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Q.

A block of mass 0.50 kg is moving with a speed of $2.00 {ms}^{- 1}$ on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

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a

0.67 J

b

0.34 J

c

1.00 J

d

0.16 J

answer is D.

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Detailed Solution

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Applying the law of conservation of momentum,

$(1.00 + 0.50) ν = 0.50 \times 2.00 \Rightarrow ν = \frac{2}{3} m / s$

$Initial energy K_{i} = \frac{1}{2} \times 0.5 \times {(2.00)}^{2} = 1.00 J$

$Final energy K_{f} = \frac{1}{2} \times 1.50 \times \frac{4}{9} = 0.33 J Loss of energy = K_{i} - K_{f} = 0.67 J$

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is