Welcome to NRICH.

 
Collision question
By Anonymous on Tuesday, April 24, 2001 - 07:42 pm:

Hello,

I am a little stuck as to how to do this question.

A small smooth sphere of mass 3kg moving on a smooth horizontal plane with speed 8 ms-1 collides directly with a sphere of mass 12kg which is at rest. Given that the spheres move in opposite directions after the collision, obtain the inequality satified by e.

I know the fact that we can use
1. The coefficient of restitution formula, e = (separation speed)/(approach speed)
2. Conservation of momentum

But not sure of how to put it so that I can answer the question. A step by step guide to setting up the question would be helpful indeed. Thanks in advance.
By Kerwin Hui (Kwkh2) on Tuesday, April 24, 2001 - 08:08 pm:

Take the direction in which the 3kg sphere initially travels as the positive direction. Let u be the velocity of 3kg sphere after the collision, and the v be the velocity of the 12kg sphere after the collision. Use Newton's Law of Restitution and Conservation of linear momentum to give you the value of u and v in terms of e. v is definitely positive, so you only need to impose the condition u<0 and a bit of simplifying yields the solution.

Kerwin

By Anonymous on Tuesday, April 24, 2001 - 08:19 pm:

Ok, here is what I tried.

Using restitution
e = (v - u)/8 (Is this right?)

Using conservation of momentum
8(3) + 0 = 3u + 12v
So u = 8 - 4v

Subbing u into the restitution, I got e = (5v - 8)/8

This is incorrect, right?
What I am I doing wrong? And how does the inequality for e come about? I am a tad confused.

Is this right?

By Kerwin Hui (Kwkh2) on Tuesday, April 24, 2001 - 09:46 pm:

Solve for u in terms of e (NOT e in terms of v, which is what you have done) from the simultaneous equation

v-u=8e
u+4v=8

and then use the inequality u<0.

Kerwin.

By Anonymous on Tuesday, April 24, 2001 - 10:17 pm:

Ok, this is what I followed on from what you showed me,

8 = u + 4(8e +u)
u = (8 - 4(8)e)/5

But if u<0 then e>(1/4)

Is that right?

Thanks for your patience and kind help Kerwin.

By Kerwin Hui (Kwkh2) on Tuesday, April 24, 2001 - 10:22 pm:

That's correct

By Anonymous on Wednesday, April 25, 2001 - 02:58 pm:

Thanks again for your help Kerwin.