A neon atom (m = 20.0 u) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a 55.2° angle from its original direction and the unknown atom travels away at a 51.8° angle. What is the mass of the unknown atom? [Hint: You could use the law of sines.]
4 answers
Idk to be honest
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 58.5
= 17.05 v1 where v1 is its velocity after collision.
In the x direction its momentum is 20 v1 cos 58.5.=10.45 v1
The momentum of the unknown atom in the x direction = m v2 cos 48.3
=0.67 m v2 and in the y direcion is mv2 sin 48.3= 0.75 mv2
========================
17.05 v1 = 0.75 mv2
v1 = 0.044 mv2
--------------------------------------...
20u= 10.45 v1 +0.67 m v2
u= 0.52 v1 + 0.033 m v2
Substituting for v1
u= 0.52* 0.044 m v2 + 0.033 m v2
u = 0.056 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 = 0.044 mv2 and u 0.056 m v2
0.00314 m^2 v2^2 =0.001936 m^2 v2^2 + (m/20) v2^2
0.0012 m^2 v2^2 = (m/20) v2^2
0.0012 m = (1/20) = 0.05
m = 41.67 u
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Initial momentum before collision = 20 u.
After collision its momentum in the y direction is 20 v1 sin 58.5
= 17.05 v1 where v1 is its velocity after collision.
In the x direction its momentum is 20 v1 cos 58.5.=10.45 v1
The momentum of the unknown atom in the x direction = m v2 cos 48.3
=0.67 m v2 and in the y direcion is mv2 sin 48.3= 0.75 mv2
========================
17.05 v1 = 0.75 mv2
v1 = 0.044 mv2
--------------------------------------...
20u= 10.45 v1 +0.67 m v2
u= 0.52 v1 + 0.033 m v2
Substituting for v1
u= 0.52* 0.044 m v2 + 0.033 m v2
u = 0.056 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 = 0.044 mv2 and u 0.056 m v2
0.00314 m^2 v2^2 =0.001936 m^2 v2^2 + (m/20) v2^2
0.0012 m^2 v2^2 = (m/20) v2^2
0.0012 m = (1/20) = 0.05
m = 41.67 u
======================================...
The answer is wrong in the first step itself. 20 u is the mass, not the momentum
Funny's response stated that u is the initial velocity, thus 20 u is the momentum. However, he should have used another variable because it is confusing as the unit of mass in this question is also u.
1 answer