Question

Two particles at a distance 5 m apart, are thrown towards
each other on an inclined smooth plane with equal speeds
'V'. Inclined plane is inclined at an angle of 30° with the
horizontal. It is known that both particle move along the
same straight line. The particles collide at the point from
where the lower particle is thrown. Find the value of v.
(take g = 10 m/s']
(a) 1 m/s
(b) 1.5 m/s
(c) 2 m/s
(d) 2.5 m/s​

Answer 1

Given that,

Distance = 5 m

Speed = v

Angle = 30°

We know that,

Horizontal component of g is

$g = g\sin\theta$

Time at top,

At top v = 0

We need to calculate the time

Using equation of motion

$v = u-gt'$

$0=v-gt'$

$t'=\dfrac{v}{g}$

Put the value into the formula

$t'=\dfrac{v}{g\sin\theta}$

It is known that both particle move along the  same straight line.

So, the time of collision is

$t=2t'$

Put the value of t

$t=\dfrac{2v}{g\sin\theta}$

We need to calculate the value of speed

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$5=v\times\dfrac{2v}{g\sin\theta}+\dfrac{1}{2}g\sin\theta\times(\dfrac{2v}{g\sin\theta})^2$

$5=\dfrac{4v^2}{g\sin\theta}$

$v=\dfrac{5}{2}\ m/s$

$v=2.5\ m/s$

Hence, The value of v is 2.5 m/s.

(d) is correct option.