Find final velocities in the reference frame to the ground.

In summary: For part b), I do not know how to solve for it.In summary, the ball was going -35 before the collision, and is going -81.5 after the collision.
  • #1
Sneakatone
318
0
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

 

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  • #2
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?
 
  • #3
v2 reference frame?
 
  • #4
What is your question exactly?
 
  • #5
do I need to find the reference frame for v2?
 
  • #6
The reference frame for v2 and everything else you computed is the reference frame of the bat just before the strike.
 
  • #7
so how would I find find final velocities in the reference frame to the ground?
 
  • #8
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?
 
  • #9
from your example given I feel like center of mass should be used for the velocities.
 
  • #10
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?
 
  • #11
velocity would be 105 km/h.
 
  • #12
How did you find that?
 
  • #13
since the train and the person is moving a the same direction the reference frame is 95 km/h,
never mind the last post.
 
  • #14
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.
 
  • #15
I got it from the persons reference speed to the ground so i added the speed of the train and person together.
 
  • #16
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".
 
  • #17
so is 105 km/h not the speed relative to the ground?
 
  • #18
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?
 
  • #19
V+v is the velocity to the ground
 
  • #20
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?
 
  • #21
the bats velocity V=35 m/s
 
  • #22
No it is not. Signs matter.
 
  • #23
so the bats velocity would be V=-35m/s
making velocity to the ground -81.5m/s
 
  • #24
Making velocity of WHAT to the ground?
 
  • #25
making final velocity of the bat to the reference to the ground -81.5 m/s.
 
  • #26
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?
 
  • #27
37m/s- -35m/s=72m/s
 
  • #28
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?
 
  • #29
velocities after the frame
ball=-46.5
bat=25.4
 
  • #30
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?
 
  • #31
The velocity wrt the ground is already what we need.
 

1. What is the reference frame to the ground?

The reference frame to the ground is a coordinate system that is fixed to the Earth's surface. It is used to measure the motion of objects relative to the Earth's surface.

2. Why is it important to find final velocities in the reference frame to the ground?

It is important to find final velocities in the reference frame to the ground because it allows for accurate measurement and prediction of an object's motion relative to the Earth's surface. This can be useful in various fields such as physics, engineering, and navigation.

3. How do you calculate final velocities in the reference frame to the ground?

To calculate final velocities in the reference frame to the ground, you will need to know the initial velocity, acceleration, and time of the object. You can then use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

4. What are some real-life applications of finding final velocities in the reference frame to the ground?

One real-life application is in projectile motion, where the final velocity of a projectile is often calculated in the reference frame to the ground. This can also be used in analyzing the motion of vehicles, such as cars or airplanes, relative to the Earth's surface.

5. Are there any limitations to finding final velocities in the reference frame to the ground?

One limitation is that it assumes a flat and non-rotating Earth. In reality, the Earth is not perfectly flat and is constantly rotating, which can affect the final velocity calculations. Additionally, air resistance and other external factors may also impact the final velocity in the reference frame to the ground.

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