A Helium and Oxygen Momentum Problem?

Question

A helium atom (mass 4.0 u) moving at 601 m/s to the right collides with an oxygen molecule (mass 32 u) moving in the same direction at 402 m/s. After the collision, the oxygen molecule moves at 446 m/s to the right. What is the velocity of the helium atom after the collision?

Answer 1

The answerer above is RIGHT, but his cautioning the asker " its vector" is a bit shaky. That's why i bothered to write equations rather than theory for numerical.

momentum before = after (vector)
let (i) be unit vector along right (+x direction)
mh*vh (i) + mo*vo (i) = mh*vhf (n^) + mo*vof (i)
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here n^ is the direction for helium atom exiting at vhf
4*601(i) + 32*402(i) = 4*vhf (n^) +32*446(i)
(2404+12864) i - 14272 i = 4*vhf (n^)
4*vhf (n^) = 996 (i)
vhf (n^) = 249 (i)
this gives the velocity of helium atom (|vhf| = 249m/s) and its direction is still towards right n^ = i (+x)
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note : a lot of magnitude of helium atom's speed (from 601 to 249) is stolen by oxygen, but its own speed merely rose by 44. This is caused by higher mass of oxygen under conservation domain

Answer 2

Use the conservation of momentum principle. The total momemtum before the collision must be the same as the total momentum after the collision.

And remember that momentum is a vector, so be sure you remember whether your momentum values are to the "left" or the "right". That makes a difference.

Momentum before collision = (add up the two momentum vectors. You can do it.)

Momentum after collision = oxygen momentum vector (given in problem) plus Helium momentum vector (need to find).

Conservation of momentum says those two equations have the same value. So set them equal, then solve for the Helium momentum.