Conservation of momentum and mass of a ball

[Quantum Fizzics]

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

 

[Miles Whitmore]

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.

"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.

 

[haruspex]

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.

 

[Miles Whitmore]

Gravity doesn't enter into it regardless of direction.

Good point

 

[HalfLostAndSearching]

I would approach this problem by first identifying the concepts at play: conservation of momentum and conservation of kinetic energy. These principles state that in a closed system, the total momentum and total kinetic energy remain constant before and after a collision.

Using the given information, we can set up the equations:

Initial momentum (before collision) = Final momentum (after collision)
m1v1 = m2v2

Initial kinetic energy (before collision) = Final kinetic energy (after collision)
1/2m1v1^2 = 1/2m2v2^2

We are given the mass of the first ball (m1 = 0.160 kg) and the initial velocity (v1) is not specified, so we can leave it as a variable. We are also given that the second ball moves off with half the original speed of the first ball, so we can write v2 = v1/2.

a) To solve for the mass of the second ball, we can substitute these values into the momentum equation and solve for m2:
m1v1 = m2v2
0.160 kg x v1 = m2 x (v1/2)
m2 = 0.320 kg

Therefore, the mass of the second ball is 0.320 kg.

b) To solve for the fraction of kinetic energy transferred to the second ball, we can substitute the values into the kinetic energy equation and solve for the fraction:
1/2m1v1^2 = 1/2m2v2^2
1/2 x 0.160 kg x v1^2 = 1/2 x 0.320 kg x (v1/2)^2
v1^2 = 1/2 x (v1/2)^2
v1^2 = 1/4 x v1^2
1/4 = v1^2/v1^2
1/4 = 1

Therefore, 1/4 or 25% of the original kinetic energy is transferred to the second ball.

In conclusion, conservation of momentum and kinetic energy allow us to solve for the mass of the second ball and the fraction of kinetic energy transferred in this elastic collision. This approach can be applied to other scenarios involving collisions and can help us better understand the behavior of objects in motion.