If momentum of a particle is given by p → = 10 sin ( 2 t ) i ^ + 10 cos ( 2 t ) j ^ a) Find the magnitude of force at any time. b) Find the angle between momentum and force.

The force can be given by the formula F → = d p → d t a) The force is, F → = d ( 10 sin ( 2 t ) i ^ + 10 cos ( 2 t ) j ^ ) d t ⇒ F → = 20 ( cos ( 2 t ) i ^ - sin ( 2 t ) j ^ ) The magnitude of force is F = 20 cos ( 2 t ) 2 + 20 sin ( 2 t ) 2 = 20 2 N b) The angle between momentum and the force is given by cos θ = p → · F → p F = 10 sin ( 2 t ) i ^ + 10 cos ( 2 t ) j ^ · 20 cos ( 2 t ) i ^ - 20 sin ( 2 t ) j ^ 20 2 10 2 ⇒ cos θ = 0 ⇒ θ = 90 ∘

If momentum of a particle is given by p → = 10 sin ( 2 t ) i ^ + 10 cos ( 2 t ) j ^ a) Find the magnitude of force at any time. b) Find the angle between momentum and force.

The force can be given by the formula F → = d p → d t a) The force is, F → = d ( 10 sin ( 2 t ) i ^ + 10 cos ( 2 t ) j ^ ) d t ⇒ F → = 20 ( cos ( 2 t ) i ^ - sin ( 2 t ) j ^ ) The magnitude of force is F = 20 cos ( 2 t ) 2 + 20 sin ( 2 t ) 2 = 20 2 N b) The angle between momentum and the force is given by cos θ = p → · F → p F = 10 sin ( 2 t ) i ^ + 10 cos ( 2 t ) j ^ · 20 cos ( 2 t ) i ^ - 20 sin ( 2 t ) j ^ 20 2 10 2 ⇒ cos θ = 0 ⇒ θ = 90 ∘

If A , B , C are three non-collinear points with position vectors a → , b → , c → , respectively, show that the length of the perpendicular from C on A B is a → × b → + b → × c → + c → × a → b → - a →

Consider △ A B C with side A B So, length of the perpendicular from C on A B is the height of the triangle, h We know that area of the triangle = 1 2 × base × height ⇒ h = Area of the triangle 1 2 A B ⇒ h = 1 2 A B → × A C → 1 2 A B → ⇒ h = b → - a → × c → - a → b → - a → ⇒ h = b → × c → - b → × a → - a → × c → + a → × a → b → - a → ⇒ h = a → × b → + b → × c → + c → × a → b → - a → Hence proved.

EASY

Earn 100

If momentum of a particle is given by $\vec{p} = 10 \sin (2 t) \hat{i} + 10 \cos (2 t) \hat{j}$

a) Find the magnitude of force at any time.

b) Find the angle between momentum and force.

Important Questions on Systems of Particles and Rotational Motion

EASY

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P}

, the angle between

\vec{P}

and

\vec{Q}

θ (0 ° < θ < 360 °)

. The value of

θ

will be ___

°

HARD

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k}

and

\vec{c} = \hat{i} - \hat{j}

and if

6 \hat{i} + 2 \hat{j} + 3 \hat{k} = λ_{1} (\vec{a} \times \vec{b}) + λ_{2} (\vec{b} \times \vec{c}) + λ_{3} (\vec{c} \times \vec{a}),

then

(λ_{1}, λ_{2}, λ_{3}) =

MEDIUM

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

The direction ratios of the line which is perpendicular to lines

\frac{x - 1}{2} = \frac{y + 17}{- 3} = \frac{z - 6}{1}

and

\frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 4}{- 2}

are

HARD

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

Let

\vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}

and

\vec{b} = 7 \hat{i} + \hat{j} - 6 \hat{k}

\vec{r} \times \vec{a} = \vec{r} \times \vec{b}, \vec{r} \cdot (\hat{i} + 2 \hat{j} + \hat{k}) = - 3,

then

\vec{r} \cdot (2 \hat{i} - 3 \hat{j} + \hat{k})

is equal to:

MEDIUM

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

The area of the parallelogram, whose diagonals are

2 \hat{i} - \hat{j} + \hat{k}

and

\hat{i} + 3 \hat{j} - \hat{k},

is equal to

EASY

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

If the angle between the vectors

\vec{P}

and

\vec{Q}

θ,

the value of

(\vec{Q} \times \vec{P}) \cdot \vec{P}

EASY

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

Consider the vectors

\vec{A} = (\hat{i} + \hat{j} - \hat{k})

\vec{B} = (2 \hat{i} - \hat{j} + \hat{k})

and

\vec{C} = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})

. What is the value of

\vec{C} \cdot (\vec{A} \times \vec{B})

EASY

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

If the area of the parallelogram with

\vec{a}

and

\vec{b}

as two adjacent sides is

15

sq. units then the area of the parallelogram having

3 \vec{a} + 2 \vec{b}

and

\vec{a} + 3 \vec{b}

as two adjacent sides in sq. units is

MEDIUM

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

If the vertices of

Δ ABC

are

A = (2, 3, 5), B = (- 1, 3, 2), C = (3, 5, - 2),

then the area of the

Δ ABC

(in sq. units) is

HARD

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

Let

D

and

E

be the midpoints of the sides

A C

and

B C

of a triangle

A B C

respectively. If

O

is an interior point of the triangle

A B C

such that

\vec{O A} + 2 \vec{O B} + 3 \vec{O C} = \vec{0},

then the area (in sq. units) of the triangle

O D E

MEDIUM

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

\vec{a}

is a unit vector, then

{|\vec{a} \times \hat{i}|}^{2} + {|\vec{a} \times \hat{j}|}^{2} + {|\vec{a} \times \hat{k}|}^{2} =

EASY

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

The magnitude of the projection of the vector

2 \hat{i} + 3 \hat{j} + \hat{k}

on the vector perpendicular to the plane containing the vectors

\hat{i} + \hat{j} + \hat{k}

and

\hat{i} + 2 \hat{j} + 3 \hat{k},

is:

MEDIUM

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

\vec{A}

and

\vec{B}

are two vectors satisfying the relation

\vec{A} \cdot \vec{B} = | \vec{A} \times \vec{B} |

. Then the value of

| \vec{A} - \vec{B} |

will be:

HARD

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

Let

\vec{a} = \hat{i} + α \hat{j} + 3 \hat{k}

and

\vec{b} = 3 \hat{i} - α \hat{j} + \hat{k}

. If the area of the parallelogram whose adjacent sides are represented by the vectors

\vec{a}

and

\vec{b}

8 \sqrt{3}

square units, then

\vec{a} \cdot \vec{b}

is equal to ___ .

MEDIUM

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

The area of the quadrilateral

A B C D,

where

A (0, 4, 1), B (2, 3, - 1), C (4, 5, 0)

and

D (2, 6, 2)

is equal to

MEDIUM

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

A, B, C

are three non-collinear points with position vectors

\vec{a}, \vec{b}, \vec{c}

, respectively, show that the length of the perpendicular from

C

A B

\frac{|(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})|}{|\vec{b} - \vec{a}|}

EASY

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}, \vec{|b|} = 6

and the angle between

\vec{a}

and

\vec{b}

\frac{π}{6},

then the area of the triangle (in square units) with

\bar{a}

and

\bar{b}

as two of its sides is

MEDIUM

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

\vec{a} = 2 \hat{i} + \hat{j} - 3 \hat{k}, \vec{b} = \hat{i} - 2 \hat{j} + \hat{k}, \vec{c} = - \hat{i} + \hat{j} - 4 \hat{k}

and

\bar{d} = \hat{i} + \hat{j} + \hat{k},

then

| (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) | =

HARD

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

Let

\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}

and

\vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k}

be two vectors. If a vector perpendicular to both the vectors

\vec{a} + \vec{b}

and

\vec{a} - \vec{b}

has the magnitude

12

then one such vector is:

EASY

Physics>Mechanics>Systems of Particles and Rotational Motion>The Vector Product of Vectors

Let

\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + 3 \hat{j} + 5 \hat{k}

and

\vec{c} = 7 \hat{i} + 9 \hat{j} + 11 \hat{k}

.

Then, the area of the parallelogram with diagonals

\vec{a} + \vec{b}

and

\vec{b} + \vec{c}

If momentum of a particle is given by p→=10sin(2t)i^+10cos(2t)j^ a) Find the magnitude of force at any time. b) Find the angle between momentum and force.

Important Questions on Systems of Particles and Rotational Motion

Learn and Prepare for any exam you want !

If momentum of a particle is given by $\vec{p} = 10 \sin (2 t) \hat{i} + 10 \cos (2 t) \hat{j}$

a) Find the magnitude of force at any time.

b) Find the angle between momentum and force.