Impulse and momentum in two dimensions - Finding velocity

In summary: If you use the Pythagorean Theorem to solve for v2fy2, you can find that it is negative, which means that the puck B moves towards the original position of puck A.
  • #1
Specter
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8

Homework Statement


In a physics lab, 0.30 kg puck A, moving at 5.0 m/s [W], undergoes a collision with a 0.40 kg puck B, which is initially at rest. Puck A moves at 4.2 m/s [W 30 N] . Find the final velocity of puck B.

Homework Equations


Conservation of momentum
Pythagorean theorum

The Attempt at a Solution


[/B]
I understand the steps I need to take to solve this question but once again I am getting a different answer than what I find online. Any help on where I went wrong would be great.

Let north and east be positive.

My diagram of the x and y components

https://i.imgur.com/0F5MsrP.png

Using conservation of momentum for the components

X-components:
(Not sure if all of the subscripts are correct in this part)
m1v1ox+m2v2ox=m1v2fx+m2v2fx
(0.30)(-5)+(0.40)(0)=(0.30)(-3.64)+(0.40)v2fx
-1.5=-0.692
v2fx=-0.808

The answer online for this was -1.0. Was it just rounded down or did I go wrong somewhere?

For the y-component:

m1v1oy+m2v2oy=m2v1fy+m2v2fy
(0.30)(0)+(0.40)(0)=(0.30)(2.1)+(0.40)v2fy
0=1.03
v2fy=1.03

The answer online for this was -1.6, so I went wrong here too?

I didn't go beyond this part because I think my answers are wrong. I know that next I create a triangle with my answers and use pythagoreans theorem to solve for the missing side, then I use tan to find the direction and I have the final velocity with it's direction.

Also I haven't learned this yet but does the "o" in "v1o" mean initial? Does the "f" in v1f mean final?
 
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  • #2
The method is correct. Recheck your calculations because it looks like you have made calculator errors.

"0" (zero not the letter o) as a subscript usually stands for "initial" as in time t = 0. With collision it makes more sense to use "i" and "f" for "initial" and "final".
 
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  • #3
kuruman said:
The method is correct. Recheck your calculations because it looks like you have made calculator errors.

"0" (zero not the letter o) as a subscript usually stands for "initial" as in time t = 0. With collision it makes more sense to use "i" and "f" for "initial" and "final".
I double checked my calculations and I am still getting the same answers. I can't figure out where I have gone wrong.
 
  • #4
Specter said:
(0.30)(-5)+(0.40)(0)=(0.30)(-3.64)+(0.40)v2fx
This is correct.
Specter said:
-1.5=-0.692
This directly below the correct expression is nonsense and does not follow.
Specter said:
(0.30)(0)+(0.40)(0)=(0.30)(2.1)+(0.40)v2fy
This is correct.
Specter said:
0=1.03
This directly below the correct expression is nonsense and does not follow.

Since when is -1.5 is equal to -0.692 and 0 is equal to 1.03 anyway? Perhaps you need to improve your algebra skills.
 
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  • #5
Please post the details of this step. I get a different result.
Specter said:
(0.30)(-5)+(0.40)(0)=(0.30)(-3.64)+(0.40)v2fx
v2fx=-0.808
 
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  • #6
kuruman said:
This is correct.

This directly below the correct expression is nonsense and does not follow.

This is correct.

This directly below the correct expression is nonsense and does not follow.

Since when is -1.5 is equal to -0.692 and 0 is equal to 1.03 anyway? Perhaps you need to improve your algebra skills.

Wow that was a stupid mistake. Here's what I think are the correct answers.

m1v1ox+m2v2ox=m1v2fx+m2v2fx
(0.30)(-5)+(0.40)(0)=(0.30)(-3.64)+(0.40)v2fx
-1.5+0=-1.092+0.40v2fx
-0.408/0.40
v2fx=1.02

m1v1oy+m2v2oy=m2v1fy+m2v2fy
(0.30)(0)+(0.40)(0)=(0.30)(2.1)+(0.40)v2fy
0=0.63+0.40v2fy
-0.63=0.40v2fy
vfy2=-1.575
 
  • #7
haruspex said:
Please post the details of this step. I get a different result.
I think I figured it out in my latest reply.
 

Related to Impulse and momentum in two dimensions - Finding velocity

1. What is impulse and momentum in two dimensions?

Impulse and momentum in two dimensions refer to the measurement of the amount of force and motion of an object in a two-dimensional space. It takes into account the mass, velocity, and direction of an object to determine its momentum.

2. How is velocity calculated in two dimensions?

In two dimensions, velocity is calculated using the formula: v = √(vx² + vy²), where vx is the velocity in the x-direction and vy is the velocity in the y-direction. This formula takes into account the horizontal and vertical components of an object's velocity.

3. What is the relationship between impulse and momentum in two dimensions?

The relationship between impulse and momentum in two dimensions is described by the law of conservation of momentum. This states that in a closed system, the total momentum before and after a collision will remain constant. In other words, the impulse applied to an object will result in a change in its momentum.

4. How does the angle of impact affect the velocity in two dimensions?

The angle of impact can affect the velocity in two dimensions by changing the direction of the velocity vector. When an object collides at an angle, the x and y components of its velocity will change, resulting in a change in its overall velocity.

5. What are some real-life applications of impulse and momentum in two dimensions?

Impulse and momentum in two dimensions have various real-life applications, such as in sports like billiards and baseball, where the angle of impact and velocity of an object can affect its trajectory. They are also used in engineering and design, such as in the development of car airbags and the calculation of rocket trajectories.

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