\theta }_1=v_{2}sin{\theta }_2$ (v)
Also, conservation of kinetic energy gives $\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$
or $u^2=v_1^2+v_2^2$ (vi)
Also Equation (i) above becomes
$mu=m{v}_{1}cos{\theta }_1+m{v}_{2}cos{\theta }_2$
or $v_{2}cos{\theta }_2=u-v_{1}cos{\theta }_1$ (vii)
Squaring (v) and (vii) and adding we get $v_2^2=u^2+v_1^2-2u{v}_{1}cos{\theta }_1$ (viii)
Eliminating $v_2$ between (vi) and (vii), we get $cos{\theta }_1=\frac{v_2}{u}$ (ix)
Eliminating $v_2$ between (v) and (vii) we get $tan{\theta }_2=\frac{sin{\theta }_1}{(\frac{u}{v_1}-cos{\theta }_1)}$ (x)
Using (ix) in (x), we get
$tan{\theta }_2=\frac{sin{\theta }_1}{(\frac{1}{cos{\theta }_1}-cos{\theta }_1)}=\frac{sin{\theta }_{1}cos{\theta }_1}{(1-co{s}^2{\theta }_1)}$ =$cot{\theta }_1=tan({90}^{\circ }-{\theta }_1)$
or ${\theta }_2={90}^{\circ }-{\theta }_{1}or{\theta }_1+{\theta }_2={90}^{\circ }$.
Hence the correct choice is (c).