Answer Check: A golf ball is hit and hits a bee. find the work done on the bee

[joe426]

Homework Statement

Homework Equations

F=ma
W= Fdcosθ

The Attempt at a Solution

I first found the acceleration:
t= d/v
t= 37m/28m/s = 1.32s
a = v/t
a = 28m/s / 1.32s
a= 21.21 m/ss

Second, the force:
F_club - F_gsinθ = ma
F_club = a +gsinθ
F_club = 21.21m/ss + 9.8m/ss sin84
F_club = 30.96N

Then I plugged in:
W=(30.96N)(37m)cos84
W=119.75J

I'm pretty sure this is correct. only thing I'm worried about really is the acceleration.

 

[frogjg2003]

Using energy considerations

You have to use potential and kinetic energy conservation to solve this problem.

 

[haruspex]

I first found the acceleration:
t= d/v
t= 37m/28m/s = 1.32s
a = v/t
a = 28m/s / 1.32s
a= 21.21 m/ss

That calculation is completely wrong. 28m/s is the initial speed of the ball, and it is at some angle to the vertical. The flight time to reach the bee would be 1.32s if the ball were traveling straight up at 28m/s and not slowed by gravity along the way.
What acceleration do you think this is calculating?

Second, the force:
F_club - F_gsinθ = ma
F_club = a +gsinθ
F_club = 21.21m/ss + 9.8m/ss sin84
F_club = 30.96N

It is impossible to calculate the force from the club since you do not know how long the impact lasted. You can calculate the impulse (momentum), but that doesn't help with this question.

 

[joe426]

You have to use potential and kinetic energy conservation to solve this problem.

So, since we can ignore resistance, the energy is conserved and the potential energy at the beginning equals the kinetic energy at the end.

PE_i = KE_f
mgy = 1/2mv²
v = sqrt(2gy)
v = sqrt(2 * 9.8m/s² * 37m)
v= 26.93m/s

 

[haruspex]

So, since we can ignore resistance, the energy is conserved and the potential energy at the beginning equals the kinetic energy at the end.

PE_i = KE_f

This isn't a stone being dropped from rest. Total energy is conserved here means PE_i + KE_i= PE_f + KE_f
What is KE_i?

 

[joe426]

This isn't a stone being dropped from rest. Total energy is conserved here means PE_i + KE_i= PE_f + KE_f
What is KE_i?

KE_i = 0 because the initial velocity is 0. And would make 1/2mv² = 0

And PE_f = 0 because y is the vertical distance to move the object. we don't need to move the object because its already at its finaly position of 37m. If wrong on this and I should use 37m for y and solve for v, I end up with 0

 

[frogjg2003]

KE_i = 0 because the initial velocity is 0. And would make 1/2mv² = 0

The initial kinetic energy is not zero. The ball has some initial speed, namely 28 m/s.

And PE_f = 0 because y is the vertical distance to move the object. we don't need to move the object because its already at its finaly position of 37m. If wrong on this and I should use 37m for y and solve for v, I end up with 0

It's usually more convenient to have the starting, not final position have position and potential energy 0. This means that the final position is at y=37m and a potential energy of mgy.

You know the initial potential and kinetic energies, and the final potential energy. The only thing you don't know is the final kinetic energy. Solve for this and then determine the speed.

 

[joe426]

The initial kinetic energy is not zero. The ball has some initial speed, namely 28 m/s.


It's usually more convenient to have the starting, not final position have position and potential energy 0. This means that the final position is at y=37m and a potential energy of mgy.

You know the initial potential and kinetic energies, and the final potential energy. The only thing you don't know is the final kinetic energy. Solve for this and then determine the speed.

Ok, I understand this and I've come up with:
1/2mv_i² = mgy_f + 1/2mv_f²
v_f = sqrt [ v_i² - 2(gy_f) ]
v_f = 7.67m/s

 

[frogjg2003]

There you go. You've got it.