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

• A. $2:3$
• B. $5:3$
• C. $1:2$
• 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.