The Inequality for e in Newton's Law of Restitution Collision?

  • Thread starter Thread starter FeDeX_LaTeX
  • Start date Start date
  • Tags Tags
    Law Newton's law
Click For Summary
SUMMARY

The discussion centers on the inequality for the coefficient of restitution (e) in a collision between two spheres, one with a mass of 3 kg moving at 8 m/s and another stationary sphere of mass 12 kg. The participant initially derived that e < 0.25, contradicting the textbook answer of e > 0.25. Through the conservation of momentum and the definition of e as the ratio of the speed of separation to the speed of approach, the participant confirmed their calculations but later resolved the issue independently.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of the coefficient of restitution
  • Knowledge of conservation of momentum
  • Basic algebra for rearranging equations
NEXT STEPS
  • Study the derivation of the coefficient of restitution in elastic and inelastic collisions
  • Explore examples of momentum conservation in two-dimensional collisions
  • Learn about the implications of different values of e in real-world collisions
  • Investigate the effects of mass ratios on collision outcomes
USEFUL FOR

Students of physics, particularly those studying mechanics, as well as educators and anyone interested in understanding collision dynamics and the coefficient of restitution.

FeDeX_LaTeX
Science Advisor
Messages
436
Reaction score
13

Homework Statement


"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."

Homework Equations


e = v/u

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.
 
Physics news on Phys.org
Solved it, never mind.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 42 ·
2
Replies
42
Views
6K
Replies
10
Views
1K
Replies
13
Views
3K
Replies
7
Views
1K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K