Perfectly Elastic Collision

[Dorney]

[SOLVED] Perfectly Elastic Collision

1. A particle of mass m and speed V collides at a right angle with a very massive wall in a perfectly elastic collision. What is the magnitude of the change in momentum of the particle?


2. P before colision = P after collision.


3. I think the answer is zero. Is this correct?

Thanks in advance! :)

 

[shawshank]

sorry, wrong

 

[ddrtrinity]

Well, it does change direction. :) So it's -1 x P (for the new velocity).

 

[Bill Foster]

Momentum prior to collision: $$m\vec{v_1}$$
Momentum after collision: $$m\vec{v_2}$$

$$\vec{v_1}=v_x \hat{x}$$
$$\vec{v_2}=-v_x \hat{x}$$

$$\Delta{p}=m(\vec{v_1}-\vec{v_2})$$
$$\Delta{p}=m(v_x\hat{x}- - v_x\hat{x})$$
$$\Delta{p}=2m v_x \hat{x}$$

Since it asks for magnitude:

$$2mv$$

 

[shawshank]

Sorry, bill is right, i was wrong.
Perfectly elastic collision occurs in isolated systems tho no? so you can't really consider this a collision, this is like magnitude of velocity in equals to magnitude of velocity out.

 

[Dorney]

Momentum prior to collision: $$m\vec{v_1}$$
Momentum after collision: $$m\vec{v_2}$$

$$\vec{v_1}=v_x \hat{x}$$
$$\vec{v_2}=-v_x \hat{x}$$

$$\Delta{p}=m(\vec{v_1}-\vec{v_2})$$
$$\Delta{p}=m(v_x\hat{x}- - v_x\hat{x})$$
$$\Delta{p}=2m v_x \hat{x}$$

Since it asks for magnitude:

$$2mv$$

Thank you! That makes so much more sense!