Centre of Mass 3 Question 4
4. A body of mass $2 kg$ makes an elastic collision with a second body at rest and continues to move in the original direction but with one-fourth of its original speed. What is the mass of the second body?
(2019 Main, 9 April I)
(a) $1.5 kg$
(b) $1.2 kg$
(c) $1.8 kg$
(d) $1.0 kg$
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Answer:
Correct Answer: 4. (b)
Solution:
Key Idea For an elastic collision, coefficient of restitution (e), i.e. the ratio of relative velocity of separation after collision to the relative velocity of approach before collision is 1 .
Given, mass of small body, $m=2 kg$
Given situation is as shown
Using momentum conservation law for the given system, (Total momentum)
(Total momentum $) _{\text {after collision }}$
$$ \Rightarrow \quad m(v)+M(0)=m \frac{v}{4}+M\left(v^{\prime}\right) \ldots(i) $$
$\because e=1$ and we know that,
$$ \begin{aligned} & e & =-\frac{v _2-v _1}{u _2-u _1} \\ \Rightarrow \quad & 1 & =-\frac{v^{\prime}-v / 4}{0-v} \\ \Rightarrow & v & =v^{\prime}-v / 4 \\ \text { or } & v^{\prime} & =5 v / 4 \end{aligned} $$
Using value from Eq. (ii) into Eq. (i), we get
$$ \begin{aligned} & m v=\frac{m v}{4}+M \frac{5 v}{4} \\ & \Rightarrow \quad m v-\frac{v}{4}=M \frac{5 v}{4} \\ & \Rightarrow \quad \frac{3}{4} m v=\frac{5}{4} M v \\ & M=\frac{3}{5} m=\frac{3}{5} \times 2=1.2 kg \end{aligned} $$
Key Idea Total linear momentum is conserved in all collisions, i.e. the initial momentum of the system is equal to final momentum of the system.
Given,
$m _2=0.5 m _1 \quad \Rightarrow \quad m _1=2 m _2$
Let $m _2=m$, then, $m _1=2 m$
Also, $v _3=0.5 v _1$
Given situation of collinear collision is as shown below Before collision,
After collision,
$\therefore$ According to the conservation of linear momentum, Initial momentum $=$ Final momentum
$$ \begin{aligned} & m _1 v _1 \hat{\mathbf{i}}+m _2 v _2 \hat{\mathbf{i}}=m _1 v _3 \hat{\mathbf{i}}+m _2 v _4 \hat{\mathbf{i}} \\ & \Rightarrow 2 m v _1 \hat{\mathbf{i}}+m v _2 \hat{\mathbf{i}}=2 m\left(0.5 v _1\right) \hat{\mathbf{i}}+m v _4 \hat{\mathbf{i}} \\ & \Rightarrow v _4=v _1+v _2 \Rightarrow v _1=v _4-v _2 \end{aligned} $$
