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A 45.0-kg girl stands on a 13.0-kg wagon holding two 18.0-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 6.5 m/s relative to herself . Assuming that the wagon was at rest initially, what is the speed of the girl relative to the ground after she throws both weights at the same time

Sagot :

cjrodriguez89

Answer:

v = 4.0 m/s

Explanation:

  • Assuming no external forces acting during the instant that the girl throws the weights, total momentum must be conserved.
  • Since all the masses at rest initially, the initial momentum must be zero.
  • So, due to momentum must keep constant, final momentum must be zero too, as follows:

       [tex]p_{f} = m_{w} * v_{w} + m_{g+w} *v_{g+w} = 0 (1)[/tex]

  • Assuming the direction towards the back of the wagon as positive, and replacing the masses in (1), we can solve for vg, as follows:

       [tex]v_{g+w} =- \frac{m_{w} *v_{w}}{m_{g+w} } = - \frac{36.0kg *6.5m/s}{58.0kg } = -4.0m/s (2)[/tex]

  • This means that the girl (along with the wagon on she is standing) will move at a speed of 4.0 m/s in an opposite direction to the one she threw the weights.
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