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 find the constant force exerted on the ball by the wall, we can use Newton's second law of motion, which states that the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a).

In this case, the ball strikes the wall and bounces off, so its velocity changes. The change in velocity can be calculated using the formula:

Δv = v2 - v1

Where Δv is the change in velocity, v2 is the final velocity, and v1 is the initial velocity.

Let's calculate the change in velocity:
Δv = 6.7 m/s - (-7.9 m/s)
Δv = 6.7 m/s + 7.9 m/s
Δv = 14.6 m/s

Since the time of contact between the ball and the wall is given as 0.21 s, we can calculate the acceleration using the formula:

a = Δv / t

Where a is the acceleration, Δv is the change in velocity, and t is the time.

Let's calculate the acceleration:
a = 14.6 m/s / 0.21 s
a ≈ 69.52 m/s^2

Now that we know the acceleration, we can find the force using the formula:

F = m * a

Where F is the force, m is the mass of the ball, and a is the acceleration.

Given that the mass of the ball is 1.5 kg, let's calculate the force:
F = 1.5 kg * 69.52 m/s^2
F ≈ 104.28 N

Therefore, the constant force exerted on the ball by the wall is approximately 104.28 Newtons.