A boy m = 79 kg travels at a velocity vi = 1.55 m/s before jumping on a skateboard. After jumping on the board the boy has a velocity vf = 1.41 m/s.

Question

A boy m = 79 kg travels at a velocity v_i = 1.55 m/s before jumping on a skateboard. After jumping on the board the boy has a velocity v_f = 1.41 m/s.

Part (a)  Write an expression for the weight of the skateboard W. 

Part (b)  What is the mass of the skateboard in kilograms? 

Part (c)  The boy soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in this process, what is the skateboard's new velocity in meters per second? [Note: this may not be a very good assumption, as there can be significant friction in the ball bearings of the skateboard] 

Answer 1

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_sg [ where g = 9.81m/s², is acceleration due to gravity]

b) From conservation of momentum

m_bv_bi + m_sv_si = (m_b + m_s)v, but v_si =0 (since skateboard is stationary)

==> (m_bv_bi -m_bv)/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_bv_bf = (79 kg)(1.0m/s) = 79kgm/s = -p_s = m_sv_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!