A ball of mass $m$ moving with a velocity of $6 \mathrm{~m/s}$ collides head-on with another station

Question

A ball of mass $m$ moving with a velocity of $6 \mathrm{~m/s}$ collides head-on with another stationary ball of double the mass ($2m$). If the coefficient of restitution is $1/3$, then the velocities of the first and second balls respectively after the collision (in $\mathrm{m/s}$) will be

Professor PC Thomas Classes · Answer

(c) : Let the velocities after collision be $v_1$ and $v_2$. By conservation of linear momentum: $m(6) + 2m(0) = m v_1 + 2m v_2 \implies v_1 + 2v_2 = 6$. From the coefficient of restitution definition: $e = \frac{v_2 - v_1}{u_1 - u_2} \implies \frac{1}{3} = \frac{v_2 - v_1}{6} \implies v_2 - v_1 = 2 \implies v_1 = v_2 - 2$. Substituting this into the momentum equation gives $(v_2 - 2) + 2v_2 = 6 \implies 3v_2 = 8 \implies v_2 = 8/3 \mathrm{~m/s}$. Then, $v_1 = 8/3 - 2 = 2/3 \mathrm{~m/s}$.