Question

A proton, moving with a velocity of viî, collides elastically with another proton that is initially at rest. If the speed of the initially moving proton is 1.90 times the speed of the initially at rest proton, find the following.

(a) the speed of each proton after the collision in terms of vi
(b) the direction of the velocity vectors after the collision

Answer 1

We can approach this problem using the conservation of momentum and the conservation of kinetic energy.

Conservation of momentum:

The momentum before the collision is given by:

p1 = m1v1 = m1viî

where m1 is the mass of each proton (which we assume to be the same) and v1 is the velocity of the initially moving proton.

The momentum after the collision is given by:

p2 = m1v1' + m1v2'

where v1' and v2' are the velocities of the two protons after the collision.

Since the collision is elastic, the total momentum is conserved:

p1 = p2

m1viî = m1v1' + m1v2'

We can simplify this equation by dividing both sides by m1:

viî = v1' + v2'

Conservation of kinetic energy:

The kinetic energy before the collision is given by:

K1 = (1/2)m1 We need the value of the velocity v1, which is not given in the problem statement.