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 94 kg. The mass of the rock is 0.27 kg. Initially the wagon is rolling forward at a speed of 0.49 m/s. Then the person throws the rock with a speed of 15 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.
the momentum of the system stays constant ___ 94 * .49
forward throw
(94 - .27)x + (.27 * 15) = 94 * .49
backward throw
(94 - .27)x - (.27 * 15) = 94 * .49
To solve this problem, we can start by using 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, assuming no external forces act on the system.
Let's denote the initial speed of the wagon as Vw, the final speed of the wagon after the rock is thrown forward as Vf (case a), and the final speed of the wagon after the rock is thrown backward as Vb (case b).
The initial momentum of the system (wagon, rider, and rock) is given by:
Initial momentum = (Total mass) × (Initial speed)
= (Mass of wagon + Mass of rider + Mass of rock) × (Initial speed)
= 94 kg × 0.49 m/s
= 46.06 kg·m/s
Now, let's consider the two cases:
(a) When the rock is thrown forward:
In this case, the momentum of the rock in the forward direction is given by:
Momentum of rock = (Mass of rock) × (Speed of rock)
= 0.27 kg × 15 m/s
= 4.05 kg·m/s
As there are no external forces acting on the system, the total momentum after the event is the sum of the momenta of the wagon and the rock. Therefore:
Total momentum = Final momentum of wagon + Momentum of rock
= (Mass of wagon + Mass of rider) × (Final speed of wagon) + Momentum of rock
= 94 kg × Vf + 4.05 kg·m/s
Since the total momentum before and after the event should be equal, we can write:
Total initial momentum = Total final momentum
46.06 kg·m/s = 94 kg × Vf + 4.05 kg·m/s
Simplifying this equation, we can solve for Vf:
Vf = (46.06 kg·m/s - 4.05 kg·m/s) / 94 kg
= 0.455 m/s
Therefore, the speed of the wagon after the rock is thrown forward (case a) is 0.455 m/s.
(b) When the rock is thrown backward:
In this case, the momentum of the rock in the backward direction is given by:
Momentum of rock = (Mass of rock) × (Speed of rock)
= 0.27 kg × (-15 m/s) (since the rock is thrown backward)
= -4.05 kg·m/s
Using the same principle of conservation of momentum, we can write:
Total initial momentum = Total final momentum
46.06 kg·m/s = 94 kg × Vb - 4.05 kg·m/s
Simplifying this equation, we can solve for Vb:
Vb = (46.06 kg·m/s + 4.05 kg·m/s) / 94 kg
= 0.527 m/s
Therefore, the speed of the wagon after the rock is thrown backward (case b) is 0.527 m/s.