Please see the attachment.© BrainMass Inc. brainmass.com March 4, 2021, 9:06 pm ad1c9bdddf
Please see the attachment.
This type of problems has a common approach, simply the masses and the initial velocities are different. Let us first derive the equations in common.
Let the masses of the particles be m1 and m2 and their initial velocities (along line joining) are u1 and u2 respectively (u1 must be > u2 for collision), and after collision moving with velocities v1 and v2 respectively (v2 >v1 ; separation)
1) If the collision is elastic. According to law of conservation of linear momentum and of kinetic energy m1u1 + m2u2 = m1v1 +m2v2 ............ (1) and
(1/2) m1u12 + (1/2) m2u22 = (1/2) m1v12 + (1/2) m2v22,
Gives m1u12 + m2u22 = m1v12 + m2v22 ......... (2)
To avoid square terms we can solve ...
Oblique collision of a proton with nucleus. The atomic mass of the nucleus and the loss of energy is calculated.