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 92.5 kg. The mass of the rock is 0.310 kg. Initially the wagon is rolling forward at a speed of 0.550 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.
Answer in m/s.
Find the speed of the wagon after the rock is thrown directly backward.
Answer in m/s.
m1+m2+m3 = 92.5 kg.
m1= 0.31 kg
m2+m3 =92.19 kg
v=0.55 m/s
v1= - 17 m/s
v2=?
(m1+m2+m3) •v= - m1•v1+(m2+m3) •v2
v2= {(m1+m2+m3) •v +m1•v1}/(m2+m3)=…
To solve this problem, we will use the principle of conservation of momentum. According to this principle, the total momentum before and after an event remains the same, provided there are no external forces acting on the system.
Let's denote the initial velocity of the wagon as Vw, the final velocity of the wagon after the rock is thrown forward as Vwf, and the final velocity of the wagon after the rock is thrown backward as Vwb.
1. Find the initial momentum:
The initial momentum of the system (wagon + rider + rock) is given by:
P_initial = (mass_wagon + mass_rider + mass_rock) * Vw
P_initial = (92.5 kg) * (0.550 m/s)
P_initial = 50.875 kg·m/s
2. Find the final momentum when the rock is thrown forward:
The final momentum of the system (wagon + rider + rock) after the rock is thrown forward is given by:
P_final_forward = (mass_wagon + mass_rider) * Vwf + mass_rock * rock_velocity_forward
3. Find the final momentum when the rock is thrown backward:
The final momentum of the system (wagon + rider + rock) after the rock is thrown backward is given by:
P_final_backward = (mass_wagon + mass_rider) * Vwb - mass_rock * rock_velocity_backward
Since momentum is conserved, we can equate the initial momentum to the final momentum in both cases:
P_initial = P_final_forward and P_initial = P_final_backward
Substituting the values we know, we get:
50.875 kg·m/s = (92.5 kg) * Vwf + (0.310 kg) * 17.0 m/s
50.875 kg·m/s = (92.5 kg) * Vwb - (0.310 kg) * 17.0 m/s
Simplifying, we can solve for Vwf and Vwb:
Vwf = (50.875 kg·m/s - (0.310 kg) * 17.0 m/s) / (92.5 kg)
Vwb = (50.875 kg·m/s + (0.310 kg) * 17.0 m/s) / (92.5 kg)
Vwf = 0.539 m/s (approximately)
Vwb = 0.563 m/s (approximately)
Therefore, the speed of the wagon after the rock is thrown directly forward is approximately 0.539 m/s, and the speed of the wagon after the rock is thrown directly backward is approximately 0.563 m/s.