(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

e = v/u

3. The attempt at a solution

I am sure I have the method right but I am just getting the wrong sign in my answer.

Textbook answer is e > 0.25. I'm getting e < 0.25...

Diagram:

---> = positive direction

8 m/s..............0 m/s

->

(3 kg).............(12 kg)

<-.....................->

A m/s..............B m/s

By the conservation of momentum:

24 = 12B - 3A

⇒8 = 4B - A

e = (speed of separation)/(speed of approach)

Speed of approach is 8 m/s.

Speed of separation is A + B.

⇒ e = (A + B)/8

⇒ 8e = A + B

So we have:

4B - A = 8

A + B = 8e

Adding both equations gets us:

5B = 8(1 + e)

so e = (5B - 8)/8

B > 0 since moving in positive direction. So e > -1.

A < 0 since moving in negative direction.

Since 8e = A + B and A < 0, we can say:

8e - B < 0

We say that e = (5B - 8)/8, so re-arranging in terms of B, we have B = (8e + 8)/5.

So:

8e - (8e + 8)/5 < 0

⇒ (32e - 8)/5 < 0

⇒ 32e < 8

∴ e < 0.25.

But textbook's answer is e > 0.25. Why?

Thanks.

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