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Standard XII

Physics

Work Energy Theorem Application

A bullet of m...

Question

A bullet of mass $m_{1}$ is fired into a block mass $m_{2}$ suspended like a pendulum. The block, initially at rest, is raised to a height h, after the impact. The velocity of the bullet is

(1 + \frac{m_{1}}{m_{2}}) \sqrt{g h}

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\frac{m_{1}}{m_{2}} \sqrt{2 g h}

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(1 + \frac{m_{1}}{m_{2}}) \sqrt{2 g h}

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(1 + \frac{m_{2}}{m_{1}}) \sqrt{2 g h}

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Solution

The correct option is D $(1 + \frac{m_{2}}{m_{1}}) \sqrt{2 g h}$
Final velocity of bullet and the block $= \sqrt{2 g h}$
If initial velocity of the bullet is u, then
$m_{1} u + 0 = (m_{1} + m_{2}) \sqrt{2 g h}$
$\Rightarrow u = (1 + \frac{m_{2}}{m_{1}}) \sqrt{2 g h}$

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A bullet of mass m1 is fired into a block mass m2 suspended like a pendulum. The block, initially at rest, is raised to a height h, after the impact. The velocity of the bullet is

The correct option is D (1+m2m1)√2ghFinal velocity of bullet and the block =√2ghIf initial velocity of the bullet is u, thenm1u+0=(m1+m2)√2gh⇒u=(1+m2m1)√2gh

A bullet of mass $m_{1}$ is fired into a block mass $m_{2}$ suspended like a pendulum. The block, initially at rest, is raised to a height h, after the impact. The velocity of the bullet is

The correct option is D $(1 + \frac{m_{2}}{m_{1}}) \sqrt{2 g h}$
Final velocity of bullet and the block $= \sqrt{2 g h}$
If initial velocity of the bullet is u, then
$m_{1} u + 0 = (m_{1} + m_{2}) \sqrt{2 g h}$
$\Rightarrow u = (1 + \frac{m_{2}}{m_{1}}) \sqrt{2 g h}$