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
To find the average force exerted by the wall on the ball, you correctly used the equation:
F = −2mv sin(θ)/∆t
where:
F is the force exerted by the wall on the ball,
m is the mass of the ball,
v is the velocity of the ball,
θ is the angle of incidence between the ball's velocity and the plane of the wall, and
∆t is the duration of contact between the ball and the wall.
Let's calculate the magnitude of the force using the given data:
m = 3.30 kg (mass of the ball)
v = 10.0 m/s (velocity of the ball)
θ = 60.0° (angle of incidence between the ball's velocity and the plane of the wall)
∆t = 0.204 s (duration of contact between the ball and the wall)
Plugging in these values, we get:
F = −2 * 3.30 kg * 10.0 m/s * sin(60.0°) / 0.204 s
Calculating this expression, we find:
F ≈ -285.7 N
The negative sign indicates that the force is directed in the opposite direction of the ball's initial velocity. The magnitude of the force, ignoring the negative sign, is approximately 285.7 N.
Please note that there might be rounding errors in the calculations, so it is always good to verify the solution with the correct formula and data.