Question

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?

• A. 12.5 m/s
• B. 25.0 m/s
• C. 35.7 m/s
• D. 83.3 m/s
• E. 100 m/s

Answer 1

Answer: (B)

25.0 m/s

A constant force acts on the luge over a period of time. Hence the momentum attained by the luge=$F\Delta t$

$=500N\times 5s=2500kg.m/s$

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=$(M+m)v=2500$

$⟹v=\frac{2500}{100}=25m/s$