A 275-gram ball bounces off a wall. The ball stays in a horizontal plane throughout the motion. The ball strikes the wall with a speed of 4.50 m/s at an angle of 𝜃 = 35.0o relative to the wall and rebounds with the same speed at the same angle. If the ball is in contact with the wall for 8.00 ms, calculate the average force exerted on the wall by the ball. (Determine both the magnitude and direction of this force.)
change in x momentum = 2 * mass * x velocity
= 2 * .275 * 4.5 sin 35
force on ball by wall = change in momentum /time
= change in momentum/.008
wall on ball is -x so ball on wall is + x direction
To calculate the average force exerted on the wall by the ball, we can use the impulse-momentum principle. The impulse-momentum principle states that the change in momentum of an object is equal to the impulse applied to it, where impulse is defined as the product of force and time.
First, let's calculate the initial and final momentum of the ball:
Initial momentum (p₁):
Since momentum is defined as the product of mass and velocity, we can calculate the initial momentum using the given values:
p₁ = m * v
= 0.275 kg * 4.50 m/s
Final momentum (p₂):
The ball rebounds with the same speed and angle, so the final momentum will have the same magnitude but opposite direction:
p₂ = m * v
= 0.275 kg * 4.50 m/s
Next, we need to calculate the change in momentum (Δp):
Δp = p₂ - p₁
Now, we can use the impulse-momentum principle to calculate the average force (Favg):
Impulse = Force * Time
Δp = Favg * Δt
Rearranging the equation, we can solve for the average force:
Favg = Δp / Δt
Now let's plug in the values and calculate the average force exerted on the wall:
Δp = p₂ - p₁
Δp = (0.275 kg * 4.50 m/s) - (0.275 kg * 4.50 m/s)
Δt = 8.00 ms
Δt = 8.00 × 10^(-3) s
Favg = Δp / Δt
Now you can calculate the average force exerted on the wall by the ball.