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 95.0
. The mass of the rock is 0.300
. Initially the wagon is rolling forward at a speed of 0.500
. Then the person throws the rock with a speed of 16.0
. Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown directly forward in one case and directly backward in another.
The Correct Answer and Explanation is:
This problem involves the principle of conservation of momentum. Since friction is negligible, there are no external forces acting on the system, and the momentum of the system (wagon, rider, and rock) must be conserved. The total momentum before and after the rock is thrown must be equal.
Case 1: Rock thrown directly forward
- Given data:
- Total mass of the wagon, rider, and rock: Mtotal=95.0 kgM_{\text{total}} = 95.0 \, \text{kg}Mtotal=95.0kg
- Mass of the rock: mrock=0.300 kgm_{\text{rock}} = 0.300 \, \text{kg}mrock=0.300kg
- Initial velocity of the wagon: vwagon, initial=0.500 m/sv_{\text{wagon, initial}} = 0.500 \, \text{m/s}vwagon, initial=0.500m/s
- Speed of the rock after being thrown: vrock, thrown=16.0 m/sv_{\text{rock, thrown}} = 16.0 \, \text{m/s}vrock, thrown=16.0m/s (forward, relative to the ground)
- Before the rock is thrown:
The total momentum of the system is the sum of the momentum of the wagon (which includes the rider and rock) and the rock itself: ptotal, initial=(Mtotal)vwagon, initial=95.0 kg×0.500 m/s=47.5 kg⋅m/sp_{\text{total, initial}} = (M_{\text{total}}) v_{\text{wagon, initial}} = 95.0 \, \text{kg} \times 0.500 \, \text{m/s} = 47.5 \, \text{kg} \cdot \text{m/s}ptotal, initial=(Mtotal)vwagon, initial=95.0kg×0.500m/s=47.5kg⋅m/s - After the rock is thrown:
After the rock is thrown, the total momentum must be conserved. The momentum is divided between the wagon and the rock: ptotal, final=(Mwagon)vwagon, final+(mrock)vrock, finalp_{\text{total, final}} = (M_{\text{wagon}}) v_{\text{wagon, final}} + (m_{\text{rock}}) v_{\text{rock, final}}ptotal, final=(Mwagon)vwagon, final+(mrock)vrock, final where Mwagon=Mtotal−mrock=95.0 kg−0.300 kg=94.7 kgM_{\text{wagon}} = M_{\text{total}} – m_{\text{rock}} = 95.0 \, \text{kg} – 0.300 \, \text{kg} = 94.7 \, \text{kg}Mwagon=Mtotal−mrock=95.0kg−0.300kg=94.7kg. Substituting the known values: 47.5 kg⋅m/s=94.7 kg×vwagon, final+0.300 kg×16.0 m/s47.5 \, \text{kg} \cdot \text{m/s} = 94.7 \, \text{kg} \times v_{\text{wagon, final}} + 0.300 \, \text{kg} \times 16.0 \, \text{m/s}47.5kg⋅m/s=94.7kg×vwagon, final+0.300kg×16.0m/s Simplifying the right side: 47.5=94.7vwagon, final+4.847.5 = 94.7 v_{\text{wagon, final}} + 4.847.5=94.7vwagon, final+4.8 Solving for vwagon, finalv_{\text{wagon, final}}vwagon, final: 47.5−4.8=94.7vwagon, final47.5 – 4.8 = 94.7 v_{\text{wagon, final}}47.5−4.8=94.7vwagon, final 42.7=94.7vwagon, final42.7 = 94.7 v_{\text{wagon, final}}42.7=94.7vwagon, final vwagon, final=42.794.7=0.451 m/sv_{\text{wagon, final}} = \frac{42.7}{94.7} = 0.451 \, \text{m/s}vwagon, final=94.742.7=0.451m/s
Thus, after the rock is thrown forward, the speed of the wagon will be 0.451 m/s0.451 \, \text{m/s}0.451m/s.
Case 2: Rock thrown directly backward
- In this case:
The rock is thrown in the opposite direction (backward), so its velocity relative to the ground is negative. - After the rock is thrown backward:
Using the same conservation of momentum principle, the equation becomes: 47.5 kg⋅m/s=94.7 kg×vwagon, final+0.300 kg×(−16.0 m/s)47.5 \, \text{kg} \cdot \text{m/s} = 94.7 \, \text{kg} \times v_{\text{wagon, final}} + 0.300 \, \text{kg} \times (-16.0 \, \text{m/s})47.5kg⋅m/s=94.7kg×vwagon, final+0.300kg×(−16.0m/s) Simplifying: 47.5=94.7vwagon, final−4.847.5 = 94.7 v_{\text{wagon, final}} – 4.847.5=94.7vwagon, final−4.8 Solving for vwagon, finalv_{\text{wagon, final}}vwagon, final: 47.5+4.8=94.7vwagon, final47.5 + 4.8 = 94.7 v_{\text{wagon, final}}47.5+4.8=94.7vwagon, final 52.3=94.7vwagon, final52.3 = 94.7 v_{\text{wagon, final}}52.3=94.7vwagon, final vwagon, final=52.394.7=0.552 m/sv_{\text{wagon, final}} = \frac{52.3}{94.7} = 0.552 \, \text{m/s}vwagon, final=94.752.3=0.552m/s
Thus, after the rock is thrown backward, the speed of the wagon will be 0.552 m/s0.552 \, \text{m/s}0.552m/s.
Final Results:
- If the rock is thrown forward: The wagon’s speed decreases to 0.451 m/s0.451 \, \text{m/s}0.451m/s.
- If the rock is thrown backward: The wagon’s speed increases to 0.552 m/s0.552 \, \text{m/s}0.552m/s.
