Question

LEVEL 2
Objective Questions
Single Correct Option
When a man moves down the inclined plane with a constant speed 5 ms which makes an angle
37° with the horizontal, he finds that the rain is falling vertically downward. When he moves
up the same inclined plane with the same speed, he finds that the rain makes an angle
otan with the horizontal. The speed of the rain is
(a) V116 ma-
(c) 6 ms
(b) 32 ma-
(d) 73 mg​

Answer 1

Answer:

1)

When the man is moving down the incline plane, the relative velocity of the rain is in the -y direction. When the man is moving up the plane, the relative velocity of the rain is tan⁻¹ (7/8), or 41.2º (with the horizontal, I assume).

Relative velocity is the actual velocity of the rain minus the velocity of the man. So the actual velocity is the sum of the relative velocity and the man's velocity.

If we break up each velocity into its x and y components:

vman = (-5 m/s cos 37º) î + (-5 m/s sin 37º) ĵ

vrel₁ = -vrel₁ ĵ

vactual = (-5 m/s cos 37º) î + (-5 m/s sin 37º - vrel₁) ĵ

If we repeat this for when the man is going up the inclined plane:

vman = (5 m/s cos 37º) î + (5 m/s sin 37º) ĵ

vrel₂ = (-vrel₂ cos 41.2º) î + (-vrel₂ sin 41.2º) ĵ

vactual = (5 m/s cos 37º - vrel₂ cos 41.2º) î + (5 m/s sin 37º - vrel₂ sin 41.2º) ĵ

We now have two equations for the actual velocity of the rain. If we match the x and y components of each and set them equal, we can solve for the magnitudes:

-5 m/s cos 37º = 5 m/s cos 37º - vrel₂ cos 41.2º

vrel₂ cos 41.2º = 10 m/s cos 37º

vrel₂ = (10 m/s cos 37º) / (cos 41.2º)

vrel₂ = 10.6 m/s

We can repeat in the y direction, but we already have enough information to answer the question. Plugging in the value into our second equation for vactual:

vactual = (5 m/s cos 37º - vrel₂ cos 41.2º) î + (5 m/s sin 37º - vrel₂ sin 41.2º) ĵ

vactual = (5 m/s cos 37º - 10.6 m/s cos 41.2º) î + (5 m/s sin 37º - 10.6 m/s sin 41.2º) ĵ

vactual = -3.99 m/s î + -3.98 ĵ

So the magnitude is:

|| vactual || = √( (-3.99 m/s)² + (-3.98 m/s)² )

|| vactual || = 5.64 m/s

Or approximately √32 m/s

Answer 2

Thus the speed of train is √ 32 m/s

Option (b) is correct.

Explanation:

We are given that:

• Angle = 37° with horizontal
• Speed of man = 5 m/s

Solution:

Velocity of train = V  R  =a  i^  −b  j ^

In vertical  direction

a + 4 = 0 or a = −4

In second case:

V M  = (a−4)  i ^  +(−b−3  j ^  )

=−8  i ^  +(−b−3  j ^  )

θ = tan^−1  (   7/8  )

− b − 3 = −7

- b = - 7 + 3

- b = - 4

Minus sign get cancelled on both sides.

Speed of train = √ a^2 + b^2 = √ 32 m/s

Thus the speed of train is √ 32 m/s