Path Retraced: Find Condition for Elastic Collision

[prabhat rao]

A)find the condition such that a "projectile fired at an angle theta from an inclined plane of angle of inclination "alpha" on to another inclined plane of angle of inclination "beta" retraces its path after the first collision"?

NOTE : the collision is elastic, alpha not equal to beta. i hav attached the fig.

B)I used the conservation of momentum to do the problem
My try at the problem

c)Now consider the particle to be ejected at a velocity v at an angle of inclination theta to the plane. For the particle to retrace its on upon striking the inclined surface beta the direction of the velocities should be exactly reversed then itself it is going to retrace the path it came by.

Now we have velocities along the x and y direction to be
v_x = vcos (theta-alpha)

v_y = vsin(theta-alpha)

Now there is force acting on the particle is mg then how come it is a elastic collision. I will through by the energy aspect first consider that the ball goes to a maximum height, it gains potential energy and now it comes down to the same height then it loses potential energy. Therefore the net change is zero. Now the momentum aspect. Since the particle returns to its own speed . Now there is a momentum change but here e have to consider the system as the earth+configuartion present for the energy and the momentum to be conserved.

As explained earlier that the particle upon colliding with the inclined plane gets its direction (velocity) reversed. Now for the particle to come back and follow its own path

v_2 sin(m-beta) = v_y

v_2 cos(m-beta) = v_x

By energy conservation we have
1/2mv^2 = 1/2mv_2^2
Now we have v= v_2

and by the definition of an elastic collision if there is no any loss of translation K.E into any other form of energy. Otherwise the collision is not an elastic one

Now the momentum conservation
v cos(theta-alpha) = v_2 cos (m-beta)

Now we have
theta – alpha = m-beta
That gives m = theta-alpha+beta

So therefore the particle must strike the plane at an angle
m = theta- alpha+beta

So the neceesary condition is that
m = theta – alpha+beta

 

[mukundpa]

It is simple if we say that for the path to be retraced, the particle should collide normal to the surface of incline because if the particle hits in any other way the direction of velocity will make same angle with the normal like reflection of light (in case of elastic collision only)