Finding Mass and Velocity in an Elastic Collision at an Angle

[maxalador]

Homework Statement

Ball A with mass 4 kg is moving with a velocity of 8 m/s North when it crashes into Ball B with unknown mass moving with a velocity of 6 m/s West. This collision is perfectly elastic. If Ball A ends up moving with a velocity of 6.5 m/s @ 120° after the collision, find the mass and final speed of Ball B if it moves at an angle of 160° after the collision.

Homework Equations

m1v1 + m2v2 = m1v1’ + m2v2’

The Attempt at a Solution

I'm not sure how to start the problem but I think you take the starting velocities and angles and use sine and cosine to find their x and y components

 

[tiny-tim]

hi maxalador!

(try using the X₂ button just above the Reply box )

the collision is perfectly elastic …

that means you also have conservation of energy ​

 

[maxalador]

I know that means Pi=Pf so Ball A's P=32, but does it mean that ball A initial = ball B initial? or just than Ball A initial = Ball A final?

 

[technician]

momentum is conserved in all collisions.
Kinetic energy is conserved in Elastic collisions.
It is important to realize that it is KINETIC energy conservation
(energy is always conserved but it does not always appear as Kinetic energy !)

 

[maxalador]

momentum is conserved in all collisions.
Kinetic energy is conserved in Elastic collisions.
It is important to realize that it is KINETIC energy conservation
(energy is always conserved but it does not always appear as Kinetic energy !)

I know that KEi=KEf, but how will i get the mass or velocity needed to use this equation? This is the main problem I have with this equation

 

[technician]

Start by looking at conservation of momentum.
Momentum is a VECTOR quantity so you need to take directions into account as well as value.
KE is not a vector quantity...what you start with... you end up with.
Some tricky maths involved sorting this out

 

[tiny-tim]

… when it crashes into Ball B with unknown mass moving with a velocity of 6 m/s West

I know that means Pi=Pf so Ball A's P=32, but does it mean that ball A initial = ball B initial? or just than Ball A initial = Ball A final?

no

call the mass of B "M"

then the momentum of B is 6M west …

the conservation of energy, and conservation of momentum in the x and the y directions will give you three equations, which should enable you to find M ​

 

[technician]

I know that KEi=KEf, but how will i get the mass or velocity needed to use this equation? This is the main problem I have with this equation

This is another question which, at first sight, looks a bit strange !
Are the angles you have given compass bearings?
or are they relative to some other direction?

 

[barryj]

Since you have two unknowns, mass of B and velocity of B you can solve just using conservation of momentum, treating momentum as a vector quantity as technician pointed out.

 

[PeterO]

Homework Statement

Ball A with mass 4 kg is moving with a velocity of 8 m/s North when it crashes into Ball B with unknown mass moving with a velocity of 6 m/s West. This collision is perfectly elastic. If Ball A ends up moving with a velocity of 6.5 m/s @ 120° after the collision, find the mass and final speed of Ball B if it moves at an angle of 160° after the collision.

Homework Equations

m1v1 + m2v2 = m1v1’ + m2v2’

The Attempt at a Solution

I'm not sure how to start the problem but I think you take the starting velocities and angles and use sine and cosine to find their x and y components

How are these angles referenced? 120° to what?

The original momentum included a Northerly component (due to A) and a Westerly component (due to B) - giving an initial momentum in the North West quadrant.

If the 120° is a true bearing, then A has some "South-Easterly" momentum, but if B is traveling at a true bearing 160° then its momentum is also in a "South Easterly" quadrant, and those two things together cannot equal the original ?

Now if A was originally traveling from the North, while B was traveling from the West there is some chance.

EDIT: Just noticed "technician" had a similar question.

 

[barryj]

The problem seems to work if you assume North is 90 deg, West is 180 deg, and then the other angles as given.

 

[PeterO]

The problem seems to work if you assume North is 90 deg, West is 180 deg, and then the other angles as given.

Sort of like a polar co-ordinate reference.

The numbers may work, but the physical situation is curious, as in a real collision I am pretty sure the bodies don't "cross over" during collision.

Now if A was going at 160° and B going at 120° I could visualise the collision.

 

[barryj]

I think it will work.

 

[PeterO]

I think it will work.

If A is finally traveling at 120° (compared to its original 90°) It has been hit from the East - OK if B is further east than A when the collision happens. (simple)

If B is finally traveling at 160° (compared to its original 180°) It has been hit from the South - OK if A is further South than B when the collision happens. (also simple)

That physical situation, I think, requires that after collision, A's direction will be further South and further West than B - a situation only consistent with A traveling at 160° while B travels at 120° - if we must use the directions 160° and 120°.

As I said - numerically the originally described situation may work, but conceptually it is a bit dodgy.

 

[barryj]

Picture this. Let the mass of B be larger than the mass of A. Let B be moving slower than A. Let A strike a glancing blow as it passes just in front of B. Then, A will continue mostly north east, and B will be deflected slightly north. Pcture A striking B at about 8 O'clock. ?

 

[PeterO]

Picture this. Let the mass of B be larger than the mass of A. Let B be moving slower than A. Let A strike a glancing blow as it passes just in front of B. Then, A will continue mostly north east, and B will be deflected slightly north. Pcture A striking B at about 8 O'clock. ?

What magic shape of the particles enables them to pass each other in this way?

If they are spheres, it is not possible for A to pass B in that way. Draw the common tangent at the points of contact. A will be continuing on its initial side of that tangent, while B will be continuing on its side of that tangent - unless A can mysteriously move through the body of B and emerge from the other side.

 

[barryj]

First assume that B is stationary and A hits B at say the 8:30 position. A will be deflected to the West but continuing mostly North. If B is moving West, the colision at 8:30 should push B in the North direction also. See my figure.

Bedtime :-)

 

[PeterO]

First assume that B is stationary and A hits B at say the 8:30 position. A will be deflected to the West but continuing mostly North. If B is moving West, the colision at 8:30 should push B in the North direction also. See my

And how did A get through the bulge of B to go in that direction?

I stress - draw the common tangent at the point of collision. Each sphere has to stay on their respective sides of that common tangent - unless they can somehow travel through each other.

 

[barryj]

I'm not seeing it. Would you not agree that if B is stationary, locked into position and if A hits B at 8:30, would not A move north and slightly west after the glancing blow?

Now if B were to be moving at a slow velocity west when struck by A, would not B be nudged slightly to the north as A deflects off of it?

I am not sure about your tangent point. If I draw a tangent at the point of impact, it seems that both masses will deflect at the point of tangent but I don't see the problem? I have attached a clearer diagram, I hope.

The numbers seem to work out unless I made a calculation error.

It is early in the morning here and maybe my brain is still sleeping.

 

[PeterO]

I'm not seeing it. Would you not agree that if B is stationary, locked into position and if A hits B at 8:30, would not A move north and slightly west after the glancing blow?

Now if B were to be moving at a slow velocity west when struck by A, would not B be nudged slightly to the north as A deflects off of it?

I am not sure about your tangent point. If I draw a tangent at the point of impact, it seems that both masses will deflect at the point of tangent but I don't see the problem? I have attached a clearer diagram, I hope.

The numbers seem to work out unless I made a calculation error.

It is early in the morning here and maybe my brain is still sleeping.

The problem is if they both deflect away from the impact tangent line. That has mass B traveling to the right, rather than the left, of A.
If the mass of B is so large as to not deflect much - and follow the 160^o direction, A will be deflected below that direction - perhaps even South of West.
if the particles are spherical, A cannot get around the edge of B to travel in a more Northerly direction than B.

 

[barryj]

I may be dense but I still don't agree with you as yet. If A hit B at say the 7 o'clock position of B, then I would agree that A would travel South West and B would be deflected north. But let's take an extreme case where B is large and moving very very slowly and that A strikes B at the 8:59 position of B. Surely you would agree that A would be deflected only slightly west of north. B would experience a slight bump in the northly direction as A hit it and would be slightly deflected to say WNW

I worked the numbers using conservation of momentum and got answers. However when I used the answers I got to see if KE was conserved, it wasn't. SO, you are probably correct and I need to rethink the problem. Bummer, I was hoping for a victory here :-)

 

[haruspex]

I may be dense but I still don't agree with you as yet. If A hit B at say the 7 o'clock position of B, then I would agree that A would travel South West and B would be deflected north. But let's take an extreme case where B is large and moving very very slowly and that A strikes B at the 8:59 position of B. Surely you would agree that A would be deflected only slightly west of north. B would experience a slight bump in the northly direction as A hit it and would be slightly deflected to say WNW

From the movement of A, you can calculate (with this interpretation of the angles) that at the moment of impact the centre of B was at an angle of about 60 degrees from that of A (zero degrees being East). In that direction, A moves off with a positive component, while moves off with a negative one. Thus the two balls would have to move through each other.
maxalador - are you still on this thread? Is this question a translation? Is it possible that the given angles are deflections to the original courses?

 

[PeterO]

From the movement of A, you can calculate (with this interpretation of the angles) that at the moment of impact the centre of B was at an angle of about 60 degrees from that of A (zero degrees being East). In that direction, A moves off with a positive component, while moves off with a negative one. Thus the two balls would have to move through each other.
maxalador - are you still on this thread? Is this question a translation? Is it possible that the given angles are deflections to the original courses?

Deflections would certainly make a lot more sense!

 

[maxalador]

Unfortunately, no. the document i pulled this from is on my flash drive, so i just copy pasted the question.