A 3.42-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. If the ball is in contact with the wall for 0.170 s, what is the average force exerted by the wall on the ball?
magnitude N
direction
LOL - That same old impulse thing again.
find change of momentum
= final m v - initial mv
(doing only component perpendicular to wall, v = 10 sin 60 inbound
v = - 10 sin 60 outbound
change in v = 2 * 10 sin 60
Force = change in momentum/time
It only acted perpendicular to the wall. We know that because the velocity component parallel to the wall did not change.
To find the average force exerted by the wall on the ball, we can use the impulse-momentum principle. According to this principle, the change in momentum of an object is equal to the impulse applied to it.
The momentum of an object is given by the product of its mass and velocity: p = m * v
The impulse applied to an object can be calculated as the change in momentum: J = Δp = m * Δv
In this case, since the ball bounces off the wall with the same speed and angle, the change in velocity (Δv) is equal to 2 times the initial velocity (v). Therefore, Δv = 2 * v
The impulse can also be calculated as the average force (F) multiplied by the time of contact (Δt): J = F * Δt
Setting these two equations equal, we can find the average force: F * Δt = Δp
Now we can substitute the values given in the problem:
Mass of the ball (m) = 3.42 kg
Initial velocity of the ball (v) = 10.0 m/s
Time of contact (Δt) = 0.170 s
First, let's calculate the change in velocity (Δv):
Δv = 2 * v = 2 * 10.0 m/s = 20.0 m/s
Then, we can calculate the change in momentum (Δp):
Δp = m * Δv = 3.42 kg * 20.0 m/s = 68.4 kg·m/s
Finally, we can calculate the average force (F):
F * Δt = Δp
F = Δp / Δt = 68.4 kg·m/s / 0.170 s = 402.35 N
Therefore, the average force exerted by the wall on the ball is 402.35 N.