A 40 kg skateboarder (Ms) on a 3 kg board (Mb) is training with two 5 kg weights

Question

A 40 kg skateboarder (Ms) on a 3 kg board (Mb) is training with two 5 kg weights (Mw). Beginning at rest (with respect to the ground) he throws the weights together horizontally form the board. The velocity of each weight is 7 m/s relative to him and the board after they are thrown. Assume that the board rolls without friction. Assume the positive x direction is to the right.

A) what is the initial momentum? Why?

B) What is the velocity of the weights with respect to the ground if his velocity after throwing them is Vr (with respect to the ground). Just write it down.

C) Write out the final momentum of Ms, Mb, Mw, and Vf.

D) How fast is he propelled in the opposite direction after the weights are thrown?

Explanation / Answer

intial momentum when all were at rest = 0

so final momentum will also be zero

weight move towards right

boy moves towards left

velocity of boy = v

velocity of weight relative to boy = V weight - Vboy

= 7 - - v

= 7 + v

thus momentum after throwing = 10(7+v) - (40+ 3) x v

70 + 10v - 43v

= 70 - 33v

as from conservation of momentum this must be zero

thus

70 - 33v = 0

v = 2.12 m/s

final momentum

boy = 40 x 2.12 = 84.8 kgm/s

skate = 3 x 2.12 = 6.36 kgm/s

weight each = 5 x (7 + 2.12) = 45.6 kgm/s

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