Question

Three ball 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 $4m/s,1m/s$ and $3/4m/s$. 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)

$V_A=0,V_B=0$

Initial momentum $\left(p_i\right)=m_A\left(4\right)+m_B\left(1\right)=12kgm/s$

Final momentum $\left(p_f\right)=m_A{v}_A+m_B{v}_B$

Now coefficient of restitution $\left(e\right)=\frac{v_B-v_A}{3}=1$

Applying linear momentum conservation,

$p_i=p_f\Rightarrow 12=2\left(v_A\right)+4(3+v_A)$

$\Rightarrow v_A=0\Rightarrow v_B=3m/s$

$2^{nd}$ Collision:-

$p_i=4\left(3\right)+8\left(\frac{3}{4}\right)=18kgm/s$

$e=1=\frac{v_C-v_B}{9/4}\Rightarrow v_C=v_B+9/4$

$p_f=4\left(v_B\right)+8(v_B+9/4)=12v_B+18$

Now $p_i=p_f$

$\Rightarrow 18=12v_B+18\Rightarrow v_B=0$ and $v_A=0$ (D)