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## #1 2017-03-15 03:57:17

markosheehan
Member
Registered: 2016-06-15
Posts: 51

### collision

two smooth spheres of masses 2kg and 4kg collide obliquely . the 2 kg mass is brought to rest. the coefficient of restitution is 1/2
prove before the impact the spheres were moving in perpendicular directions to each other and find their velocities before and after the collision.

using conservation of momentum. 4x+2y=4p+2(0) where x and y are velocities before impact. p is velocity after impact of 4kg

the velocity in the j direction does not change as its along the i axis

newtons experimental law p-0/x-y =-1/2 dont know what to do next

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## #2 2017-03-15 06:22:29

bob bundy
Registered: 2010-06-20
Posts: 8,084

### Re: collision

hi markosheehan

There's good news and bad news.. I'll deal with the bad first.

Each sphere has a line of direction before the impact and these meet at a point.  Two lines meeting at a point define a plane so the action must take place in a 2-D plane.  That means you cannot just use a single variable for the velocities.  You'll have to have 2 components for each before the impact and 2 after.  That seems like a lot of unknowns.

But now for the good news.

I'll take the x axis as along the line of centres.  So for the 4Kg before impact I'll use components Ux and Uy.

And  Vx and Vy for the 4 Kg afterwards.

And I'll use Wx and Wy for the 2 Kg before.  I don't even need anything for afterwards because we're told it is brought to rest.

So now for some equations:

conservation of momentum in the x direction:

..................1

y direction:

...................2

I think the experimental law only applies along the line of centres as there is no 'bounce' in the tangential direction.  If that seems strange consider a sphere bouncing on a surface.  If we drop the ball onto the surface, it bounces according to the coefficient of restitution.  But if you roll the sphere across the surface there's no bounce at all.

When the spheres touch they share a common tangential plane at the point of impact  The radius of each at that point is at right angles to the tangential plane. So the line of centres is the x axis.   So apply the experimental law to the x components:

......................3

I'm a bit unsure about this next bit but it does yield the required result and it won't if it isn't true so here goes:

Perpendicular to the line of centres I think the relative velocities are equal so:

..............................4

1 + 3 tell us about Ux and 2 + 4 tell us about Wy.  That will give you the first required result.

Off to do some more research on this now.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #3 2017-03-16 04:31:46

markosheehan
Member
Registered: 2016-06-15
Posts: 51

### Re: collision

i can prove that they were moving  in  perpendicular directions before  the impact .   so the velocity of the 2kg after the impact is 0i +0j  .  the collision is along the i axis,  so the j component does not change. this means that before the collision the 2kg was moving along with say pi+oj.  this means that the 4kg must of been moving with 0i+xj   and if they were moving perpindicular.  i know this doesnt really prove they were moving before but ill use this information in working out there velocities

so    2p+4(0)=4y         and   y-o/0-p  =-1/2    these equations dont work either

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## #4 2017-03-16 21:40:48

bob bundy
Registered: 2010-06-20
Posts: 8,084

### Re: collision

hi markosheehan

and find their velocities before and after the collision

There is not enough information to say what these are absolutely.  eg cannot say p = 5 m/s.

But you can plug in the zero components into the momentum equations to say what the after-velocities are,  in terms of p and x.  That's as far as this problem will go without more information, I think.

Where did this question come from?

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #5 2017-03-17 03:11:26

markosheehan
Member
Registered: 2016-06-15
Posts: 51

my textbook.

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## #6 2017-03-17 03:17:01

bob bundy
Registered: 2010-06-20
Posts: 8,084

### Re: collision

That's handy.  Have you got a 'book answer'?

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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