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A bullet of mass 0.25 kg is fired with v...

A bullet of mass 0.25 kg is fired with velocity 302m/s into a block of wood of mass 37.5kg.It gets embedded into it. The block `m_(1)` is resting on a long block and the horizontal surface on which it is placed. The coefficient of friction between `m_(1)` and `m_(2)` is 0.5. Find the displacement of on and the common velocity of `m_(1)` and `m_(2)`. Mass `m^(2)`= 12.5kg.

Text Solution

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The correct Answer is:
`[0.2m,1.51 m//s]`
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  • A bullet of mass m moving with velocity v strikes a block of mass M at rest and gets embedded into it. The kinetic energy of the composite block will be

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    `(1)/(2) mv^(2) xx (m)/((m + M))`
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