Two masses m1 and m2(m2 > m1) are positioned as shown in figure, m1 being on the ground and m2 at a height h above the ground. When m2 is released, the speed at which it hits the ground will be

Since, a=m2−m1m2+m1g and v2−u2=2a s ⇒v2−02=2m2−m1m2+m1gh⇒v=2m2−m1m2+m1gh

Two masses m1 and m2(m2 > m1) are positioned as shown in figure, m1 being on the ground and m2 at a height h above the ground. When m2 is released, the speed at which it hits the ground will be

Since, a=m2−m1m2+m1g and v2−u2=2a s ⇒v2−02=2m2−m1m2+m1gh⇒v=2m2−m1m2+m1gh

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Two masses m₁ and m₂(m₂ > m₁) are positioned as shown in figure, m₁ being on the ground and m₂ at a height h above the ground. When m₂ is released, the speed at which it hits the ground will be

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$\sqrt{\frac{2 {ghm}_{1}}{m_{2}}}$

$\sqrt{\frac{2 gh (m_{1} - m_{2})}{(m_{1} + m_{2})}}$

$\sqrt{\frac{2 gh (m_{1} + m_{2})}{(m_{1} - m_{2})}}$

$\sqrt{\frac{2 gh (m_{2} - m_{1})}{(m_{1} + m_{2})}}$

answer is D.

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$Since, a = (\frac{m_{2} - m_{1}}{m_{2} + m_{1}}) g and v^{2} - u^{2} = 2 a s$

$\begin{array}{l} \Rightarrow v^{2} - 0^{2} = 2 (\frac{m_{2} - m_{1}}{m_{2} + m_{1}}) gh \\ \Rightarrow v = \sqrt{2 (\frac{m_{2} - m_{1}}{m_{2} + m_{1}}) gh} \end{array}$

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Two masses m1 and m2(m2 > m1) are positioned as shown in figure, m1 being on the ground and m2 at a height h above the ground. When m2 is released, the speed at which it hits the ground will be

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Two masses m₁ and m₂(m₂ > m₁) are positioned as shown in figure, m₁ being on the ground and m₂ at a height h above the ground. When m₂ is released, the speed at which it hits the ground will be