An athlete of mass 70.0 kg applies a force of 500 N to a 30.0 kg luge, which is initially at rest, over a period of 5.00 s before jumping onto the luge. Assuming there is no friction between the luge and the track on which it runs, what is its velocity after the athlete jumps on?

12.5 m/s

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25.0 m/s

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35.7 m/s

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83.3 m/s

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100 m/s

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Solution

The correct option is D 25.0 m/s
A constant force acts on the luge over a period of time. Hence the momentum attained by the luge= $F Δ t$
$= 500 N \times 5 s = 2500 k g . m / s$
When the man jumps on the luge, no external force acts on the system, hence momenutm remains conserved,.
Hence the momentum of system after man jumps on the luge= $(M + m) v = 2500$
$⟹ v = \frac{2500}{100} = 25 m / s$

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