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 30.8 ? angle from its original direction and the unknown atom travels away at a -54.7 ? angle. What is the mass (in u) of the unknown atom?

Answer 1

let mass =k u

let velocity of neon atom before collision be v1.

let velocity of neon atom after collision is v2.

let velocity of unknown mass after collision be v3.

conserving momentum along perpendicular to the original direction:

0=20*v2*sin(30.8)+k*v3*sin(-54.7)

==>v2=0.08*k*v3...(1)

conserving momentm along the direction of motion:

20*v1=20*v2*cos(30.8)+k*v3*cos(54.7)

==>v1=0.0976*k*v3...(2)

conserving kinetic energy:

0.5*20*v1^2=0.5*20*v2^2+0.5*k*v3^2

==>0.19055*k^2*v3^2=0.128*k^2*v3^2+0.5*k*v3^2

dividing by k^2*v^2

0.19055=0.128+(0.5/k)

==>k=7.9936

hencemass of the unknown atom is 8 u