To solve this problem, we can apply the principles of conservation of momentum and conservation of kinetic energy in an elastic collision.
Let's denote the lighter particle as particle A and the heavier particle as particle B.
Given:
Mass of particle A (ma) = 9.109e-31 kg
Initial velocity of particle A (va) = 20 m/s (in the x-direction)
Mass of particle B (mb) = 1.672e-24 kg
Initial velocity of particle B (vb) = 12 m/s (making an angle of 110 degrees with the x-direction)
Final angle made by particle A with the x-direction = 100 degrees
Since momentum is conserved in the x-direction, we have:
(ma * va) + (mb * vb) = (ma * vaf) + (mb * vbf)
(9.109e-31 kg * 20 m/s) + (1.672e-24 kg * 12 m/s) = (9.109e-31 kg * vaf) + (1.672e-24 kg * vbf) --(1)
Now, let's find the x and y components of the velocity of particle B:
vb_x = vb * cos(110 degrees) --> x-component of velocity of particle B
vb_y = vb * sin(110 degrees) --> y-component of velocity of particle B
vb_x = 12 m/s * cos(110 degrees) = -5.699 m/s (negative sign indicates velocity opposite to the x-direction)
vb_y = 12 m/s * sin(110 degrees) = 10.935 m/s (positive sign indicates velocity in the positive y-direction)
Since kinetic energy is also conserved in an elastic collision, we have:
(1/2 * ma * va^2) + (1/2 * mb * vb^2) = (1/2 * ma * vaf^2) + (1/2 * mb * vbf^2) --(2)
Substituting the x and y components of the velocities into equation (1), we have:
(9.109e-31 kg * 20 m/s) + (1.672e-24 kg * (-5.699 m/s)) = (9.109e-31 kg * vaf) + (1.672e-24 kg * vbf_x) --(3)
(9.109e-31 kg * 20 m/s) + (1.672e-24 kg * 10.935 m/s) = (9.109e-31 kg * vaf) + (1.672e-24 kg * vbf_y) --(4)
Now, using the information about the final angle made by particle A with the x-direction, we can calculate the final velocity components of particle A as follows:
vaf_x = vaf * cos(100 degrees) --> x-component of velocity of particle A
vaf_y = vaf * sin(100 degrees) --> y-component of velocity of particle A
Finally, we can solve equations (3) and (4) simultaneously to find vaf and vbf.