First slide

Connected bodies

Question

If 4 kg block is held after 2 s of start then how high 2 kg block will rise from the beginning before coming to rest momentarily?

Moderate

A

g/9 m
B

8g/9 m
C

g/3 m
D

2g/3 m

Solution

$\large a = \left( {\frac{{4 - 2}}{{4 + 2}}} \right)g = \frac{g}{3}$
velocity acquired by each block after 2 s is $v = \frac{2 g}{3}$
Height attained by 2 kg in 2 s,
$\large {h_1} = \frac{1}{2}a{t^2} = \frac{{2g}}{3}$
At the instant the 4 kg block is held, the 2 kg block has attained a velocity of 2g/3 and this block will continue to move upward with constant retardation 'g'. So the further rise of 2 kg block till it comes to rest is given by
h₂ = V² / 2g = (2g/3)² / 2g = 2g / 9.
The total height = h₁ + h₂ = 2g / 3 + 2g / 9
= 8g / 9

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