What is the right formula to use in this context? (momentum conservation)

  • #1
El foolish Phenomeno
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Homework Statement
A pool ball X of mass 0.3 kg moving with velocity 5 m/ s hit a stationary ball Y of mass 0.4 kg . Y moves off with a velocity of 2 m/ s at 30 ° to the initial direction of X . Find the X and it's direction after hitting it.
Relevant Equations
momentum formulas
First i think the correct solution to the problem is

1000078137.jpg
But my friends argue that it is not what i did , i am confused we didn't see the whole chapter on momentum in class, (Youtube thank you)

here is what my friends say :

(0.3×5) + 0 = (V×0.3)+(0.4×2)
and they get they a final speed of 2.33 m/s , with negative direction.
 
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  • #2
Hi,

I think you are correct. Can you guess what kind of collision your friends' equation describes ?

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  • #3
@kuruman : ball Y moves off with x component velocity in the positive direction ...

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  • #4
BvU said:
@kuruman : ball Y moves off with x component velocity in the positive direction ...

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Oops! I misread the problem. I deleted my post to avoid confusion. Thanks for the heads up.
 
  • #5
Sin
BvU said:
Hi,

I think you are correct. Can you guess what kind of collision your friends' equation describes ?

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since there is no mention of conservation of energy in that equation , i'd say inelastic collision.
 
  • #6
Consider that it is momentum conservation in one single direction: a central collision. And elastic (check !)

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  • #7
BvU said:
Consider that it is momentum conservation in one single direction: a central collision. And elastic (check !)

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Lol. In an elastic collision , both Energy and Momentum are conserved. I was distracted 😅 .
BvU said:
Consider that it is momentum conservation in one single direction: a central collision. And elastic (check !)

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  • #8
El foolish Phenomeno said:
Lol. In an elastic collision , both Energy and Momentum are conserved. I was distracted 😅 .
You may be missing the point of @BvU's question.
You asked how come your friends did not need energy conservation. The reason is that they treated it as a head-on collision. On that basis they did not need to consider energy; there was enough information to find X's final velocity (and to calculate the energy change).
As a 2D collision, we have one more unknown, so a second equation is needed.
 
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  • #9
I'm coming back to post #1, checking the math and such :rolleyes:

El foolish Phenomeno said:
here is what my friends say :

(0.3×5) + 0 = (V×0.3)+(0.4×2)
and they get they a final speed of 2.33 m/s , with negative direction.
Doing the math gets me 2.33 m/s in the positive direction....

El foolish Phenomeno said:
the correct solution to the problem ...
1700315891309.png
I get -26.36##^\circ## and 3.00 m/s. Nitpicking ?

BvU said:
And elastic (check !)
Check: ##T_{in} = {1\over 2}m_x v_x^2## = 3.75 J and
##T_{out} = {1\over 2}m_x \; 2.33^2 + {1\over 2}m_y u_y^2## = 1.62 J, so not elastic (my bad suggesting that in post #6 o:) ).

and for the correct result ##T_{out} = ## 2.15 J, not elastic either.

If you want to get some practice, determine ##u_y## for the case of an elastic collision with ##\theta = 30^\circ## :smile:
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  • #10
BvU said:
I get -26.36##^\circ##
Yes. The mistake in the given answer was in rounding 0.49555.. to 0.5 in the preceding line.
 
  • #11
So it's nitpicking allright. Given data are only one decimal ...

However, my rule is not to round off intermediate results unnecessarily.

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  • #12
BvU said:
So it's nitpicking allright. Given data are only one decimal ...

However, my rule is not to round off intermediate results unnecessarily.

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Not nitpicking at all. The answer should have been given as either 26° or 26.4°, not 26.6°.
 
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FAQ: What is the right formula to use in this context? (momentum conservation)

1. What is the basic formula for momentum conservation?

The basic formula for momentum conservation is \( p_{\text{initial}} = p_{\text{final}} \), or \( m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \), where \( m \) represents mass, \( v \) represents velocity, and the primes (') denote the velocities after the interaction.

2. How do you apply momentum conservation in a collision problem?

To apply momentum conservation in a collision problem, you need to calculate the total momentum of all objects before the collision and set it equal to the total momentum after the collision. This can be expressed as \( m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \) for a two-object system.

3. What is the difference between elastic and inelastic collisions in terms of momentum conservation?

In both elastic and inelastic collisions, the total momentum is conserved. However, in elastic collisions, both momentum and kinetic energy are conserved. In inelastic collisions, only momentum is conserved, while kinetic energy is not necessarily conserved.

4. How do you handle momentum conservation in a system with external forces?

If external forces act on the system, momentum is not conserved unless the net external force is zero. In such cases, you need to account for the impulse provided by external forces over the time interval of interest. The formula becomes \( p_{\text{initial}} + \text{Impulse} = p_{\text{final}} \).

5. Can momentum conservation be applied in two or three dimensions?

Yes, momentum conservation can be applied in two or three dimensions. You need to consider the vector components of momentum in each direction. For example, in two dimensions, you would use \( m_1 \vec{v}_1 + m_2 \vec{v}_2 = m_1 \vec{v}_1' + m_2 \vec{v}_2' \), where \( \vec{v} \) represents the velocity vector.

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