AR U. answered • 10/20/19

Experienced Physics and Math Tutor [Edit]

This a conservation of momentum problem! Which states that momentum before collision equals momentum after collision.

Known quantities:

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mass of boy, m_{b} = 79kg

initial velocity of boy (before jumping onto skateboard) , v_{bi} =1.55 m/s

final velocity (boy + skateboard), v = 1.41 m/s

mass of skateboard, m_{s} = ?

a) Assuming no friction, the weight of the skateboard, W = m_{s}g [ where g = 9.81m/s^{2}, is acceleration due to gravity]

b) From conservation of momentum

m_{b}v_{bi} + m_{s}v_{si} = (m_{b} + m_{s})v, but v_{si} =0 (since skateboard is stationary)

==> (m_{b}v_{bi} -m_{b}v)/v = m_{s} = ((79kg)(1.55m/s) - (79kg)(1.41m/s))/1.41 m/s = 7.8kg

the mass of skateboard is 7.8kg

c) Recalling your Newton's third law, we can say that the momentum of the boy,p_{b} is equal but opposite the momentum of the skateboard, p_{s} when the boy falls backward with velocity,v_{bf }= 1.0m/s

===> p_{b} = -p_{s,}

But p_{b} = m_{b}v_{bf }= (79 kg)(1.0m/s) = 79kgm/s = -p_{s } = m_{s}v_{sf}

Solving for the velocity of the stakeboard, v_{sf} = (- 79kgm/s)/7.8kg = -10 m/s

NB: We do not need the negative sign because the skateboard moves forward!