Conservation of Energy in Elastic Collisions: Solving for Initial Velocity

[joejoe333]

Hello all,

On our test today the question said, "An 2 Kg object is moving along the x-axis and explodes, with particle 1, 1 Kg, shooting north at 3 m/s, and particle 2, 1 Kg, traveling 30 degrees north of east at 5 m/s. What was the initial velocity of the object?"

Now I was talking with a friend, and we each came up with two answers:
Would you treat it like an elastic collision, and if so, would you use the kinetic energy or momentum equations?

kinetic energy: ((m1v1i)^2)/2 + ((m2v2i)^2)/2 = ((m1v1f)^2)/2 + ((m2v2f)^2)/2

(2)(v1^2)/2 = (4.5) + (12.5)

v1 = 4.12

OR

m1v1i = m1v1f + m2v2f
2vi = 3 + 5
vi = 4

OR

Would you turn the problem into a triangle; since the velocity is 3 m/s going straight north, and 5 m/s 30 degrees north of east, you can turn it into a right triangle, and the missing side would have to be 4 m/s.

This question has been bugging me all day, until I finally decided to find this site, register, and ask you guys! haha. The answer of 4.00 m/s seems more reasonable, but it's weird that when you use the kinetic energy conservation equation that it ends up as 4.12 and not 4.00. Thank you in advance!

-Joe

 

[radou]

There is no collision. It's a momentum conservation problem. Simply use that fact for both the x and y direction.

 

[HallsofIvy]

Except that, if the original object was moving on the x-axis, with no northward component of velocity, the two products of the explosion cannot have a net northward component as here.

Your calculation:
m1v1i = m1v1f + m2v2f
2vi = 3 + 5
vi = 4
is incorrect- you are treating this as if all velocities were due east. The velocities are not in the same direction.

 

[joejoe333]

So is 4.11 a wrong answer, as well? Sorry, I'm kind of a n00b, I'm just in 1081 and my professor is straight from China and has an extrememly heavy accent so I have no clue what he says when I talk to him. What you guys are saying, though, is that I would have to solve the both the y and x directions of momentum?

 

[OlderDan]

So is 4.11 a wrong answer, as well? Sorry, I'm kind of a n00b, I'm just in 1081 and my professor is straight from China and has an extrememly heavy accent so I have no clue what he says when I talk to him. What you guys are saying, though, is that I would have to solve the both the y and x directions of momentum?

There was no need to mention an x-axis in the problem. We tend to think of the x-axis pointing East because of the way we draw maps, but in this problem the x-axis could not possibly have been East. The answer is found, as others have said, by finding both components of momentum before and after the explosion, but since the final velocies are given relative to compass points, do the problem in terms of compass points and forget about x and y.