Question

Two blocks of masses \(3m\) and \(2m\) are in contact on a smooth table. A force \(P\) is first applied horizontally on block of mass \(3m\) and then on mass \(2m\). The contact forces between the two blocks in the two cases are in the ratio



3m _

• A. $2:3$
• B. $1:2$
• C. $5:3$
• D. $3:2$

Answer 1

The correct option is A $2:3$

Case $\left(1\right)$: Force $P$ is applied on $3m$ block,

$P-N_1=3ma...\left(1\right)$
$N_1=2ma...\left(2\right)$
equation $\left(1\right)+\left(2\right)$
$P=5ma$
$\Rightarrow a=\frac{P}{5m}$

$\therefore N_1=2m\times \frac{P}{5m}$
$\Rightarrow N_1=\frac{2P}{5}$

Case $\left(2\right)$: Force $P$ is applied on $2m$ block,


$P-N_2=2ma....\left(3\right)$
$N_2=3ma....\left(4\right)$
equation $\left(3\right)+\left(4\right)$
$P=5ma\Rightarrow a=\frac{P}{5m}$
$N_2=\frac{3mP}{5m}\Rightarrow N_2=\frac{3P}{5}$
Now the required ratio,

$\frac{N_1}{N_2}=\frac{2P/5}{3P/5}=\frac{2}{3}$

Hence, option $\left(B\right)$ is correct.