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°).

Answer 2

A) The speed of each proton after the collision in terms of vi is; v_f = (vi)/√2

B) The direction of the velocity vectors after the collision is;

45° from the i^ vector direction

Let the initial velocity of the first proton be u1.

Since the other proton is initially at rest then it's initial velocity is 0. Thus, u2 = 0

We are told that both protons have equal speed after collision which we will call v_f.

Let mass of proton be m.

This is an inelastic collision and we will use conservation of energy which is;

Kinetic energy before collision = kinetic energy after collision

Formula for kinetic energy is; K.E=½mv²

Thus;

½mu1² + ½m(0)² = ½(m + m)v_f²

>> ½mu1² = ½(2m)v_f²

½mu1² = mv_f²

Divide through by m to get;

½u1² = v_f²

u1² = 2v_f²

v_f = (u1)/√2

B) Let v1 be the final velocity for the first proton and let v2 be the final velocity for the second proton.

Using principle of conservation of momentum in the j^ vector direction gives;

Pi_y = Pf_y = 0

Now,

Pf_y = m•v1y +m•v2y

Thus;

m•v1y +m•v2y = 0

v1y = −v2y

We are told that the speed of the protons are equal and as such the magnitude of their speeds in the i^ vector direction must be equal. Thus;

v1x = v2x = v_f since v_f is the final equal speed.

Similar to the j^ direction, we apply conservation of momentum in the i^ vector direction to get;

Pi_x = Pf_y

Pi_x = m•u1

Pf_x = m•v1x + m•v2x

Thus;

m•u1 = m•v1x + m•v2x

Since v1x = v2x, then;

m•u1 = 2mv_f,x

So;

v_f,x = ½u1

From Pythagorean theorem, we can get the magnitude of vector vfy using the formula;

v_f² = (v_f,x)² + (v_f,y)²

Plugging in the relevant values gives;

½(u1)² = ¼(u1)² + (v_f,y)²

Thus;

v_f,y = ½u1

Thus;

v_f,y = v_f,x

Since v_f,y and v_f,x which are the i^ and j^ vector components of v_f are equal, then we can say that both protons will be deflected away at an angle of 45° from the i^ vector direction, with opposite j^ components.

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