Conservation of momentum and mass of a ball

AI Thread Summary
In an elastic head-on collision, a 0.160 kg ball collides with a second ball initially at rest, which then moves off at half the original speed of the first ball. The conservation of momentum principle states that the total momentum before the collision equals the total momentum after the collision, allowing for the calculation of the second ball's mass. Additionally, since the collision is elastic, kinetic energy is also conserved, providing a second equation to solve for unknowns. Gravity does not factor into the calculations as the motion is horizontal and occurs over a short time interval. Understanding these principles is crucial for solving the problem effectively.
Quantum Fizzics
Messages
22
Reaction score
0

Homework Statement


A 0.160 kg ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball.
a) what is the mass of the ball
b) what fraction of the original kinetic energy gets transferred to the second ball.

Homework Equations


p=mv

The Attempt at a Solution


I honestly have no idea how to start this. But I assume that this would be a pendelum. So gravity comes in. Since its asking for the mass of the 2nd ball I was thinking of using the equation above. "p" as 10m/s & of course "m" as 0.160kg. Since it says on the question the 2nd ball moves off with half the original speed of the first ball. I would divide it by 2 then use whatever the V of ball 2 & use the same equation except this time I calculate the mass. Idk what do u guys think? I'll do it right now
 
Physics news on Phys.org
Quantum Fizzics said:
But I assume that this would be a pendelum. So gravity comes in.
Assuming the balls are moving horizontally, gravity will not come into play here.
Quantum Fizzics said:
"p" as 10m/s
I'm not sure where you're getting this number and it does not have the correct units for momentum.

Total momentum is conserved in all collisions, so start with that. Using your relevant equation, write down an expression for the momenta of each ball before and after the collision. Since total momentum is conserved, the sum of momenta before the collision equals the sum of momenta after the collision. That gives you one equation, but you will still have too many unknowns to solve for the unknown mass and velocity.

But you are also told this is an elastic collision, so by definition another quantity is also conserved and you can write another conservation equation. Then solve for unknowns.
 
Miles Whitmore said:
Assuming the balls are moving horizontally, gravity will not come into play here.
Gravity doesn't enter into it regardless of direction. The question only concerns events in an arbitrarily short interval of time. The momentum change contributed by gravitational forces is therefore arbitrarily small.
 
haruspex said:
Gravity doesn't enter into it regardless of direction.

Good point
 
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 '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