Question

In a two block system shown in figure, match the following.



$\begin{array}{cc}\text{Table -1}& \text{Table -2}\\ \text{(A) Velocity of centre of}& \text{(P) Keep on changing all}\\ \text{mass}& \text{the time}\\ \text{(B) Momentum of centre of}& \text{(Q) First decreases then}\\ \text{mass}& \text{becomes zero}\\ \text{(C) Momentum of 1 kg block with respect to 2 kg block}& \text{(R) zero}\\ & \text{(S) constant}\end{array}$

• A. $\text{A}\to \text{P},\text{B}\to \text{P, Q},\text{C}\to \text{Q}$
• B. $\text{A}\to \text{P},\text{B}\to \text{R, S},\text{C}\to \text{Q}$
• C. $\text{A}\to \text{R, S},\text{B}\to \text{R, S},\text{C}\to \text{R}$
• D. $\text{A}\to \text{R, S},\text{B}\to \text{R, S},\text{C}\to \text{Q}$

Answer 1

The correct option is D $\text{A}\to \text{R, S},\text{B}\to \text{R, S},\text{C}\to \text{Q}$

Velocity of centre of mass

$\left(V_{cm}\right)=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$

From the date given in the figure,

$V_{cm}=\frac{1\times 10+2\times (-5)}{3}=0\text{m/s}$
$\therefore$ Momentum of centre of mass $\left(P_{cm}\right)=m{V}_{cm}=0\text{kg m/s}$

Net force on the system is zero, hence $V_{cm},P_{cm}$ will remain constant.

Relative velocity of $1\text{kg}$ block with respect to $2\text{kg}$ block is $5\text{m/s}.$

Since friction always opposes the relative motion, velocity of $1\text{kg}$ block decreases and becomes zero.

Thus, momentum of $1\text{kg}$ block decreases and becomes zero.

Option (d) gives correct matching.