A metal ball of mass $4 \mathrm{~kg}$ moving with a speed of $18 \mathrm{~km/h}$ has a head-on colli

Question

A metal ball of mass $4 \mathrm{~kg}$ moving with a speed of $18 \mathrm{~km/h}$ has a head-on collision with a stationary ball of mass $6 \mathrm{~kg}$. If after the collision both balls stick together and move as a single mass, then the loss in kinetic energy due to the collision is

Professor PC Thomas Classes · Answer

(c) : The speed of the metal ball is $v_1 = 18 \mathrm{~km/h} = 18 	imes \frac{5}{18} \mathrm{~m/s} = 5 \mathrm{~m/s}$. The collision is perfectly inelastic since they stick together. The formula for the loss in kinetic energy in a perfectly inelastic collision is $\Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} v_{rel}^2$. Here, $m_1 = 4 \mathrm{~kg}$, $m_2 = 6 \mathrm{~kg}$, and the relative approach velocity is $v_{rel} = 5 - 0 = 5 \mathrm{~m/s}$. Thus, $\Delta K = \frac{1}{2} 	imes \frac{4 	imes 6}{4 + 6} 	imes (5)^2 = \frac{1}{2} 	imes \frac{24}{10} 	imes 25 = 1.2 	imes 25 = 30 \mathrm{~J}$.