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Standard XII

Physics

Com in Perfectly Inelastic Collision

A block of ma...

Question

A block of mass 1kg is pushed towards another block of mass 2kg from 6m distance as shown in figure. Just after collision velocity of 2 kg block becomes 4 m/s. coefficient of restitution between the blocks is

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Solution

The correct option is A 1
Final velocity of 2nd body after inelastic collision is
$v_{2} = \frac{(1 + e) m_{1} u_{1}}{m_{1} + m_{2}} ∵ u_{2} = 0$

$4 = \frac{(1 + e) \times 1 \times 6}{(1 + 2)}$

$\Rightarrow e = 1$

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A block of mass 1kg is pushed towards another block of mass 2kg from 6m distance as shown in figure. Just after collision velocity of 2 kg block becomes 4 m/s. coefficient of restitution between the blocks is

The correct option is A 1Final velocity of 2nd body after inelastic collision is v2=(1+e)m1u1m1+m2∵u2=0 4=(1+e)×1×6(1+2) ⇒e=1

The correct option is A 1
Final velocity of 2nd body after inelastic collision is
$v_{2} = \frac{(1 + e) m_{1} u_{1}}{m_{1} + m_{2}} ∵ u_{2} = 0$

$4 = \frac{(1 + e) \times 1 \times 6}{(1 + 2)}$

$\Rightarrow e = 1$