Question

Particles *P* and *Q* move on a line of greatest slope of a smooth inclined plane. *P* is released from rest at a point *O* on the line and 2s later passes through the point *A* with speed `3.5ms^-1`. (i) Find the acceleration of *P* and the angle?

Particles P and Q move on a line of greatest slope of a smooth inclined plane. P is released from rest at a point O on the line and 2s later passes through the point A with speed `3.5ms^-1`.
(i) Find the acceleration of P and the angle of inclination of the plane.

At the instant that P passes through A the particle Q is released from rest at O . At time t s after Q is released from O, the particles P and Q are 4,9m apart.
(ii) Find the value of t

Answer 1

acceleration of P = `1.75 m/s^2` ;
Angle of inclination, `theta` = `sin^-1 (1.75 / g)`
t= 0.4 s

Explanation:

P is released from rest on the incline making an angle `theta`
with horizontal.

Initial velocity u=0 final velocity v = 3,5 m/s

time elapsed = 2 s

v= u + a. t

3.5 m/s = 0 + a . 2 s

acceleration along the incline a = (3.5/2) = 1.75 `m/s^2`

as P is travelling along the incline
the component of mg along incline mg sin`(theta)`

therefore the acceleration along the incline a = `g .sin(theta)`

g.sin`(theta)` = 1.75 `m/s^2`

therefore the angle of incline `Theta`= `sin ^-1(1.75/g)`

After 2 s the particle Q is released from O when particle P was at A;

OA = S = (1/2) .a.`t^2` = (1/2) . 1.75 . 4 = 3.5 m

at A the particle P is moving with speed 3.5 m/s and if Q is released at this instant ..both are moving with same acceleration

and after t sec the separation is 4.9 m
Additional distance covered = 4.9 - 3.5 = 1.4 m

therefore in t sec the particle P must have travelled the additional distance 1.4 m with the same speed 3.5 m/s ; therefore

t= 1.4 / 3.5 = 0.4 sec.

*as the two particle are moving with same acceleration
any additional distance covered by them due to accelerated motion
would be same and would not contribute in separation distance* .