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A particle of mass 1 kg has velocity $v_{1} = (2 t) \hat{i}$ and another particle of mass 2 kg has velocity $v_{2} = (t^{2}) \hat{j}$ . Match the following columns and mark the correct option from the codes given below.
Column I Column II
(A) Net force on centre of mass at 2 s (p) $\frac{20}{9}$ unit
(B) Velocity of centre of mass at 2 s (q) $\sqrt{68} unit$
(C) Displacement of centre of mass in 2s (r) $\sqrt{80} / 3 unit$
(s) None
Codes
A B C

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q p r

s q r

p r s

q r p

answer is A.

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Detailed Solution

$F_{CM} = F_{1} + F_{2} = m_{1} a_{1} + m_{2} a_{2} = (2 \hat{i} + 8 \hat{j})$

$∴ |F_{CM}| = \sqrt{4 + 64} \Rightarrow |F_{CM}| = \sqrt{68}$ unit

$v_{CM} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = \frac{(1) (4 \hat{i}) + (2) (4 \hat{j})}{3} = \frac{4 \hat{i} + 8 \hat{j}}{3}$

$∴ |v_{CM}| = \frac{1}{3} \sqrt{16 + 64} = \frac{\sqrt{80}}{3}$ unit

$s_{1} = \int_{0}^{2} v_{1} d t = (4 \hat{i})$

$s_{2} = \int_{0}^{2} v_{2} d t = (\frac{8}{3} \hat{j})$

Now, $s_{CM} = \frac{m_{1} s_{1} + m_{2} s_{2}}{m_{1} + m_{2}} = \frac{(1) (4 \hat{i}) + 2 (\frac{8}{3} \hat{j})}{3} = (\frac{4}{3} \hat{i} + \frac{16}{9} \hat{j})$

$∴ |s_{CM}| = \sqrt{\frac{16}{9} + \frac{256}{81}} = \frac{20}{9}$ unit

Hence, $A \to q, B \to r, C \to p .$

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