Centre of Mass 3 Question 1
1. A man (mass $=50 kg$ ) and his son (mass $=20 kg$ ) are standing on a frictionless surface facing each other. The man pushes his son, so that he starts moving at a speed of $0.70 ms^{-1}$ with respect to the man. The speed of the man with respect to the surface is
(2019 Main, 12 April I)
(a) $0.28 ms^{-1}$
(b) $0.20 ms^{-1}$
(c) $0.47 ms^{-1}$
(d) $0.14 ms^{-1}$
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Answer:
Correct Answer: 1. (b)
Solution:
- The given situation can be shown as below
Using momentum conservation law,
$=(\text { Total momentum }) _{\text {after collision }}$
$$ \begin{gathered} \left(m _1 \times 0\right)+\left(m _2 \times 0\right)=m _1 \mathbf{v} _1+m _2 \mathbf{v} _2 \\ 0=m _1\left(-v _1\right) \hat{\mathbf{i}}+m _2 v _2 \hat{\mathbf{i}} \end{gathered} $$
$$ \begin{aligned} \Rightarrow & m _1 v _1 & =m _2 v _2 \\ \Rightarrow & 50 v _1 & =20 v _2 \\ \Rightarrow & v _2 & =2.5 v _1 \end{aligned} $$
Again, relative velocity $=0.70 m / s$
But from figure, relative velocity $=v _1+v _2$
$$ \therefore \quad v _1+v _2=0.7 $$
From Eqs. (i) and (ii), we get
$$ v _1+2.5 v _1=0.7 $$
$$ \Rightarrow \quad v _1(3.5)=0.7 $$
$$ v _1=\frac{0.7}{3.5}=0.20 m / s $$
