In the figure shown, both the blocks are released from rest. The time (in s) to cross each other is

Question

In the figure shown, both the blocks are released from rest. The time (in s) to cross each other is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

4 g-T=4 a T-g=a 3 g=5 a ⇒ a=(3 g/5) Relative acceleration =(6 g/5) By second equation of motion, s=u t+(1/2) a t2 6=(1/2) × (6 g/5) × t2 ⇒ t2=1 ⇒ t=1 s (∵ u =0)

Q. In the figure shown, both the blocks are released from rest. The time (in $s$ ) to cross each other is

$4 g - T = 4 a$
$T - g = a$
$3 g = 5 a$
$\Rightarrow a = \frac{3 g}{5}$
Relative acceleration $= \frac{6 g}{5}$
By second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
$6 = \frac{1}{2} \times \frac{6 g}{5} \times t^{2}$
$\Rightarrow t^{2} = 1$
$\Rightarrow t = 1 s (∵ u = 0)$

Q. In the figure shown, both the blocks are released from rest. The time (in s) to cross each other is

4g−T=4a T−g=a 3g=5a ⇒a=53g​ Relative acceleration =56g​ By second equation of motion, s=ut+21​at2 6=21​×56g​×t2 ⇒t2=1 ⇒t=1s(∵u=0)

Q. In the figure shown, both the blocks are released from rest. The time (in $s$ ) to cross each other is

$4 g - T = 4 a$
$T - g = a$
$3 g = 5 a$
$\Rightarrow a = \frac{3 g}{5}$
Relative acceleration $= \frac{6 g}{5}$
By second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
$6 = \frac{1}{2} \times \frac{6 g}{5} \times t^{2}$
$\Rightarrow t^{2} = 1$
$\Rightarrow t = 1 s (∵ u = 0)$