A 36-kg girl stands on a 13-kg wagon holding two 15.0-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 3.0 m/s relative to herself. Assuming that the wagon was at rest initially, what is the speed with which the girl will move after she throws the weights one at a time, each with a speed of 3.0 m/s relative to herself?
According to the law of conservation of momentum, the total momentum before and after the weights are thrown should be the same.
The initial momentum of the system can be calculated by adding the momentum of the girl, wagon, and weights before any of the weights are thrown.
The momentum of the girl is given by:
P_girl = m_girl * v_girl
P_girl = 36 kg * 0 m/s
P_girl = 0 kg·m/s
The momentum of the wagon is given by:
P_wagon = m_wagon * v_wagon
P_wagon = 13 kg * 0 m/s
P_wagon = 0 kg·m/s
The momentum of the weights is given by:
P_weights = m_weights * v_weights
P_weights = 15 kg * 3 m/s
P_weights = 45 kg·m/s
The initial total momentum of the system is therefore:
P_initial = P_girl + P_wagon + 2 * P_weights
P_initial = 0 kg·m/s + 0 kg·m/s + 2 * 45 kg·m/s
P_initial = 90 kg·m/s
After each weight is thrown, the momentum of the system will change. The momentum change depends on the mass and velocity of the weight.
When the first weight is thrown, the momentum change is given by:
ΔP_1 = m_weights * (-v_weights)
ΔP_1 = 15 kg * (-3 m/s)
ΔP_1 = -45 kg·m/s
When the second weight is thrown, the momentum change is given by:
ΔP_2 = m_weights * (-v_weights)
ΔP_2 = 15 kg * (-3 m/s)
ΔP_2 = -45 kg·m/s
Since the momentum change of the system must be equal and opposite to the momentum change of the weight, the final momentum of the system can be calculated:
P_final = P_initial + ΔP_1 + ΔP_2
P_final = 90 kg·m/s - 45 kg·m/s - 45 kg·m/s
P_final = 0 kg·m/s
Since the initial momentum of the system was zero, and the final momentum of the system is also zero, the speed of the girl after throwing the weights will also be zero. Therefore, the girl will not move after she throws the weights.