Bullet Lodges In Baseball (vertical)

[Schoomy]

Homework Statement

A 0.028 kg bullet is fired vertically at 185 m/s into a 0.15 kg baseball that is initially at rest. The bullet lodges in the baseball and, after the collision, the baseball/ bullet rise to a height of 37 m.

(a) What was the speed of the baseball/bullet right after the collision?

(b) What was the average force of air resistance while the baseball/bullet was rising? (Consider the positive direction to be upwards.)

Homework Equations

conservation of momentum.

The Attempt at a Solution

m1v1=m2v2 yeilding v2 (velocity of ball/bullet combo) of 29.1 m/s.
This was correct.

However, I can't figure out b.

I tried using Force of Air Resistance = KE - PE, but that doesn't work.

I also tried using Force Air Resitance =KE/h - m(total)-g

That didn't work either (found it on yahoo answers).

What am I missing here?

 

[LowlyPion]

A 0.028 kg bullet is fired vertically at 185 m/s into a 0.15 kg baseball that is initially at rest. The bullet lodges in the baseball and, after the collision, the baseball/ bullet rise to a height of 37 m.

(a) What was the speed of the baseball/bullet right after the collision?

(b) What was the average force of air resistance while the baseball/bullet was rising? (Consider the positive direction to be upwards.)

Homework Equations

conservation of momentum.

The Attempt at a Solution

m1v1=m2v2 yeilding v2 (velocity of ball/bullet combo) of 29.1 m/s.
This was correct.

However, I can't figure out b.

I tried using Force of Air Resistance = KE - PE, but that doesn't work.

I also tried using Force Air Resitance =KE/h - m(total)-g

That didn't work either (found it on yahoo answers).

What am I missing here?

I think you need to consider that KE = PE_final + W_air

Where

W_air = AvgForce_air * distance.

 

[Schoomy]

Yes, but I still don't get the right answer even then:

KE = PE_final + W_air

Thus, W_air = KE - PE

specifically:
Wair = ((.5)(.178)(29.1^2)) - ((.178)(9.8)(37)) = 75.366-64.543 = 10.823

Then,

W_air = AvgForce_air * distance

becomes:

AvgForce_air = w_air / distance

AvgForce_air = 10.823 / 37 = 0.293 (which isn't accepted)

Thoughts?

 

[LowlyPion]

Yes, but I still don't get the right answer even then:

KE = PE_final + W_air

Thus, W_air = KE - PE
specifically:
Wair = ((.5)(.178)(29.1^2)) - ((.178)(9.8)(37)) = 75.366-64.543 = 10.823

Then,

W_air = AvgForce_air * distance

becomes:

AvgForce_air = w_air / distance

AvgForce_air = 10.823 / 37 = 0.293 (which isn't accepted)

Thoughts?

Look at it another way.

Your initial velocity is 29.1.

Figure acceleration against the upward motion.

V² = 2*a*x

a = 29.1²/(2*37) = 11.44 m/s²

Well you know g was 9.8 so you apparently had an additional force that affected a 1.64 m/s² downward drag.

1.64*.178 = ... oops ... looks like .293N again.

 

[Schoomy]

So I am still confused?

 

[LowlyPion]

So I am still confused?

Or maybe .293 is correct?

 

[Schoomy]

http://dl.getdropbox.com/u/119186/Picture%201.png

 

[Schoomy]

Webassign sucks...this is for practice and it gives no feedback as to what we may be doing wrong...

 

[LowlyPion]

http://dl.getdropbox.com/u/119186/Picture%201.png

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They want -.293 I think. The force is against the direction of motion which is up.