Question

A neon atom (m=20.0u) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a 23.6 ?angle from its original direction and the unknown atom travels away at a -57.2 ? angle.

What is the mass (in u) of the unknown atom? [Hint: You can use the law of sines.]

Answer 1

Let u be the initial velocity of neon atom. in the x direction
Initial momentum before collision = 20 u.
After collision its momentum in the y direction is 20 v1 sin 23.6
= 8.006 v1 where v1 is its velocity after collision.

In the x direction its momentum is 20 v1 cos 23.6.=18.327 v1

The momentum of the unknown atom in the x direction = m v2 cos 57.2
=0.541 m v2 and in the y direcion is mv2 sin 23.6= 0.400 mv2
========================
8.006 v1 = 0.400 mv2
v1 = 0.0499 mv2
--------------------------------------...
20u= 18.327 v1 +0.541 m v2
u= 0.916 v1 + 0.02705 m v2
Substituting for v1
u= 0.916* 0.0499 m v2 + 0.02705 m v2
u = 0.0727 m v2
--------------------------------------...

Using conservation of kinetic energy
1/2* 20 u^2 =1/2 *20 v1^2 + 1/2* m v2^2
u^2 = v1^2 + ( m/20) v2^2
Substituting for v1 = 0.0499 mv2 and u 0.0727 m v2
0.03136 m^2 v2^2 =0.02490 m^2 v2^2 + (m/20) v2^2
0.00646 m^2 v2^2 = (m/20) v2^2
0.00646 m = (1/20) = 0.05
m = 77.399 u