Ball $A$ of mass $1 \mathrm{~kg}$ moving along a straight line with a velocity of $4 \mathrm{~ms}^{-1}$ hits…
Ball $A$ of mass $1 \mathrm{~kg}$ moving along a straight line with a velocity of $4 \mathrm{~ms}^{-1}$ hits another ball $B$ of mass $3 \mathrm{~kg}$ which is at rest. After collision, they stick together and move with the same velocity along the same straight line. If the time of impact of the collision is $0.1 \mathrm{~s}$ then the force exerted on $B$ is
$30 \mathrm{~N}$
$24 \mathrm{~N}$
$36 \mathrm{~N}$
$27 \mathrm{~N}$
Solution
For ball $A$,
$\begin{aligned} & m_A=1 \mathrm{~kg}, v_A=4 \mathrm{~ms}^{-1} \\ & \text { for ball } B, m_B=3 \mathrm{~kg}, v_B=0\end{aligned}$
$\therefore$ Total momentum before collision,
$\begin{aligned} p_i & =m_A v_A+m_B v_B \\ & =1 \times 4+3 \times 0=4 \mathrm{~kg}-\mathrm{ms}^{-1}\end{aligned}$
Since, after collision, both body stick together, therefore, it is the case of perfectly enelastic collision. Let $v$ be the common velocity, then total momentum after collision,
$\begin{aligned} p_f & =\left(m_A+m_B\right) v \\ & =(1+3) v=4 v\end{aligned}$
According to law of conservation of linear momentum,
$p_i=p_f$
$\Rightarrow \quad 4=4 v \Rightarrow v=1 \mathrm{~ms}^{-1}$
Time of impact, $\Delta t=0.1 \mathrm{~s}$
Force exerted on the body $B$,
$F_B=\frac{\Delta p_B}{\Delta t}=\frac{m_B\left(v_B-u_B\right)}{\Delta t}=\frac{3(v-0)}{01}$
$\Rightarrow \quad F_B=\frac{3(1-0)}{01}=\frac{3}{01}=30 \mathrm{~N}$