Question

A proton moving at a speed of 5 kilometre per second makes an elastic collision with an other proton initially at rest .given that after collision the moving proton moves at an angle 37 degree with speed v1 and proton at rest moves at angle of theta2 in downward direction with x axis with speed v-2 to find v1 ,v2 and theta 2​

Answer 1

Answer:

Apply principle of conservation of momentum along x-direction,

mu=mv

1

cos45+Mv

2

cos45

mu=

2

1

(mv

1

+Mv

2

) ...(1)

along y−direction,

o=mv

1

sin45−Mv

2

sin45

o=(mv

1

−Mv

2

)

2

1

...(2)

Coefficient of e=1=

ucos45

v

2

−v

1

cos90

Restution

⇒

2

u

v

2

=1

⇒u=

2

v

2

...(3)

solving eqn (1), (2), & (3) we get

M=m