Test: Motion in a Plane (June 25) - NEET MCQ

Given, 

 = 3

= 4

= 5

We have a relation,  

Squaring both side, we get 

⇒ (  )²  = ()²

⇒( )²+ ()² + 2 Cosθ = ()²

⇒ (3)² + (4)² + 2.3.4 Cosθ = (5)²

⇒ 9 + 16 + 24 Cosθ = 25

⇒ 24 Cosθ = 25

⇒ 24Cosθ = 25 - 25

⇒ θ = Cos^-1 (0)

⇒ θ = 90°

i.e π/2

 = 50 km h^-1 due North;

 = 50 km h^-1 due West. Angle between

= 50 km h^-1  due South

∴ Change in velocity

= 70.7 km/h

The direction of this change in velocity is in South-West.

Horizontal range is same when angle of projection is θ or (90° – θ).

When a vector is multiplied with a scalar, the result is a vector.

Horizontal range is same wh en an gle of projection with the horizonatal is θ and (90° – θ).

Since displacement is along the y-direction, hence displacement  

Work done

On circular motion, the force acts along the radius and displacement at a location is perpendicular to the radius i.e., θ = 90° As work done = 

A vector quantity has both magnitude and direction. In the given options, speed has only magnitude, therefore, it is non- vector or scalar quantity.

Kinetic energy is E

Velocity at highest point

[∴ at the highest point there is only horizontal velocity]

K.E. at highest point

Let after 1 sec angle become 60°. When the end A moves by 10 m left, the end B moves upward by BB' =    = 17.3 m/s

We know that, R² = A² + B² + 2 AB cosθ
(24)² = (12)² + (18)² + 2(12)(18) cos θ

Minimum speed with which the string is rotating in a vertical circle (v) = The minimum speed of stone is independent of mass of stone.

Given, u₁ = u₂ = u, θ₁ = 60º, θ₂ = 30º In 1^st case, we know that range

In 2^nd case, when θ₂ = 30º, then

[we get same value of ranges].

In the case of a body describing a vertical circle,

Tension is maximum when cosθ = +1 and velocity is maximum Both conditions are satisfied at θ = 0º (i.e. at lowest point B]