Three masses are positioned on a frictionless surface, as shown. Initially, mass m1 (1.0 kg) moves with a velocity of 2.0 m/s to the right, mass m2 (2.0 kg) is at rest, and mass m3 (3.0 kg) moves to the left with a velocity of 0.50 m/s. First, mass m1 collides elastically with mass m2 and recoils to the left. Afterwards, mass m3 collides with mass m2 and sticks. Calculate,
- the speeds of masses m1 and m2 after the first collision.
- the speeds of masses m2 and m3 after the second collision.
M1V1 + M2V2 = M1V1' + M2V2'
1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2[/B]
The Attempt at a Solution
First I put in the values and narrowed the conservation of momentum equation to:
2 = v1' + (2) * v2' which rearranges to v1' = 2 - (2)v2' (equation 1.1)
For the Conservation of Energy equation I did the same and got,
2 = (0.5) v1'^2 + v2'^2 (equation 1.2)
I then substituted the value of v1' from equation 1.1 into 1.2 and simplified to,
v2' (3v2' -4) = 0
v2' = 0 or 1.33
since it is not 0 it must be 1.33 m/s
I then substitute this answer back into equation 1.1 and get
v1' = -0.667
Therefore, the velocity of m1 after the collision is 0.67 m/s to the left and the velocity of m2 = 1.3 m/s to the right.
for this part it is an inelastic collision and therefore only momentum is conserved.
m2v2 + m3v3 = (m2 +m3) v2and3
m2 = 2.0 kg
v2 = 1.3 m/s
m3 = 3.0 kg
v3 = -0.50 m/s
but these into the above equation and i get
((2.66 kg * m/s) + (-1.5 kg * m/s) / 5.0 kg) = v2and3
v2and3 = 0.23 m/s (positive indicates to the right)
Does this look correct? I made an earlier attempt but with a different answer but it didn't sit right so I retried this way.