what are their velocities afterwards?

Question

If we have 2 masses both moving towards the right along the x axis, and mass 1 (1kg) has a velocity of 4m/s and to its right is mass 2 (2kg) with a velocity of 1m/s. After the two masses collide elastically, what are their velocities afterwards? Also compute the velocity of the center of mass for both before and after the collision. (Hint they should be the same).

Answer 1

OK, lexrunne… gave it a good beginning but I disagree with the advice about ignoring units. The confusion came about because of a minor error with units. If you carefully watch the units, they can help keep you from going off track.

I'll revise lexrunne'…'s answer because I liked the beginning. I'll star '*' lines that I have changed.

Conserve energy and momentum, since this is an elastic collision.

Conserving momentum:
m1v1 + m2v2 = m1v'1 + m2v'2
where v' represents final velocities.
1 kg (4 m/s) + 2 kg (1 m/s) = 1 kg (v'1) + 2 kg (v'2)
* 6 kg m/s = 1 kg v'1 + 2kg (v'2)
*This error (left kg out of 2nd term) is why things later got confusing for lexrunne….
*Cancel out the kgs
*[6 m/s = 1 (v'1) + 2 (v'2)]

Conserving energy:
v1 - v2 = v'2 - v'1
[ 4 m/s - 1 m/s = v'2 - v'1 ]

Now, take the 2 equations in brackets and solve by simulataneous equations.

*6 m/s = 1 (v'1) + 2 (v'2)
4 m/s - 1 m/s = v'2 - v'1

* Add. You're left with 9 = 3v'2. Therefore, v'2 = 3 m/s.

Substitute this back into 4 m/s - 1 m/s = v'2 - v'1 and solve for v'1.
3 = 3 - v'1, so v'1 = 0 m/s.

You also asked for velocity of the center of mass. The c-of-m was at the point of impact. Mass m1 stopped. 1 second later m2 is 3 meters to the right of the collision point. To find the c-of-m:
Xc-of-m = (m1*x1 + m2*x2) / (m1 + m2)
Xc-of-m = (0 + 2 kg*3 m) / (1 kg + 2 kg)
Xc-of-m = 6 kg m / 3 kg = 2 m

Answer 2

Conserve energy and momentum, since this is an elastic collision.

Conserving momentum:
m1v1 + m2v2 = m1v'1 + m2v'2
where ' represents final velocities.
1 kg (4 m/s) + 2 kg (1 m/s) = 1 kg (v'1) + 2 kg (v'2)
[ 6 kg m/s = v'1 + 2kg (v'2) ]

Conserving energy:
v1 - v2 = v'2 - v'1
[ 4 m/s - 1 m/s = v'2 - v'1 ]

Now, take the 2 equations in brackets and solve by simulataneous equations.

6 kg m/s = v'1 + 2kg (v'2)
4 m/s - 1 m/s = v'2 - v'1

Add. You're left with 9 = 3v'2 (ignore units... they get too confusing here). Therefore, v'2 = 3 m/s.

Substitute this back into 4 m/s - 1 m/s = v'2 - v'1 and solve for v'1.
3 = 3 - v'1, so v'1 = 0 m/s.