A 0.0600 kg tennis ball, moving with a speed of 2.50 m/s, has a head-on collision with a 0.0900 kg ball initially moving away from it at a speed of 1.00 m/s. Assuming a perfectly elastic collision, what is the speed and direction of each ball after the collision?
-I was given that the answer for the speeds of the balls are .70m/s and 2.2m/s.
equations:
m1v1+m2v2 = m1v1'+m2v2'
V1-V2=V2' - V1'
My attempt:
with the given, I know that
m1= .06 kg
m2=.09 kg
v1= 2.5 m/s
v2= 1.0 m/s
after this, I got to .135=m1v1'+m2v2'
and 1.5= V2'-V1'
I agree with this
m1v1+m2v2 = m1v1'+m2v2'
I disagree with this:
V1-V2=V2'-V1'
where in the world did that come from?
I asked someone else for help on this problem and they told me to use that equation. That's even what they wrote down for me. I agree it doesn't make sense.
Anyways the initial and final mass will stay the same right? So I will be at...
135=(.06)v1'+(.09)v2'
Now you have one other equation you can use, the magic equation conservation of energy is true
Intial KE= final KE
You know initial KE. Final ke, 1/2 m1*V1'^2 + 1/2 m2 v2'^2 is equal to that initial.
Now, in your 135=(.06)v1'+(.09)v2'
solve for v2' in terms of all the other stuff, then put that v2' into the conservationof energy equation (yes square v2') It is messy algebra, you will get a quadratic you can solve (quadratic equation).
To solve for the velocities after the collision, you can use the equations you mentioned.
Equation 1: m1v1 + m2v2 = m1v1' + m2v2'
Equation 2: v1 - v2 = v2' - v1'
Given values:
m1 = 0.0600 kg
m2 = 0.0900 kg
v1 = 2.50 m/s
v2 = -1.00 m/s (converted to negative since it's moving away)
Let's substitute these values into the equations:
0.0600 kg * 2.50 m/s + 0.0900 kg * (-1.00 m/s) = 0.0600 kg * v1' + 0.0900 kg * v2'
0.135 kg·m/s = 0.0600 kg * v1' + 0.0900 kg * v2' [Equation 1]
2.50 m/s - (-1.00 m/s) = v2' - v1'
3.50 m/s = v2' - v1' [Equation 2]
Now, we have a system of two equations with two variables (v1' and v2'). We can solve this system of equations to find the values.
From Equation 2, we can express v2' in terms of v1':
v2' = 3.50 m/s + v1'
Substitute this into Equation 1:
0.135 kg·m/s = 0.0600 kg * v1' + 0.0900 kg * (3.50 m/s + v1')
0.135 kg·m/s = 0.0600 kg * v1' + 0.315 kg·m/s + 0.0900 kg * v1'
0.135 kg·m/s - 0.315 kg·m/s = 0.0600 kg * v1' + 0.0900 kg * v1'
Simplifying:
-0.180 kg·m/s = 0.150 kg * v1'
v1' = -0.180 kg·m/s / 0.150 kg
v1' = -1.20 m/s
Now, substitute this back into Equation 2 to find v2':
3.50 m/s = v2' - (-1.20 m/s)
3.50 m/s = v2' + 1.20 m/s
v2' = 3.50 m/s - 1.20 m/s
v2' = 2.30 m/s
So, the speed and direction of the first ball (m1) after the collision is 1.20 m/s in the opposite direction (negative value).
The speed and direction of the second ball (m2) after the collision is 2.30 m/s in the same direction as before the collision (positive value).
However, these calculated values do not match the given speeds of 0.70 m/s and 2.2 m/s. There may have been an error in either the given values or the calculation process. Please double-check your inputs and recalculate to confirm the correct answer.