Collision between two particles.

In Summary, Coconut did not use the proper equation for velocity of separation and still got the correct answer. Coconut should redo the first part.Coconut may assign right as positive and left as negative.Its all the game of two equations: conservation of linear momentum and that velocity of separation one.BTW, are you a coconut fan ? :tongue:In any case, the speed of approach of the particles is (2u) - (-u) = 3u, not 2u.I thought both particles are coming together with the same speed?BTW, are you a coconut fan ? :tongue:What ._.No, coconut is derived from a word in a language directly translated from some english words which
  • #1
coconut62
161
1

Homework Statement



Question: First image
My working: Second image

I used Va+Vb for the speed of separation. I got the answer for (a), but couldn't get the answer for (b).

I know the proper expression is "Vb-Va" with Vb being negative.

But if I use Vb-Va, (a) would be wrong.

How come?

Homework Equations


Principle of conservation of momentum, Newton's Experimental Law.

The Attempt at a Solution



As shown in the second image.
 

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  • #2
Realize that when you post an image of your work, instead of typing it in, that it's difficult to make comments. (You'll get quicker help if you make it easy to comment. :wink:)

In any case, the speed of approach of the particles is (2u) - (-u) = 3u, not 2u. (Wrong!)

And the correct expression is Vb - Va, not Vb + Va.

Edit: My statement about speed of approach was incorrect; it should be (u) - (-u) = 2u, just as you had it. I misread the attachment.
 
Last edited:
  • #3
coconut62 said:

Homework Statement



Question: First image
My working: Second image

I used Va+Vb for the speed of separation. I got the answer for (a), but couldn't get the answer for (b).

I know the proper expression is "Vb-Va" with Vb being negative.

But if I use Vb-Va, (a) would be wrong.

How come?

Homework Equations


Principle of conservation of momentum, Newton's Experimental Law.

The Attempt at a Solution



As shown in the second image.

Coconut (:tongue:) just lucked out. Coconut did not use the correct equation for velocity of separation and still got the correct answer. Coconut should redo the first part.

Coconut may assign right as positive and left as negative.
Its all the game of two equations: conservation of linear momentum and that velocity of separation one.

BTW, are you a coconut fan ? :tongue:
 
  • #4
Doc Al said:
In any case, the speed of approach of the particles is (2u) - (-u) = 3u, not 2u.

I thought both particles are coming together with the same speed?

sankalpmittal said:
BTW, are you a coconut fan ? :tongue:

What ._.
No, coconut is derived from a word in a language directly translated from some english words which have the same pronunciation as the abbreviation of my real name being directly translated into english.

(wow, that was fun)
 
  • #5
coconut62 said:
I thought both particles are coming together with the same speed?
Oops! I misread your post. My bad! :redface:

In any case, the best way to approach this is algebraically without plugging in numbers for e until a later step.

And the relative velocity is Va - Vb, so the key relation would be Vb' - Va' = e(Va - Vb).
 
  • #6
Doc Al said:
...the key relation would be Vb' - Va' = e(Va - Vb).

u = 2Va + Vb (from principle of conservation of momentum)

Vb-Va = e (Ua-Ub)

Vb-Va = e　(u- -u)

Vb-Va = e (2u)

Vb-Va = e (4Va + 2Vb)

When e > 1/2,

Vb-Va < 1/2 (4Va + 2Vb)

< 2Va+Vb

3Va > 0

Still not reversed :uhh:
 
  • #7
coconut62 said:
u = 2Va + Vb (from principle of conservation of momentum)
Good. Call this equation 1.

Vb-Va = e (Ua-Ub)

Vb-Va = e　(u- -u)

Vb-Va = e (2u)
Call this equation 2.

Now use equation 1 to eliminate Vb from equation 2, thus expressing Va in terms of u and e only.

Then you'll be able to draw some conclusions about Va.

(Then do the same for Vb.)
 
  • #8
u = 2Va + Vb--1
Vb-Va = e(2u)--2

Solution:

u-2Va-Va = e(2u)
u-3Va= 2eu

Va= (u-2eu)/3

=u(1-2e)/3 (is this sufficient to jump to the next statement?)

when e> 1/2, Va<0
 
  • #9
coconut62 said:
u = 2Va + Vb--1
Vb-Va = e(2u)--2

Solution:

u-2Va-Va = e(2u)
u-3Va= 2eu

Va= (u-2eu)/3

=u(1-2e)/3 (is this sufficient to jump to the next statement?)

when e> 1/2, Va<0
Good. The sign of 1-2e changes when e > 1/2.
 
  • #10
Ok thanks.
 

1. What is a collision between two particles?

A collision between two particles is a physical event where two particles come into contact with each other, resulting in a change in their motion or properties. It is a fundamental concept in physics and is often studied in the context of Newton's laws of motion.

2. What factors determine the outcome of a collision between two particles?

The outcome of a collision between two particles is determined by a few key factors, including the masses of the particles, their velocities, and the angle at which they collide. Additionally, the type of interaction between the particles (e.g. elastic or inelastic) will also affect the outcome.

3. What is the difference between an elastic and inelastic collision?

An elastic collision is one where the total kinetic energy of the particles before and after the collision remains the same. In contrast, an inelastic collision is one where the total kinetic energy is not conserved, and some energy is lost in the form of heat or sound.

4. How do you calculate the momentum of two particles after a collision?

The momentum of two particles after a collision can be calculated using the equation:

pf = m1v1f + m2v2f

where pf is the final total momentum, m1 and m2 are the masses of the particles, and v1f and v2f are their final velocities.

5. What real-life examples can demonstrate a collision between two particles?

Collisions between particles are happening all around us, from the impact of a ball hitting a bat in a game of baseball, to the collision of atoms during a chemical reaction. Other examples include car crashes, billiard balls colliding, and the collision of subatomic particles in a particle accelerator.

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