A 1.5 kg ball strikes a wall with a velocity of 7.9 m/s to the left. The ball bounces off with a velocity of 6.7 m/s to the right. If the ball is in contact with the wall for 0.21 s, what is the constant force exerted on the ball by the wall?
To determine the constant force exerted on the ball by the wall, you can use Newton's second law of motion, which states that the force applied to an object is equal to the rate of change of its momentum.
First, let's calculate the change in momentum of the ball:
Δp = m * Δv
Where Δp is the change in momentum, m is the mass of the ball, and Δv is the change in velocity.
By using the given information, we can find the change in momentum:
Initial velocity (u) = -7.9 m/s (to the left)
Final velocity (v) = 6.7 m/s (to the right)
Time (t) = 0.21 s
Mass (m) = 1.5 kg
Δv = v - u = 6.7 m/s - (-7.9 m/s) = 6.7 m/s + 7.9 m/s = 14.6 m/s
Δp = m * Δv = 1.5 kg * 14.6 m/s = 21.9 kg * m/s
Next, we can use the formula for force:
F = Δp / t
Where F is the force exerted on the ball and t is the time of contact.
Plugging in the values:
F = 21.9 kg * m/s / 0.21 s
Calculating the force:
F = 104.29 N
Therefore, the constant force exerted on the ball by the wall is approximately 104.29 Newtons.