finding average force?

Question

A tennis ball of mass 0.0614 kg is served. It strikes the ground with a velocity of 52.3 m/s (117 mi/h) at an angle of 23.5° below the horizontal. Just after the bounce it is moving at 48.3 m/s at an angle of 17.6° above the horizontal. If the interaction with the ground lasts 0.0731 s, what average force did the ground exert on the ball?

Answer 1

The force exerted on the ball over a given time period (impulse) on the ball changes the ball’s momentum.

Momentum = mass * velocity

We are told that the ball has a mass of .0614 kg,
m = .0614 kg

Initially, the ball is traveling at 52.3 m/s at an angle of 23.5° to the horizontal.
V = 52.3 m/s at an angle of 23.5° to the horizontal
We can translate this velocity into its X and Y components to be,
V_x = (52.3 m/s) * cos (23.5°) = 47.96 m/s in the horizontal direction, and
V_y = (52.3 m/s) * sin (23.5°) = -20.85 m/s in the upward direction (= 20.85 m/s in downward direction).

After hitting the ground, we are told that the ball is now traveling at a velocity of 48.3 m/s at an angle of 17.6° above the horizontal.
V = 48.3 m/s at an angle of 17.6° above the horizontal
Or, translating this into X and Y components,
V_x = (48.3 m/s) * cos (17.6°) = 46.04 m/s in the horizontal direction, and
V_x = (48.3 m/s) * sin (17.6°) = +14.60 m/s in the upward direction

ΔV_x = V_x_final – V_x_initial
ΔV_x =(46.04 m/s) – (47.96 m/s) = -1.92 m/s

ΔV_y = V_y_final – V_y_initial
ΔV_y =(+14.60 m/s) – (-20.85 m/s) = +35.45 m/s


To find the change in momentum,
Δp = m * ΔV
Since the mass is constant.

We can find the change in momentum is both the X and Y directions.
ΔP_x = m * ΔV_x
ΔP_y = m * ΔV_y

ΔP_x = (.0614 kg) * (-1.92 m/s) = –.118 kg m/s
ΔP_y = (.0614 kg) * (+35.45 m/s) = +2.18 kg m/s

This change in momentum in each direction is equal to the total impulse exerted on the ball in each direction by the force applied to the ball over the given time period.
We are told that the ball is in contact with the ground for a time of .0731 seconds,
t = .0731 s

Impulse = Force * time = Δp
Force = Δp / t

So we can now find the force exerted on the ball in both the X and the Y direction,
F_x = ΔP_x / t
F_y = ΔP_y / t

F_x = (–.118 kg m/s) / (.0731 s) = -1.61 Newtons
F_y = +2.18 kg m/s) / (.0731 s) = 29.8 Newtons

We now know the average force acting on the ball in the horizontal and vertical directions; we can combine these two vectors to get the overall force acting on the ball.

F = sqrt (F_x^2 + F_y^2)
F = sqrt ((-1.61 N)^2 + (29.8 N)^2) = sqrt (890.6 N^2)
F = 29.84 Newtons
This is just the magnitude; however, we still need to find the direction the force is acting.

We know that the tangent of the angle which the force is acting at is equal to F_y / F_x,
Tan (θ) = F_y / F_x
We can solve for θ by taking the arctangent of both sides,
θ = arctan (F_y / F_x) = tan^-1 (F_y / F_x)
θ = arctan (29.8 N / -1.61 N) = arctan (-18.5)
θ = 3.1° to the left right of the normal vector to the ground (the ‘vertical’), or 93.1° to the usual horizontal.

If we check this by multiplying the magnitude of the force by the cosine and sine of the angle we just found (with respect to the 0° horizontal), we should get the forces applied in the X and Y direction.
F_x = F * cos (93.1°) = -1.61 N
F_y = F * sin (93.1°) = 29.8 N
Which is what we found earlier.

It is important that the force in the X direction be negative (since the ball is slowing down in that direction) and the force in the Y direction be large and positive (since the ball was originally traveling downward but had to reverse directions as it bounced).

Answer 2

By definition
momentum p=mv
m - mass
v velocity

Force equal to the change of momentum with respect to time
or mass times the change of velocity with respect to time. The force averaged over time F is

F=m dv/dt=m(v1-v2)/(t1-t2)

F=0.0614 (52.3 -48.3 )/ 0.0731=
F=3.36 N