!!!!MOMENTUM COLLISION INELASTIC AND ELASTIC HELP!!!!

[DOpEfrshKID]

Hi,

i i have a few questions on momentum and collisions.

 

1)in a inelastic collision are both objects moving or can 1 be stationary.

2) In a elastic collision why is kinecetic energy conserved?

 

3) if kinectic energy is conserved in a elastic collision does it mean that it travels with the same velocity it had prior to the collision?

 

 

And lucky last,

 

a collision invovling dogems or bumper cars whould be what type of collision.

 

Thanks,

 

 

 

[Cap'n Refsmmat]

1. You can have collisions where one object is stationary. You can always switch to the perspective of one of the colliding objects, because from its own perspective, it's stationary.
2. That's the definition of an elastic collision. It wouldn't be elastic otherwise.
3. No. One of the objects can give its energy to the other. For example, imagine shooting a pool ball into a stationary ball. The collision could be perfectly elastic, meaning the stationary ball ends up moving away with some speed and the cue ball slows down or stops. Newton's cradle also works with elastic collisions.

[DOpEfrshKID]

so lets use this example,

 

 

bumper car A has a mass of 200kg and travels at 5ms-1 east while bumper car b weighing 150kg travels 5ms-1 West.

 

if this is a elastic collision , would the bumper cars rebound back at the same speed?

[Cap'n Refsmmat]

No. If they did, momentum wouldn't be conserved. The total momentum vector would point east before and west afterwards.

[King Thando Mathe]

The momentum and KE are shared between the two bodies in any collision.But in the 'idealised' situation of perfectly elastic collisions,the approaching mass will donate all its KE to the other body or will impart some.The basic rule z ENERGY WILL NOG BE LOST AS OTHER FORMS.A body can thus continue to move (or after rebound) with a reduced velocity because it has shared its KE.

 

The momentum and KE are shared between the two bodies in any collision.But in the 'idealised' situation of perfectly elastic collisions,the approaching mass will donate all its KE to the other body or will impart some.The basic rule z ENERGY WILL NOG BE LOST AS OTHER FORMS.A body can thus continue to move (or after rebound) with a reduced velocity because it has shared its KE.

[kidnate98]

I also have a question but this subject requires a missing velocity

 

 

if you had a velocity of 9 m/s accompanied by a mass of 5kg and 10kg how would you find the unknown velocity.

 

its homework but I thought Id try a forum to see if I could get an answer.

[swansont]

I also have a question but this subject requires a missing velocity

 

 

if you had a velocity of 9 m/s accompanied by a mass of 5kg and 10kg how would you find the unknown velocity.

 

its homework but I thought Id try a forum to see if I could get an answer.

 

You have two unknown velocities — both particles after the collision. So you need to know some additional information: Is the collision elastic? Or completely inelastic? Those are the two solutions we can do with the given setup.

[kidnate98]

completely inelastic

[DimaMazin]

m₁=200kg m₂=150kg

u₁=5 m/s u₂= -5 m/s

v₁ is speed of car A after collision

v₂ is speed of car B after collision

Can you use some equations?:http://en.wikipedia.org/wiki/Elastic_collision

http://en.wikipedia.org/wiki/Inelastic_collision

Edited January 13, 2015 by DimaMazin

[swansont]

So momentum is conserved, and if particle 2 is initially at rest and they go off with the same final speed, how would you write that down?

[DimaMazin]

So momentum is conserved, and if particle 2 is initially at rest and they go off with the same final speed, how would you write that down?

v=m_au_a / (m_a+m_b)

 

I also have a question but this subject requires a missing velocity

 

 

if you had a velocity of 9 m/s accompanied by a mass of 5kg and 10kg how would you find the unknown velocity.

 

its homework but I thought Id try a forum to see if I could get an answer.

 

For collision they should have different speeds, for example: u_a=9 m/s u_b= -9 m/s

m_a=5kg m_b=10kg

 

v=(m_aU_a+m_bu_b) / (m_a+m_b)

 

Edited January 14, 2015 by swansont

[swansont]

I was't asking because I don't know. This is HW help. The point is to get the student to answer, and not just answer it for them

 

I've modified the post.