A 44-kg girl stands on a 7-kg wagon holding two 15.0-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 9.5 m/s relative to herself. Assuming that the wagon was at rest initially, what is the speed of the girl after she throws both weights at the same time?
To find the speed of the girl after she throws both weights, we can apply the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, provided there are no external forces acting on the system.
Before the girl throws the weights, the total momentum of the system is zero since both the girl and the wagon are initially at rest. After she throws the weights, the momentum of the girl and the wagon will still be zero because the girl's forward momentum is canceled out by the backward momentum of the thrown weights.
However, in order to find the speed of the girl, we need to consider the mass of the girl, the wagon, and the thrown weights. Let's denote the speed of the girl after throwing the weights as Vg.
The total momentum of the system can be calculated using the equation:
Total momentum before = Total momentum after
Since the initial momentum is zero, we can write:
0 = (Mass of the girl + Mass of the wagon) * Vg
Substituting the given masses:
0 = (44 kg + 7 kg) * Vg
0 = 51 kg * Vg
Solving for Vg:
Vg = 0 / 51 kg
Thus, the speed of the girl after throwing both weights is zero m/s.