Calculate Velocity of Gliders After Collision: -1m/s Left

[MrPickle]

Homework Statement

There are two 'gliders' X and Y on an air track. The mass of X is 0.2kg and it's velocity is 1.5m/s to the right. The mass of Y is 0.3kg and it is stationary. When 'glider' X collides with trolley Y they move off together.

Calculate the velocity of the 'gliders' after the collision and give their direction.

Homework Equations

mass x velocity = momentum
mass Y x velocity Y = -(mass X x velocity X)

The Attempt at a Solution

mass Y x velocity Y = -(mass X x velocity X)
0.3 x velocity Y = -(0.2 x 1.5)
$$Velocity Y = \frac{-0.3kgm/s}{0.3kg} = -1$$

So it's moving 1m/s to the left? I'm not sure whether that's the velocity of both of the gliders, or just velocity of Y because that's what my equation shows.

 

[Doc Al]

The Attempt at a Solution

mass Y x velocity Y = -(mass X x velocity X)

This is incorrect. What would be true is that the change in Y's momentum will be equal and opposite to the change in X's momentum.

Treat this as a completely inelastic collision (the carts stick together and thus end up with the same final velocity) and write an equation for momentum conservation.

 

[MrPickle]

My book says that's the equation for conservation of momentum.

 

[Doc Al]

My book says that's the equation for conservation of momentum.

What book are you using?

I suspect that the equation in your book is more like:
$$M_y \Delta V_y = - M_x \Delta V_x$$

For inelastic collisions, I recommend this version:
$$m_1v_1 + m_2v_2 = (m_1 + m_2)v_f$$

 

[MrPickle]

My book is AQA Science GCSE Physics

Conservation of momentum in an explosion
In the trolley examples:

• momentum of A after the explosion = (mass of A x velocity of A)
• momentum of B after the explosion = (mass of B x velocity of B)
• total momentum before the explosion = 0 (because both trolleys were at rest).

Using conservation of momentum gives:
(mass of A x velocity of A) + (mass of B x velocity of B) = 0
Therefore
(mass of A x velocity of A) = -(mass of B x velocity of B)
This tells us that A and B move apart with equal and opposite amounts of momentum.

 

[borgwal]

You have to learn to distinguish an equation that is generally valid (such as conservation of momentum when there is no external force), and equations that apply to specific situations (such as the one you are using, even though that equation follows from the more general law).

 

[MrPickle]

There is no other equations to do with conservation of energy in my book though :(

I hate my stupid Physics/Chemistry teachers, he teaches us jack all. He told us out loud he's not going to teach us the theory. WTF is he going to teach us then.

 

[borgwal]

Wow, that's sad...I suppose he is "teaching" you how to mindlessly put numbers into equations.

In any case, the general equation for conservation of momentum (assuming the masses stay constant!) is that:
[tex]
\sum_i m_i \vec{v}_i^{{\rm b}} = \sum_i m_i \vec{v}_i^{{\rm a}}
[/itex]
thus relating velocities before (b) and after (a) a collision between particles with masses m_i.

 

[MrPickle]

I thought mass was constant O.O

What's that big E thing mean?

 

[borgwal]

I was just having troubles with LaTeX: it's fine now. There's no big E anymore.

 

[MrPickle]

It's still there, this thing: $$\sum$$

 

[borgwal]

Oh, that's a summation sign: you sum over all particles in your system. In your case you have two gliders, X and Y, so you'd sum over i taking the values X and Y.

 

[borgwal]

Written in a different way, you have:

$$m_X v_X^b+m_Yv_Y^b=m_X v_X^a+m_Yv_Y^a$$

 

[MrPickle]

Now that, I understand apart from the ^a and ^b, where has a and b come from or is that what I need to work out?

 

[borgwal]

a=after
b=before

 

[MrPickle]

So:
$$0.2\times1.5 + 0.3\times0 = (m_Xv_X + m_Yv_Y)^a$$
$$\frac{0.3}{1.5 + 0.3} = (v_X + v_Y)^a$$

and that gets me the final velocity?

 

[borgwal]

Not quite, but you're on the right track:

first you should note that after the collision X and Y end up traveling with the same velocity, so v_X^a=v_Y^a.

Second, I don't know why you divided by 1.5+0.3: the masses are 0.2 and 0.3, not 1.5 and 0.3!

 

[Doc Al]

Don't forget that after the collision the trolleys "move off together"--that means they have a single velocity after the collision.

It might be simpler for you to use the equation I gave in post #4. (The 2nd equation. The left side represents the momentum before the collision; the right side, after the collsion.)

 

[MrPickle]

I think I've got it now:
$$Velocity = \frac{0.3}{0.5}$$

So the velocity is 0.15m/s to the left?

 

[Doc Al]

I think I've got it now:
$$Velocity = \frac{0.3}{0.5}$$

Good.

So the velocity is 0.15m/s

No. (That fraction doesn't equal this.)

to the left?

Why to the left?

 

[MrPickle]

Oops, 0.18m/s I must of hit * on the calculator instead of /

and I think it might be to the right because that's the direction in which glider X hits glider Y and they'll carry on moving together.

 

[Doc Al]

Oops, 0.18m/s I must of hit * on the calculator instead of /

Still not right. Do it one more time.

and I think it might be to the right because that's the direction in which glider X hits glider Y and they'll carry on moving together.

Much better.

 

[MrPickle]

Ahhh, 0.6m/s

3rd time lucky?

 

[Doc Al]

Ahhh, 0.6m/s

3rd time lucky?

Good. 3rd time's the charm.

I recommend that you review the entire calculation to make sure you understand everything covered in this thread.

(I find it hard to believe that your book covers momentum conservation for explosions, but not collisions. Better double check that.

Of course, an inelastic collision can be considered the mirror image of an explosion.)

 

[MrPickle]

Could you give me an example question or point me to a site which can provide me with several samples please?

Thanks for the help BTW.

 

[Doc Al]

Explore these links to learn more about conservation of momentum:

http://hyperphysics.phy-astr.gsu.edu/HBASE/conser.html#conmom"

http://hyperphysics.phy-astr.gsu.edu/HBASE/colcon.html#c2"