PQ = (u 2 sin 2 α )/ g cos θ

Sol.

Accelerations of particle and block are shown in figure.

Acceleration of particle with respect to block

= acceleration of particle – acceleration of block

= (g sin θ + g cos θ ) – ( g sin θ ) = g cos θ

Now motion of particle with respect to block

will be a projectile as shown.

The only difference is, g will be replaced by g cos θ

∴ PQ = Range (R) =

PQ = Ans.

Horizontal displacement of particle with respect to ground is zero. This implies that initial velocity of particle with respect to ground is only vertical, or there is no horizontal component of the absolute velocity of the particle.

Let v be the velocity of the block down the plane.

Velocity of particle with respect to block = u cos ( α + θ ) + u sin ( α + θ )

Velocity of block = – v cos θ – v sin θ

∴ Velocity of particle with respect to ground = { u cos ( α + θ ) – v cos θ } + { u sin ( α + θ ) – v sin θ }

Now as we said earlier with that horizontal component of absolute velocity should be zero.

Therefore,

u cos ( α + θ ) – v cos θ = 0

or ( down the plane )

Ans.