Find final velocities in the reference frame to the ground.

[Sneakatone]

for the 1st two I used
v is to the reference frame = 37m/s- -35m/s=72m/s
a)ball=[(m2-m1)/(m1+m2)]v



bat=v2=[(2m1)/(m1+m2)]*v

b) I do not know how to solve for part b.

The Attempt at a Solution

 

[voko]

When you solved a), you obtained the velocity of the ball in the bat's reference frame by subtracting -35 m/s from the velocity of the ball in the ground frame. What do you need to do to obtain the velocity of the ball in the ground frame from its velocity in the bat's frame?

 

[Sneakatone]

v2 reference frame?

 

[voko]

What is your question exactly?

 

[Sneakatone]

do I need to find the reference frame for v2?

 

[voko]

The reference frame for v2 and everything else you computed is the reference frame of the bat just before the strike.

 

[Sneakatone]

so how would I find find final velocities in the reference frame to the ground?

 

[voko]

You have successfully converted from a ground frame to a moving frame. You did that by subtracting the velocity of the moving frame (relative to the ground frame) from the velocities of two objects (initially taken relatively to the ground frame) to obtain their velocities relatively to the moving frame. Now you have some velocities taken relatively to the moving frame, and you want to convert them to the ground frame. Given all of the above, is it not obvious how?

If not, consider this. Say there is a train going at 100 km/h (relatively to the ground). You are in the train and walking along it at 5 km/h. What is your velocity relatively the ground if you walking in the direction of the train's motion? Against?

 

[Sneakatone]

from your example given I feel like center of mass should be used for the velocities.

 

[voko]

What could the example possibly have to do with the center of mass?

Train at 100 km/h, you at 5 km/h in the train, your velocity re the ground?

 

[Sneakatone]

velocity would be 105 km/h.

 

[voko]

How did you find that?

 

[Sneakatone]

since the train and the person is moving a the same direction the reference frame is 95 km/h,
never mind the last post.

 

[voko]

You are not answering the question. The question was "how" did you get the number. But you just replied with another number. The first number was correct. Explain how and why you got that number.

 

[Sneakatone]

I got it from the persons reference speed to the ground so i added the speed of the train and person together.

 

[voko]

The speed of the person walking in the train was relative to the train. The question was "what is his speed relative to the ground".

 

[Sneakatone]

so is 105 km/h not the speed relative to the ground?

 

[voko]

It is, but do you understand why that is so? If the train's velocity w.r.t. the ground is V, and your velocity w.r.t. the train is v, what is your velocity w.r.t. the ground?

 

[Sneakatone]

V+v is the velocity to the ground

 

[voko]

Very well.

Now let's go back to the original problem. The frame initially co-moving with the bat is the train. What is V?

The velocities you got after the collision are relative to this frame, i.e., the "train". Can you find the velocities w.r.t. the ground?

 

[Sneakatone]

the bats velocity V=35 m/s

 

[voko]

No it is not. Signs matter.

 

[Sneakatone]

so the bats velocity would be V=-35m/s
making velocity to the ground -81.5m/s

 

[voko]

Making velocity of WHAT to the ground?

 

[Sneakatone]

making final velocity of the bat to the reference to the ground -81.5 m/s.

 

[voko]

So it was going -35 before the collision, and is going -81.5 - even faster and in the same direction - after the collision. How is that possible? What velocity do you get in the moving frame?

 

[Sneakatone]

37m/s- -35m/s=72m/s

 

[voko]

That is the velocity of the ball in the moving frame BEFORE the collision. What velocities in the moving frame did you get AFTER the collision?

 

[Sneakatone]

velocities after the frame
ball=-46.5
bat=25.4

 

[voko]

Excellent. You have these and you have the velocity of the moving frame w.r.t. the ground. What prevents you from applying the formula you got in #19?

 

[Sneakatone]

The velocity wrt the ground is already what we need.