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Standard XII

Physics

Partially Inelastic Collision

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Question

From the top of a tower of the height $100 m$ , a $10 g m$ block is dropped freely, and a $6 g m$ bullet is fired vertically upwards from the foot of the tower with velocity $100 m s^{- 1}$ simultaneously. They collide and stick together. The common velocity after collision is $(g = 10 m s^{- 2})$

27.5 m s^{- 1}

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150 m s^{- 1}

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40 m s^{- 1}

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100 m s^{- 1}

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Solution

The correct option is B $27.5 m s^{- 1}$
Their relative velocity is $v_{r} = 100 m / s$ and relative acceleration $a_{r} = 0$
So they collide at $t$ seconds later then
$s$ = $v_{r} t + \frac{1}{2} a_{r} t^{2}$
$100 = 100 t + 0$
$t = 1 s e c$
After 1 sec
$V_{b l o c k} = g t$
$V_{b l o c k} = 10$
$V_{b u l l e t} = 100 - g t$
$V_{b u l l e t} = 90$
Using conservation of momentum
$m_{b l o c k} V_{b l o c k} + m_{b u l l e t} V_{b u l l e t} = m_{c o m b i n e d} V_{c o m b i n e d}$
$.01 x 10 - .006 x 90 = .016 x V_{c o m b i n e d}$
$V_{c o m b i n e d} = - 27.5$
Hence common velocity after collision will be $27.5 m / s$ upwards.

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From the top of a tower of the height 100m, a 10gm block is dropped freely, and a 6gm bullet is fired vertically upwards from the foot of the tower with velocity 100ms−1 simultaneously. They collide and stick together. The common velocity after collision is (g=10ms−2)

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