# Homework Help: Collision conservation of energy

1. Mar 23, 2016

### Johnson1704

1. The problem statement, all variables and given/known data
Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 46.0 kg ), who is initially at rest. After the collision, Jack is traveling at 5.00 m/s in a direction 34.0∘ north of east. Ignore friction.

2. Relevant equations
What is the direction of the Jill's velocity after the collision?

What is the magnitude of the Jill's velocity after the collision?
3. The attempt at a solution
use conservation of energy because it is an elastic collision (don't stick together)
.5*m_jack*v_jack_i + .5*m_jill*v_jill_i = .5*m_jack*v_jack_f + .5*m_jill*v_jill_f
.5*(59)*(8) + .5*(46)*(0) = .5*(59)*(5cos(34)) + .5*(46)*(v_jill_f)
236 + 0 = 122.283 + 23*(v_jill_f)
236 - 122.283 = 23*(v_jill_f)
v_jill_f = 4.9442 m/s in x direction
BUT WHAT ANGLE IS SHE MOVING AT?? - so i can find the hypotenuse. v_jill_f would really be 4.9442/cos(angle))

2. Mar 23, 2016

### haruspex

Just because they do not stick together does not make it a perfectly elastic collision. Even if Jack continued in the same direction, there could be some loss of KE. Given that Jack heads off in a new direction, it must have been a glancing blow. That means they would not stick together even if it had been perfectly inelastic.
(Perfectly inelastic means they have the same velocity after impact in the direction formed by the line through their mass centres at the moment of impact.)
What other law might you use?

3. Mar 23, 2016

### Johnson1704

conservation of momentum. I found another website explaining. i think im gonna get it. ill update.
but im going to use that law because jack passes off some momentum to jill.
DO FOR X AND Y DIRECTIONS:
m_jack*v_jack_i + m_jill*v_jill_i = m_jack*v_jack_f + m_jill*v_jill_f

so far i got that step and solved for jill's velocity in the x direction to be 4.94421. going to do same for y then use Pythagorean theorem
update: got -3.586 m/s in the y direction which makes sense because she is moving south of east.
hyp velocity = 6.1 m/s (CORRECT)
now using trig to find theta:
tan^-1 (y vel / x vel) = 35.95 degrees south of east (CORRECT)

Last edited: Mar 23, 2016
4. Mar 23, 2016

### haruspex

I get a slightly smaller y velocity for Jill. Otherwise, that looks right.

5. Mar 23, 2016

### Johnson1704

yep i had to redo my work. i think i did 44kg for jill on accident or something like that
When do you use momentum vs energy conservation laws?
Collisions seem to use both. Does the type of collision effect which law to use? if it is totally inelastic can you use conservation of energy?

6. Mar 24, 2016

### Biker

There are two types of collisions.
1) Perfectly Elastic or Elastic collisions: These collision conserve both energy and momentum. There is no energy wasted in deformation.
2) Inelastic collisions: This type of collision only conserve momentum and doesn't conserve energy because some of the kinetic energy is used to deform the objects a little bit (Also heat and sound). But Perfectly inelastic is the same as inelastic however the objects stick together

7. Mar 24, 2016

### haruspex

No, although elastic and perfectly elastic mean the same, inelastic and perfectly inelastic differ.

Inelastic just means that some KE was lost. Perfectly inelastic means that as much KE was lost as possible, consistent with the conservation of momentum (linear and angular). Only in a head-on collision (line of relative motion of mass centres before collision is also the line connecting the mass centres, and is normal to the surfaces making contact), does perfectly inelastic mean sticking together. Otherwise it is a glancing blow.
In most questions at this level, a glancing blow will still be with the line joining the mass centres at collision normal to the surfaces in contact. E.g., billiard balls. If this is not the case then the impulse will have a torque about the mass centres, causing some of the KE to become rotational.
You can always use conservation of momentum if there are no external forces on the system consisting of the colliding bodies in the direction in which you are assessing momentum. Actually it is a bit broader than that. An external force is ok as long as it is limited in magnitude, so imparts negligible momentum in the duration of the impact. Thus, for equal-sized balls impacting each other on a table, you can use both directions parallel to the table surface, even if the table is sloping. Gravity makes a negligible contribution.