A particle of mass m and initial velocity v collides elastically with a particle of mass M coming from opposite direction.After collision m has velocity V/2 at right angles to incident direction and M moves at an angle of 45 degree . Find M/m.

1/3

To find the ratio of M to m, we need to analyze the conservation of momentum and kinetic energy in the collision.

Let's consider the collision in the reference frame where particle M is initially at rest. In this reference frame, the equation of conservation of momentum can be written as:

m * v = M * u

where v is the magnitude of the initial velocity of particle m, and u is the magnitude of the velocity of particle m after the collision.

Now, let's analyze the motion of the particles after the collision. We are given that particle m has a velocity of V/2 at right angles to the incident direction. Let's denote this velocity as w, where w = V/2. By using Pythagorean theorem, we can find the magnitude of the velocity:

w^2 = (V/2)^2 + u^2

Simplifying the equation gives us:

u^2 = w^2 - (V/2)^2
u^2 = V^2/4 - V^2/4
u^2 = V^2/4

Now, since the collision is elastic, kinetic energy is conserved. The initial kinetic energy is given by (1/2) * m * v^2, and the final kinetic energy is the sum of the kinetic energies of the two particles:

(1/2) * m * v^2 = (1/2) * m * u^2 + (1/2) * M * w^2

Substituting the value of u^2 from the earlier equation, we get:

(1/8) * m * v^2 = (1/2) * m * (V^2/4) + (1/2) * M * (V/2)^2

Simplifying further:

v^2/8 = V^2/8 + M * V^2/8

Now, canceling common terms and rearranging the equation, we have:

v^2 = V^2 + M * V^2

Dividing both sides of the equation by V^2 and rearranging, we get:

1 = 1 + M

M = 0

So in this scenario, M/M equals zero.