Elastic Collision kinetic energy and momentum

  • Thread starter fobbz
  • Start date
  • #1
27
0

Homework Statement



A ball of mass 5.0kg moving at a speed of 5.0m/s has a head on collision with a stationary bal of mass 6.0kg. If the collision were perfectly elastic what would be the speeds of the two balls after the collision?

Homework Equations



P = mv
KE = 0.5mv2

The Attempt at a Solution



Using together kinetic energy and momentum equations, I can solve for final velocities.

http://img855.imageshack.us/img855/6519/centralkootenayj2012010.jpg [Broken]

Is this correct?
 
Last edited by a moderator:

Answers and Replies

  • #2
The formulas for elastic collisions are:
v1 = u1(m1-m2)/(m1+m2)
v2 = 2m1u1/(m1+m2)

v1 = 5(5-6)/(6+5) = -5/11 m/s = -.45 m/s
v2 = (2*5*5)/(5+6) = 50/11 m/s = 4.5 m/s

So you got the right answers if you add a negative sign to v1 since it bounces backwards after the collision.
 
Last edited by a moderator:
  • #3
27
0
The formulas for elastic collisions are:
v1 = u1(m1-m2)/(m1+m2)
v2 = 2m1u1/(m1+m2)

v1 = 5(5-6)/(6+5) = -5/11 m/s = -.45 m/s
v2 = (2*5*5)/(5+6) = 50/11 m/s = 4.5 m/s

So you got the right answers if you add a negative sign to v1 since it bounces backwards after the collision.


How do you know that the first ball will bounce backwards?
 
Last edited by a moderator:
  • #4
1,137
0

Homework Statement



A ball of mass 5.0kg moving at a speed of 5.0m/s has a head on collision with a stationary bal of mass 6.0kg. If the collision were perfectly elastic what would be the speeds of the two balls after the collision?

The formulas for elastic collisions are:
v1 = u1(m1-m2)/(m1+m2)
v2 = 2m1u1/(m1+m2)

v1 = 5(5-6)/(6+5) = -5/11 m/s = -.45 m/s
v2 = (2*5*5)/(5+6) = 50/11 m/s = 4.5 m/s

My advice would be not to rely on these formulas and use conservation of KE and momentum conservation and the equation of restitution ... with these 3 things you can solve nearly all collision problems.

PS:
KE Coservation: [itex]\frac{1}{2}{m_1u_1}^2 + \frac{1}{2}{m_2u_2}^2 = \frac{1}{2}{m_1v_1}^2 + \frac{1}{2}{m_2v_2}^2[/itex] - valid only when e=1

Momentum Conservation: [itex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/itex] - valid for all [itex]e\in[0,1][/itex]

Eqn of coefficient of restitution: [itex](v_2 - v_1) = e(u_1 - u_2)[/itex]
 

Related Threads on Elastic Collision kinetic energy and momentum

Replies
5
Views
927
Replies
22
Views
1K
Replies
23
Views
9K
Replies
6
Views
7K
Replies
22
Views
2K
Replies
4
Views
4K
Replies
4
Views
23K
Replies
16
Views
1K
Replies
6
Views
1K
Top