A 40 kg boy is running at a speed of 3.0 m/s, when he jumps on a stationary skateboard. If the system, consisting of the boy and the skateboard, begins to roll at a speed of 2.5 m/s, what is the mass of the skateboard?

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A 40 kg boy is running at a speed of 3.0 m/s, when he jumps on a stationary skateboard. If the system, consisting of the boy and the skateboard, begins to roll at a speed of 2.5 m/s, what is the mass of the skateboard?

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Answer 1 · 2017-01-20T17:44:32

8 kg

Explanation:

Momentum(P)=mass(m) times velocity(v)

The total momentum of a system will not change (P initial = P final), so momentum of this system is:
$m (b o y) \cdot v (b o y) + m (b o a r d) \cdot v (b o a r d) = [m (b o y) + m (b o a r d)] \cdot v (b o y and b o a r d)$ $m(boy) * v(boy) + m(board) * v(board) = [m(boy) + m(board)] * v(boy and board)$

$40 k g \cdot 3.0 \frac{m}{s} + m (b o a r d) \cdot 0 \frac{m}{s} = [40 k g + m (b o a r d)] \cdot 2.5 \frac{m}{s}$ $40 kg*3.0m/s +m(board)*0m/s=[40 kg+m(board)]*2.5 m/s$

$120 k g \frac{m}{s} = [40 k g + m (b o a r d)] \cdot 2.5 \frac{m}{s}$ $120 kgm/s=[40 kg+m(board)]*2.5 m/s$

$\frac{120 k g \frac{m}{s}}{2.5} \frac{m}{s} = [40 k g + m (b o a r d)]$ $(120kgm/s)/2.5 m/s=[40 kg+m(board)]$

$48 k g = [40 k g + m (b o a r d)]$ $48kg=[40 kg+m(board)]$

$m (b o a r d) = 48 k g - 40 k g = 8 k g$ $m(board)=48kg-40kg=8kg$

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A 40 kg boy is running at a speed of 3.0 m/s, when he jumps on a stationary skateboard. If the system, consisting of the boy and the skateboard, begins to roll at a speed of 2.5 m/s, what is the mass of the skateboard?

1 Answer

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