Calculating Velocity After Elastic Collision: Kinetic Energy Problem

AI Thread Summary
In a perfectly elastic collision involving a 0.8kg ball moving at 8.0m/s and a stationary 0.4kg ball, both kinetic energy and momentum must be conserved to determine the final velocities. The initial kinetic energy of the moving ball cannot simply be divided equally between the two balls after the collision. Instead, the conservation of momentum and energy equations should be set up and solved simultaneously to find the final speeds of both balls. The discussion emphasizes the importance of using both conservation laws rather than assuming equal distribution of kinetic energy. Proper application of these principles will yield the correct velocities post-collision.
ashvinthecha
Messages
7
Reaction score
0

Homework Statement


A ball of mass 0.8kg moving initially at 8.0m/s has a head on collision with a 0.4kg ball that is at rest. If the collision is perfectly elastic, what is the velocity of each ball after the collision?


Homework Equations


Ek=(mv2)/2


The Attempt at a Solution


I found the kinetic energy of the first ball, then split the knietic energy between the two balls after the collision to individually find the velocity of each.
 
Physics news on Phys.org
ashvinthecha said:
I found the kinetic energy of the first ball, then split the knietic energy between the two balls after the collision to individually find the velocity of each.
What do you mean by 'split the kinetic energy'?

Hint: What else is conserved during the collision besides energy?
 
Doc Al said:
What do you mean by 'split the kinetic energy'?

Hint: What else is conserved during the collision besides energy?

I divided the total kinetic energy by two... and is momentum the answer to your question
 
ashvinthecha said:
I divided the total kinetic energy by two...
You cannot assume that the kinetic energy divides equally.
and is momentum the answer to your question
Yes. Use conservation of momentum and energy to solve for the final speeds.
 
Doc Al said:
You cannot assume that the kinetic energy divides equally.

Yes. Use conservation of momentum and energy to solve for the final speeds.

How do I do that?
 
ashvinthecha said:
How do I do that?
Write equations for energy conservation and momentum conservation, then solve them.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top