Question

A 1-kg ball moving at a velocity of 12m/s collides head-on with a 2-kg ball moving at 24 m/s in the opposite direction. Calculate the velocity of each ball after collision if (a) e=2/3, (b) the balls stick together and (c) the collision is perfectly elastic.

Answer 1

Mass of 1^st ball(m₁) = 1 kg

Velocity of 1^st ball($u_1$) = 12 m/s

Mass of 2^nd ball(m₂) = 2 kg

Velocity of 2^nd ball($u_2$) = -24m/s

Coefficient of restitution e = $\dfrac{2}{3}$

(a) Velocity of 1^st ball(v₁) = $\dfrac{(m_1-em_2)u_1+(1+e)m_2u_2}{m_1+m_2}=-28m/s$

Velocity of 2^nd ball(v₂) = $\dfrac{(m_2-em_1)u_2+(1+e)m_1u_1}{m_1+m_2}=-4m/s$

(b) If the balls stick with each other, they both move with the same final velocity(V) after collision

$V=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}=-12m/s$

(c) For perfectly elastic collision, e = 1,

$v_1=\dfrac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}=-36m/s$

$v_2=\dfrac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2}=0m/s$