A 3.30-kg steel ball strikes a massive wall at 10.0 m/s at an angle of θ = 60.0° with the plane of the wall. It bounces off the wall with the same speed and angle (see the figure below). If the ball is in contact with the wall for 0.204 s, what is the average force exerted by the wall on the ball? find magnitude?

for magnitude i got -280.2 N which is incorrect.
i used the equation
F = −2mv sin(θ)/∆t

so can anyone help me!!!!!please

What subject is "p"? Phys Ed, Physics, Playground?

I'm a current physics student with a similar problem. However, this equation worked for me, I just removed the negative from the front.

To solve this problem, you are on the right track using the equation for average force:

F = -2mv sin(θ) / ∆t

where:
- F is the average force exerted by the wall on the ball
- m is the mass of the ball (3.30 kg)
- v is the initial speed of the ball (10.0 m/s)
- θ is the angle between the direction of motion and the plane of the wall (60.0°)
- ∆t is the time interval during which the ball is in contact with the wall (0.204 s)

However, there might be a mistake in the equation you used. The equation you wrote is missing a negative sign before the entire equation.

Here's the corrected calculation:

F = -2mv sin(θ) / ∆t
= -2(3.30 kg)(10.0 m/s) sin(60.0°) / 0.204 s

Using the given values, we can now calculate the average force:

F = -2(3.30 kg)(10.0 m/s) sin(60.0°) / 0.204 s
≈ -2(3.30)(10.0)(0.866) / 0.204 s
≈ -2(28.59) / 0.204 s
≈ -57.18 N / 0.204 s
≈ -280.88 N

Therefore, the magnitude of the average force exerted by the wall on the ball is approximately 280.88 N. The negative sign indicates that the force is acting in the opposite direction to the initial motion of the ball.