Question:
A 0.2 kg ball is kicked with an initial velocity of 8 m/s to the right. It collides with a stationary wall and rebounds in the opposite direction with a velocity of 4 m/s to the left. The collision takes 0.02 seconds.
Answer:
To calculate the change in momentum, we use the equation:
Δp = m * Δv
Where Δp is the change in momentum, m is the mass of the ball, and Δv is the change in velocity.
Given that m = 0.2 kg, Δv = 4 m/s - 8 m/s = -4 m/s, we can substitute these values into the equation:
Δp = (0.2 kg) * (-4 m/s) = -0.8 kg·m/s
Therefore, the change in momentum of the ball during the collision is -0.8 kg·m/s.
The impulse experienced by an object is given by the equation:
J = F * Δt
Where J is the impulse, F is the average force acting on the object, and Δt is the time interval over which the force is applied.
We are given the time interval, Δt = 0.02 s. We can substitute this value and the calculated change in momentum into the equation to solve for the average force:
-0.8 kg·m/s = F * (0.02 s)
F = (-0.8 kg·m/s) / (0.02 s) = -40 N
Therefore, the impulse experienced by the ball during the collision is -40 N·s.
The average force exerted on the ball during the collision can be calculated by using the equation:
F = Δp / Δt
Where F is the average force, Δp is the change in momentum, and Δt is the time interval.
We have already calculated the change in momentum as -0.8 kg·m/s and we are given the time interval Δt = 0.02 s. Substituting these values into the equation we get:
F = (-0.8 kg·m/s) / (0.02 s) = -40 N
Therefore, the average force exerted on the ball during the collision is -40 N.
Note: The negative sign indicates the direction of the force, which is opposite to the direction of motion (leftward in this case).