Two blocks A and B of masses 1.5 kg and 0.5 | vedclass

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(B) The common acceleration of the system is given by:$a = \left( \frac{m_A - m_B}{m_A + m_B} \right) g = \left( \frac{1.5 - 0.5}{1.5 + 0.5} \right) g

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) The common acceleration of the system is given by:$a = \left( \frac{m_A - m_B}{m_A + m_B} \right) g = \left( \frac{1.5 - 0.5}{1.5 + 0.5} \right) g = \frac{1}{2} g = 5 \ m/s^2$.When block $A$ is released from a height of $80 \ cm$ $(0.8 \ m)$,it moves downwards with acceleration $a = 5 \ m/s^2$. The velocity $v$ of the blocks when block $A$ hits the ground is given by $v^2 = u^2 + 2as$:$v^2 = 0 + 2(5)(0.8) = 8 \ (m/s)^2$.At this instant,block $B$ is at a height of $80 \ cm$ from the ground and has an upward velocity $v = \sqrt{8} \ m/s$.After block $A$ hits the ground,the string becomes slack,and block $B$ moves under gravity with deceleration $g = 10 \ m/s^2$.Using the third equation of motion for block $B$ to find the additional height $h$ it rises:$v_f^2 = v^2 - 2gh$$0 = 8 - 2(10)h$$h = \frac{8}{20} = 0.4 \ m = 40 \ cm$.The maximum height reached by block $B$ from the ground is the initial height plus the additional height:$H_{max} = 80 \ cm + 40 \ cm = 120 \ cm$.

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