JEE PHYSICS MCQ: MOTION IN PLANE

1 Hard

1 of 1

Application +4 | -1

A body of mass

10 \mathrm{~kg}

is projected at an angle of

45^{\circ}

with the horizontal. The trajectory of the body is observed to pass through a point

(20,10)

. If

T

is the time of flight, then its momentum vector, at time

t=\frac{T}{\sqrt{2}}

, is ______. [Take

g=10 \mathrm{~m} / \mathrm{s}^2

]

View answer and explanation

Correct option: D

100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}

Option D (

100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}

) is correct because it correctly calculates the momentum vector at the specified time using projectile motion principles.

More Questions from MOTION IN PLANE

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

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

0^{\circ}

90^{\circ}

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

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

\mathbf{a}

\mathbf{b}

\theta

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

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

50 \mathrm{~m}

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