A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is 97.0 kg. The mass of the rock is 0.310 kg. Initially the wagon is rolling forward at a speed of 0.510 m/s. Then the person throws the rock with a speed of 17.0 m/s. Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown directly forward. Find the speed of the wagon after the rock is thrown directly backward.
The Correct Answer and Explanation is:
To solve this problem, we’ll use the law of conservation of momentum, which states that the total momentum of the system (wagon + rock) before the rock is thrown must equal the total momentum after the rock is thrown. We will consider two cases: the rock being thrown forward and the rock being thrown backward.
Given Data:
- Total mass of wagon, rider, and rock: Mtotal=97.0 kgM_{\text{total}} = 97.0 \, \text{kg}Mtotal=97.0kg
- Mass of the rock: mrock=0.310 kgm_{\text{rock}} = 0.310 \, \text{kg}mrock=0.310kg
- Mass of the wagon, rider: Mwagon=Mtotal−mrock=97.0 kg−0.310 kg=96.69 kgM_{\text{wagon}} = M_{\text{total}} – m_{\text{rock}} = 97.0 \, \text{kg} – 0.310 \, \text{kg} = 96.69 \, \text{kg}Mwagon=Mtotal−mrock=97.0kg−0.310kg=96.69kg
- Initial speed of the wagon: vwagon initial=0.510 m/sv_{\text{wagon initial}} = 0.510 \, \text{m/s}vwagon initial=0.510m/s
- Speed of the rock when thrown:
- Forward: vrock=17.0 m/sv_{\text{rock}} = 17.0 \, \text{m/s}vrock=17.0m/s
- Backward: vrock=−17.0 m/sv_{\text{rock}} = -17.0 \, \text{m/s}vrock=−17.0m/s
Step 1: Momentum Conservation
The total momentum before the rock is thrown is the combined momentum of the wagon and rock moving at 0.510 m/s:Initial Momentum=(Mwagon+mrock)⋅vwagon initial=97.0 kg⋅0.510 m/s=49.47 kg⋅m/s\text{Initial Momentum} = (M_{\text{wagon}} + m_{\text{rock}}) \cdot v_{\text{wagon initial}} = 97.0 \, \text{kg} \cdot 0.510 \, \text{m/s} = 49.47 \, \text{kg} \cdot \text{m/s}Initial Momentum=(Mwagon+mrock)⋅vwagon initial=97.0kg⋅0.510m/s=49.47kg⋅m/s
After the rock is thrown, the momentum of the system should still be conserved. Therefore, the total momentum after the rock is thrown should equal the initial momentum.
Case 1: Rock is thrown forward
Let the new speed of the wagon be vwagon finalv_{\text{wagon final}}vwagon final. After the rock is thrown, the momentum is the sum of the momentum of the rock and the momentum of the wagon:Final Momentum=(Mwagon)⋅vwagon final+(mrock)⋅vrock\text{Final Momentum} = (M_{\text{wagon}}) \cdot v_{\text{wagon final}} + (m_{\text{rock}}) \cdot v_{\text{rock}}Final Momentum=(Mwagon)⋅vwagon final+(mrock)⋅vrock
Using conservation of momentum:49.47=96.69⋅vwagon final+0.310⋅17.049.47 = 96.69 \cdot v_{\text{wagon final}} + 0.310 \cdot 17.049.47=96.69⋅vwagon final+0.310⋅17.0
Simplifying:49.47=96.69⋅vwagon final+5.2749.47 = 96.69 \cdot v_{\text{wagon final}} + 5.2749.47=96.69⋅vwagon final+5.27
Solving for vwagon finalv_{\text{wagon final}}vwagon final:96.69⋅vwagon final=49.47−5.27=44.2096.69 \cdot v_{\text{wagon final}} = 49.47 – 5.27 = 44.2096.69⋅vwagon final=49.47−5.27=44.20vwagon final=44.2096.69=0.457 m/sv_{\text{wagon final}} = \frac{44.20}{96.69} = 0.457 \, \text{m/s}vwagon final=96.6944.20=0.457m/s
Thus, when the rock is thrown forward, the speed of the wagon decreases to 0.457 m/s.
Case 2: Rock is thrown backward
Similarly, when the rock is thrown backward, we have:Final Momentum=(Mwagon)⋅vwagon final+(mrock)⋅vrock\text{Final Momentum} = (M_{\text{wagon}}) \cdot v_{\text{wagon final}} + (m_{\text{rock}}) \cdot v_{\text{rock}}Final Momentum=(Mwagon)⋅vwagon final+(mrock)⋅vrock
Substitute values:49.47=96.69⋅vwagon final+0.310⋅(−17.0)49.47 = 96.69 \cdot v_{\text{wagon final}} + 0.310 \cdot (-17.0)49.47=96.69⋅vwagon final+0.310⋅(−17.0)
Simplifying:49.47=96.69⋅vwagon final−5.2749.47 = 96.69 \cdot v_{\text{wagon final}} – 5.2749.47=96.69⋅vwagon final−5.27
Solving for vwagon finalv_{\text{wagon final}}vwagon final:96.69⋅vwagon final=49.47+5.27=54.7496.69 \cdot v_{\text{wagon final}} = 49.47 + 5.27 = 54.7496.69⋅vwagon final=49.47+5.27=54.74vwagon final=54.7496.69=0.566 m/sv_{\text{wagon final}} = \frac{54.74}{96.69} = 0.566 \, \text{m/s}vwagon final=96.6954.74=0.566m/s
Thus, when the rock is thrown backward, the speed of the wagon increases to 0.566 m/s.
Conclusion:
- When the rock is thrown forward, the speed of the wagon decreases to 0.457 m/s.
- When the rock is thrown backward, the speed of the wagon increases to 0.566 m/s.
This problem demonstrates how momentum is conserved in a system where the internal forces (the person throwing the rock) do not affect the total momentum of the system.
