Question

A proton, moving with a velocity of vii⁄, collides elastically with another proton that is initially at rest. Assuming that the two protons have equal speeds after the collision, find (a) the speed of each proton after the collision in terms of vi and (b) the direction of the velocity vectors after the collision.

Answer 1

Answer:

Explanation:

Given that,

Mass of proton

Mp=m=1.627×10^-27kg

One of the proton is at rest then it velocity is 0

Let the other proton be moving at vi

Since the after collision the two proton moves together with the same velocity(i.e inelastic collision)

Then, using conservation of energy

Kinetic energy before collision = kinetic energy after collision

Given that, K.E=½mv²

Before collision = after collision

½mvi²+½m(0)²=½(m+m)vf²

½mvi²=½(2m)vf²

½mvi²=mvf²

Divide through by m

½vi²=vf²

vi²=2vf²

Take square root of both sides

√vi²=√(2vf²).

vi=√2 ×vf

Then, the final velocity is

vf = vi /√2

b. Direction of the velocity vectors after collision

Let vf1 be the final velocity for the incident proton

vf2 be the final velocity for the proton initially at rest.

Conserving momentum in the ˆj direction

Piy = 0 = Pfy = mp•vf1y +mp•vf2y

vf1y = −vf2y

So the protons have equal magnitude speeds in the ˆj direction. Because the speed of the particles are equal, the magnitude of their speeds in the ˆi direction should also be equal |vf1x| = |vf2x|.

Conserving momentum in the ˆi direction.

Pix = mp•vi = Pfx = mp•vf1x +mp•vf2x = 2mp•vfx

Then, vfx =vi/2

Using the Pythagorean theorem to solve for the magnitude of vfy

vf²=vi²/2= vfx² +vfy² =vi²/4+vfy²

vfy = vi√(½-¼)=vi/2=vfx

So because the ˆi and ˆj components of vf are the same, both protons are deflected away at an angle of θ = 45° from the ˆi direction, with opposite ˆj components (so the angle between vf1 and vf2 is 90°).