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 kg. The mass of the rock is 0.34 kg. Initially the wagon is rolling forward at a speed of 0.50 m/s. Then the person throws the rock with a speed of 17 m/s. Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown (a) directly forward in one case and (b) directly backward in another.
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To solve this problem, 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.
Let's denote the initial speed of the wagon as V_initial and the final speed of the wagon as V_final.
(a) When the rock is thrown directly forward:
The initial momentum of the system is given by:
P_initial = (mass of wagon + mass of rider + mass of rock) × V_initial
= (95 kg) × (0.50 m/s)
After the rock is thrown, the rock generates an equal and opposite momentum in the backward direction. Therefore, the final momentum of the system is given by:
P_final = (mass of wagon + mass of rider) × V_final
Since momentum is conserved, we can equate P_initial and P_final:
P_initial = P_final
(95 kg) × (0.50 m/s) = (mass of wagon + mass of rider) × V_final
Solving for V_final:
V_final = (95 kg × 0.50 m/s) / (mass of wagon + mass of rider)
= (47.5 kg m/s) / (mass of wagon + mass of rider)
(b) When the rock is thrown directly backward:
Similarly, the initial momentum of the system is given by:
P_initial = (95 kg) × (0.50 m/s)
After the rock is thrown, the rock generates an equal and opposite momentum in the forward direction. Therefore, the final momentum of the system is given by:
P_final = (mass of wagon + mass of rider) × V_final
Again, equating P_initial and P_final:
P_initial = P_final
(95 kg) × (0.50 m/s) = (mass of wagon + mass of rider) × V_final
Solving for V_final:
V_final = (95 kg × 0.50 m/s) / (mass of wagon + mass of rider)
= (47.5 kg m/s) / (mass of wagon + mass of rider)
Thus, in both cases, the speed of the wagon after the rock is thrown (a) directly forward and (b) directly backward is given by (47.5 kg m/s) / (mass of wagon + mass of rider).
To find the speed of the wagon after the rock is thrown, we can apply the law of conservation of momentum. According to this law, the total momentum before the rock is thrown should be equal to the total momentum after the rock is thrown.
(a) When the rock is thrown directly forward:
Let's assume the velocity of the wagon after the rock is thrown is v1 m/s.
Before the rock is thrown:
The initial momentum of the system is the momentum of the wagon (mass of wagon * velocity of wagon) plus the momentum of the rider (mass of rider * velocity of wagon).
Momentum before = (mass of wagon + mass of rider) * velocity of wagon
After the rock is thrown:
The final momentum of the system is the momentum of the wagon (mass of wagon * velocity of wagon) plus the momentum of the rider (mass of rider * velocity of wagon) plus the momentum of the rock (mass of rock * velocity of rock).
Momentum after = (mass of wagon + mass of rider + mass of rock) * v1
According to the conservation of momentum:
Momentum before = Momentum after
Therefore, (mass of wagon + mass of rider) * velocity of wagon = (mass of wagon + mass of rider + mass of rock) * v1
Plugging in the given values:
(95 kg) * (0.50 m/s) = (95 kg + 0.34 kg) * v1
We can solve this equation to find the value of v1. By substituting the values and solving for v1, we can find the speed of the wagon after the rock is thrown directly forward.
(b) When the rock is thrown directly backward:
Let's assume the velocity of the wagon after the rock is thrown is v2 m/s.
Using the same logic as above, we can set up another equation for momentum conservation:
(Momentum before) = (Momentum after)
Therefore, (mass of wagon + mass of rider) * velocity of wagon = (mass of wagon + mass of rider + mass of rock) * v2
By substituting the given values and solving for v2, we can find the speed of the wagon after the rock is thrown directly backward.