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.]
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
======================================...
The answer is wrong in the first step itself. 20 u is the mass, not the momentum
To find the mass of the unknown atom, we can use the law of conservation of momentum and the laws of elastic collisions.
1. Let's assume the unknown atom has a mass of m2.
2. Before the collision, only the neon atom is in motion, so its initial momentum is given by p1 = m1 * v1, where m1 is the mass of the neon atom and v1 is its initial velocity. Since the unknown atom is at rest, its initial momentum is zero.
3. After the collision, the neon atom travels away at a 55.2° angle with its original direction, and the unknown atom travels away at a 51.8° angle. We can use the law of sines to relate their momenta.
Let p2 be the momentum of the neon atom after the collision and p3 be the momentum of the unknown atom after the collision.
sin(55.2°) / sin(51.8°) = p2 / p3
4. According to the law of conservation of momentum, the total momentum before the collision must be equal to the total momentum after the collision.
p1 = p2 + p3
Substitute the expressions for p2 and p3 using the law of sines from step 3.
m1 * v1 = (sin(55.2°) / sin(51.8°)) * p3 + p3
5. Now we can solve for the mass of the unknown atom, m2.
Rearranging the equation from step 4, we get:
p3 = (m1 * v1) / ((sin(55.2°) / sin(51.8°)) + 1)
The momentum of an object is given by p = m * v, where m is the mass and v is the velocity. Since the neon atom is moving away at an angle, we need to use the component of the velocity in the direction of motion for both atoms.
v1 = v_neon * cos(55.2°) (component of velocity in the direction of motion for neon atom)
Substitute the value of v1 in the equation for p3:
p3 = (m1 * v_neon * cos(55.2°)) / ((sin(55.2°) / sin(51.8°)) + 1)
Finally, the mass of the unknown atom, m2, is given by:
m2 = p3 / v2
where v2 is the velocity of the unknown atom.
Since both atoms are moving away at angles after the collision, we need to use the component of their velocities in the direction of motion for each atom.
6. Substitute the value of p3 in the equation for m2:
m2 = ((m1 * v_neon * cos(55.2°)) / ((sin(55.2°) / sin(51.8°)) + 1)) / v2
Now, you can plug in the values of m1, v_neon, v2, sin(55.2°), and sin(51.8°) to calculate the mass of the unknown atom, m2.