How do contact forces affect collisions and object launches?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
alba
Messages
140
Reaction score
4
I have recently studied collisions, in which one or two bodies are already in motion.
I have tried to apply them to cases in which the bodies are initially at rest, like launching or kicking a ball etc, but they do not work.
Can you tell me how to google to find them, 'contact force' gives everything, I tried 'explosive forces' but I get only explosives.
Can you help me?:
suppose a girl (20 kg) on frictionless rollers launches a ball (1kg) doing work 210 J, what is the rule to find the outcome?

I imagine that if she pushes at a wall the outcome is the same of an elastic collision with a wall of a body of 20kg, , if so, she will move backward at v = √210*2/20 = - 4.58 m/s. Is that right?

Is the force on the wall equal to the force that acts on the girl? and what about momentum? the girl's is -91.6 kgm/s the wall's should be the same, but in that case it would mean it subtracted energy to the girl and her speed will be less?
How do I deal with the ball? do you have a link or a formula?
Thanks
 
Physics news on Phys.org
If you can already 'do' objects in motion then why not just put the initial velocity of one of them to zero? If you are not getting the right answer, it could be that you are not doing the right sum. Have you looked at the Hyperphysics pages on Elastic and Inelastic Collisions? All the relevant equations and ideas are presented there, although you have to learn to drive their system to get the most out of it.

The way to solve specific problems will depend upon what you are given. If the collision can be assumed to be elastic then no energy is 'lost' and you can use conservation of Energy. If not, you can always rely on Momentum being conserved. Newton 3 always applies, of course.
Have you come across the quantity 'Impulse'? That is the Change in Momentum and is given by the Force times the Time it's applied.