JEE PHYSICS MCQ: MOTION IN PLANE

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If a particle is projected with speed

{u}

from the ground at an angle

\theta

with horizontal. Radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by

View answer and explanation

Correct option: B

\frac{u^2 \cot ^2 \theta}{g \sin \theta}

Option B (

\frac{u^2 \cot ^2 \theta}{g \sin \theta}

) is correct. The radius of curvature is calculated using centripetal acceleration and the velocity at the point where the velocity vector is perpendicular to the initial velocity vector.

More Questions from MOTION IN PLANE

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Given,

\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B.

The angle between

\mathrm{\vec A}\,and\,\mathrm{\vec B}

is Assertion: The necessary condition for motion in two dimensions, velocity and acceleration vectors should have any angle between

0^{\circ}

and

90^{\circ}

between them.
Reason: For two-dimensional or three-dimensional motion, velocity and acceleration of an object works in opposite direction.

\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0

and

(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0

. Then, the angle between

\mathbf{a}

and

\mathbf{b}

is At the top of the trajectory of a projectile, the direction of its velocity and acceleration are A ball is thrown at an angle

\theta

and another ball is thrown at an angle

\left(90^{\circ}-\theta\right)

with the horizontal direction from the same point with velocity

39.2 \mathrm{~ms}^{-1}

. The second ball reaches

50 \mathrm{~m}

higher than the first ball. Find their individual heights. Take

{g}=9.8 \mathrm{~ms}^{-2}

. Assertion: The drover in a vehicle moving with a constant speed on a straight road is in a non-inertial frame of reference.

Reason: A reference frame in which Newton's laws of motion are applicable is non-inertial.