Question

A proton (mass 1 u) moving at 9.00 10^(6) m/s collides elastically and head-on with a second particle moving in the opposite direction at 2.00 10^(6) m/s. After the collision, the proton is moving opposite to its initial direction at 5.40 10^(6) m/s. Find the mass and final velocity of the second particle. [Take the proton's initial velocity to be in the positive direction.]

Answer 1

(5-(-9))/(5-2(-2.8)-9)= 8.75

m= 8.75 u

(1*(5-(-9)))/(X+2.8)......X=-1.2....

V = -1.2e6 m/s

Answer 2

Here ,

u1 = 9*10^6 m/s

m1 = 1 u

u2 = -2*10^6 m/s

v1 = -5.4 *10^6 m/s

Using conservation of momentum

9*10^6 +m2*(-2*10^6) = 1*( -5.4 *10^6) + m2*v2 ----(1)

as collision is elastic

e = 1 = (v2 + 5.4)/(9 + 2)

v2 = 2.04 *10^6 m/s

the speed of second particle is 2.04 *10^6 m/s

putting in equation 1

m2 = 3.56 u

the mass of second particle is 3.56 u

Answer 3

momentum conservation

1* (9*10^6) - m ( 2*10^6) = 1* 5.4*10^6 - mVf

5.4 *10^6 + Vf = 11*10^6

Vf = 5.6*10^6 m/sec (ans)

1* (9*10^6) - m ( 2*10^6) = 1* 5.4*10^6 - m(5.6 *10^6)

3.6*10^6 m = -3.6 *10^6

m = -1u So another mass is electron (ans)