Question

Three balls A, B and C of masses 2 kg, 4 kg and 8 kg, respectively, move along the same straight line and in the same direction, with velocities 4 m/s, 1 m/s and 3/4 m/s. If A collides with B and subsequently B collides with C, find the velocity of ball A and ball B after collision, taking the coefficient of restitution as unity.

• A. $V_A=3,V_B=9/4$
• B. $V_A=0,V_B=3$
• C. $V_A=3,V_B=0$
• D. $V_A=0,V_B=0$

Answer 1

Answer: (D)

$V_A=0,V_B=0$

Given : $u_A=4$ $m/s$ $u_B=1$ $m/s$ $u_C=\frac{3}{4}$ $m/s$

$m_A=2kg$ $m_B=4kg$ $m_c=8kg$ $e=1$

Let velocity of A after collision be $V_A$.

Also let velocities of B after Ist collision be $V_B^{\prime }$ and after 2nd collision be $V_B$

For Ist collision : $V_A=\frac{(m_A-e{m}_B)u_A+(1+e)m_B{u}_B}{m_A+m_B}$

$V_A=\frac{(2-1\times 4)4+(1+1)4\times 1}{2+4}=0$ $m/s$

$V_B^{\prime }=\frac{(m_B-e{m}_A)u_B+(1+e)m_A{u}_A}{m_A+m_B}$

$V_B^{\prime }=\frac{(4-1\times 2)1+(1+1)2\times 4}{2+4}=3$ $m/s$

For 2nd collision : $V_B=\frac{(m_B-e{m}_c)V_B^{\prime }+(1+e)m_c{u}_c}{m_c+m_B}$

$V_B=\frac{(4-1\times 8)3+(1+1)8\times \frac{3}{4}}{8+4}=0$ $m/s$