Mechanics Question :)

[turnstile]

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.


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For part (a) i am getting 2.3 m/s.. however my teacher's worked solution is giving 1.75 m/s.... now.. i dont know which one is correct...

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

Please help me out here.
Thanks
Faraz

[Doc Al]

For part (a) i am getting 2.3 m/s.. however my teacher's worked solution is giving 1.75 m/s.... now.. i dont know which one is correct...
Show how you solved it and we can take a look.

[J77]

Your teacher seems right...

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

where X is the unknown speed

[turnstile]

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....

[J77]

They're travelling in opposite directions initially...

[turnstile]

ah.. thanks a lot J77... clear now.

any idea about part (b)?
im quite sure my -0.75 is wrong

[J77]

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

[Doc Al]

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.

[turnstile]

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:

[pizzasky]

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

[turnstile]

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...
:)

[Doc Al]

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.)

[J77]

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!