Mechanics Question of two balls

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Homework Help Overview

The discussion revolves around a mechanics problem involving two colliding balls with specified masses and velocities. Participants are tasked with determining the speed of one ball after the collision and the impulse exerted on it.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of momentum and the calculations involved in determining the post-collision speed of ball B. There are conflicting results between the original poster's calculation and the teacher's solution.
  • Questions arise regarding the signs used in momentum calculations and the implications of direction on impulse.

Discussion Status

Some participants have provided guidance on the momentum conservation approach, while others are clarifying the directionality of impulse and momentum. There is an ongoing exploration of the correct values and methods, with no explicit consensus reached yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. The original poster expresses uncertainty about their calculations and seeks clarification on the correct approach.

turnstile
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Mechanics Question :)

Hi.

Two small balls A and B have masses 0.5kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when they collide directly. Immediately before the collission, the speed of A is 3m/s and the speed of B is 2 m/s. The speed of A immediately after the collision is 1.5 m/s. The direction of motion of A is unchanged as a result of the collision.

By modelling the balls as particles, Find;
(a) the speed of B immediately after the collision.

(b) the magnitude of the impulse exerted on B in the collision.


----
For part (a) i am getting 2.3 m/s.. however my teacher's worked solution is giving 1.75 m/s... now.. i don't know which one is correct...

for part B... i use Ft=mv-mu and get -0.75 ... i think I am wrong...

Please help me out here.
Thanks
Faraz
 
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turnstile said:
For part (a) i am getting 2.3 m/s.. however my teacher's worked solution is giving 1.75 m/s... now.. i don't know which one is correct...
Show how you solved it and we can take a look.
 
conservation of momentum

Your teacher seems right...

0.5*3-0.2*2=0.5*1.5+0.2*X

where X is the unknown speed
 
Last edited:
J77 said:
Your teacher seems right...

0.5*3-0.2*2=0.5*1.5+0.2*X

where X is the unknown speed

HI there..
why is it 0.5 x3 MINUS 0.2*2 ?
because my 2.3 comes from adding these two...
 
They're traveling in opposite directions initially...
 
ah.. thanks a lot J77... clear now.

any idea about part (b)?
im quite sure my -0.75 is wrong
 
Well, impulse = change in momentum...

momentum before = 0.2 * 2

momentum after = 0.2 * 1.75

implies impulse = 0.2 * (2 - 1.75) = 0.2 * 0.25 = 0.05
 
turnstile said:
any idea about part (b)?
im quite sure my -0.75 is wrong
Why do you have a minus sign? What direction is the impulse on B?

Also, regardless of direction, they asked for the magnitude of the impulse.
 
J77 said:
Well, impulse = change in momentum...

momentum before = 0.2 * 2

momentum after = 0.2 * 1.75

implies impulse = 0.2 * (2 - 1.75) = 0.2 * 0.25 = 0.05

I= 0.2 x (2+1.75)= 0.75Ns
isnt this the answer?

I just looked at the mark scheme and it confirms that... :confused:
 
  • #10
Momentum is a vector, so you must consider the direction. Based on your working for (a), B moves in the positive direction after the collision, and in the negative direction before the collision.

Do you now understand how the correct answer of 0.75 Ns is obtained?

P/S Impulse on B = Final momentum of B - Initial momentum of B
 
  • #11
pizzasky said:
Momentum is a vector, so you must consider the direction. Based on your working for (a), B moves in the positive direction after the collision, and in the negative direction before the collision.

Do you now understand how the correct answer of 0.75 Ns is obtained?

P/S Impulse on B = Final momentum of B - Initial momentum of B

Thanks a lot... i get it now...
:)
 
  • #12
turnstile said:
I= 0.2 x (2+1.75)= 0.75Ns
isnt this the answer?

I just looked at the mark scheme and it confirms that... :confused:
That's correct. Your only mistake (in your earlier answer) was including the minus sign.

PS: J77 made an error in post #7--he had the wrong sign for the initial momentum. (This is the same error that he pointed out to you in your solution to part a.)
 
  • #13
Doc Al said:
PS: J77 made an error in post #7--he had the wrong sign for the initial momentum. (This is the same error that he pointed out to you in your solution to part a.)
:redface: :redface: :redface:

oops!
 

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