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°).
