How Fast is the Wagon Going After Collision?
What happens when a basketball collides with a wagon?
The basketball hits the back of a wagon and bounces off, sending the wagon off in the original direction of the ball's travel.
Answer:
The wagon is now moving at a speed of approximately 0.086 meters per second in the original direction of the ball's travel.
When the basketball collides with the wagon, the principle of conservation of momentum comes into play. This principle states that the total momentum of an isolated system remains constant if no external forces act on it.
In this case, the system consists of the basketball and the wagon. Before the collision, the total momentum is the sum of the momentum of the basketball and the momentum of the wagon.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v), i.e., p = m * v.
Momentum of basketball = 1.2 kg * 7.5 m/s = 9 kg m/s
Momentum of wagon = 12 kg * 0 m/s = 0 kg m/s (since it's stationary)
Total initial momentum = 9 kg m/s + 0 kg m/s = 9 kg m/s
After the collision, the basketball bounces off the wagon, and the wagon gains momentum in the direction of the basketball's original travel.
Let's denote the final velocity of the wagon as "v_w" and the final velocity of the basketball as "v_b".
Momentum of basketball = 1.2 kg * 3.8 m/s = 4.56 kg m/s (in the opposite direction)
Momentum of wagon = 12 kg * v_w
Total final momentum = 4.56 kg m/s + 12 kg * v_w
According to the conservation of momentum principle, the total initial momentum equals the total final momentum:
9 kg m/s = 4.56 kg m/s + 12 kg * v_w
Solving for v_w:
12 kg * v_w = 9 kg m/s - 4.56 kg m/s
v_w = (9 kg m/s - 4.56 kg m/s) / 12 kg
v_w ≈ 0.086 m/s