abbiechantellex
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#1
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The question states
"A moving particle of mass m collides elastically with a particle of equal mass which is at rest on a smooth table. After the collision, the first particle moves at an angle of θ to its initial direction of motion and the second particle moves at an angle of 2θ to the original direction of motion of the first particle so there is an angle of 3θ between the trajectories of the first and second particles after the collision. Find θ."

I honestly have no idea where to start with this so any help will be appeciated
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DFranklin
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(Original post by abbiechantellex)
The question states
"A moving particle of mass m collides elastically with a particle of equal mass which is at rest on a smooth table. After the collision, the first particle moves at an angle of θ to its initial direction of motion and the second particle moves at an angle of 2θ to the original direction of motion of the first particle so there is an angle of 3θ between the trajectories of the first and second particles after the collision. Find θ."

I honestly have no idea where to start with this so any help will be appeciated
It will help to draw a diagram. You have 3 vectors, an initial direction of motion u and the subsequent motions of the two balls {\bf v_1}, {\bf v_2}. Use the law of conservation of momentum to see how these vectors relate. Then use conservation of energy to find the angle between v_1 and v_2.
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abbiechantellex
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(Original post by DFranklin)
It will help to draw a diagram. You have 3 vectors, an initial direction of motion u and the subsequent motions of the two balls {\bf v_1}, {\bf v_2}. Use the law of conservation of momentum to see how these vectors relate. Then use conservation of energy to find the angle between v_1 and v_2.
I have calculated the conservation of momentum and the conservation of energy, but I am unsure as to what to do now.
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ghostwalker
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(Original post by abbiechantellex)
I have calculated the conservation of momentum and the conservation of energy, but I am unsure as to what to do now.
It would help to post your working so far.


Note: This works out very nicely if you do everything in terms of vectors and dot products; and is a real pig if you look at in terms of parallel and perpendicular to the vector u (as I did initially).

Recall  v^2= |{\bf v}|^2= \bf v.v
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DFranklin
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(Original post by ghostwalker)
It would help to post your working so far.


Note: This works out very nicely if you do everything in terms of vectors and dot products; and is a real pig if you look at in terms of parallel and perpendicular to the vector u (as I did initially).

Recall  v^2= |{\bf v}|^2= \bf v.v
You don't even need that if you draw the right diagram and use Pythagoras (well, the converse to Pythagoras I guess).

But your suggestion works nicely too of course.
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old_engineer
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(Original post by DFranklin)
You don't even need that if you draw the right diagram and use Pythagoras (well, the converse to Pythagoras I guess).

But your suggestion works nicely too of course.
It's also possible to reason as follows:

a) It's an elastic collision between equal masses with the first in motion and the second stationary before the collision, so the first particle will transfer to the second particle the full component of its momentum resolved along the line joining the centres of the particles at the moment of collision;

b) From (a) the subsequent motion of the first particle must be at right angles to the line of centres (as it has stopped in the direction of the line of centres);

c) The motion of the second particle after the collision must be along the line of centres of the particles at the moment of collision, as this is the line along which they interact;

d) From (b) and (c) the particles must be moving at right angles to one another after the collision. [email protected] = 90 so @ = 30.
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DFranklin
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(Original post by old_engineer)
It's also possible to reason as follows:
Very nice, but maybe wait for the OP's response before posting a full solution?
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old_engineer
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(Original post by DFranklin)
Very nice, but maybe wait for the OP's response before posting a full solution?
Quite so. Thank you for the reminder. Will exercise self-restraint in future.
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