The vertical component of a vector is equal to its horizontal component.... | Physics

MOTION IN A PLANE

Very Short Question Answers

Q1. The vertical component of a vector is equal to its horizontal component. What is the angle made by the vector with X-axis?

Ans. Horizontal component $(𝒙)$ =Vertical component $(𝒚) = 1$

$a n g l e θ = t a n^{- 1} (\frac{y}{x})$

$θ = t a n^{- 1} (1)$

$θ = 45^{\circ}$

Q2. A vector v makes an angle ? with the horizontal. The vector is rotated through an angle ? .Does this rotation change the vector v ?

Ans. By rotating vector V through angle α its horizontal and vertical components change. Also, direction of the vector changes. So, the rotation changes the vector V.

Q3. Two forces of magnitudes 3 units and 5 units act at 60° with each other. What is the magnitude of their resultant?

Ans. $P = 3 U n i t s, Q = 5 a n d θ = 60^{\circ}$

$M a g n i t u d e R = \sqrt{P^{2} + Q^{2} + 2 P Q c o s θ}$

$R = \sqrt{3^{2} + 5^{2} + 2 \times 3 \times 5 \times c o s 60^{\circ}}$

$R = \sqrt{9 + 25 + 10}$

$R = \sqrt{49}$

$∴ R = 7$

Q4. A= i + j . W at is the angle between the vector and X-axis?

Ans. Horizontal component $(𝒙) =$ Vertical component $(𝒚) = 1$

$a n g l e θ = t a n^{- 1} (\frac{y}{x})$

$θ = t a n^{- 1} (1)$

$θ = 45^{\circ}$

Q5. When two right angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant?

Ans. $P = 7 U n i t s, Q = 24 a n d θ = 90^{\circ}$

^{$M a g n i t u d e R = \sqrt{P^{2} + Q^{2} + 2 P Q c o s θ}$}

^{$R = \sqrt{7^{2} + 24^{2} + 2 \times 7 \times 24 \times c o s 90^{\circ}}$}

^{$R = \sqrt{49 + 576 + 0}$}

^{$R = \sqrt{625}$}

^{$∴ R = 25$}

Q6. If P= 2i + 4j +14k and Q = 4i + 4j +10k find the magnitude of P + Q ?

Ans. \begin{gathered}

$\vec{P} = 2 i + 4 j + 14 k, \vec{Q} = 4 i + 4 j + 10 k$

$\vec{P} + \vec{Q} = 6 i + 8 j + 24 k$

$∴ | \vec{P} + \vec{Q} | = \sqrt{6^{2} + 8^{2} + 24^{2}}$

$= \sqrt{676}$

$| \vec{P} + \vec{Q} | = 26$

\end{gathered}

Q7. Can a vector of magnitude zero have nonzero components?

Ans. No, A vector of zero magnitude cannot have non-zero components.

Q8. What is the acceleration of projectile at the top if its trajectory?

Ans. The acceleration of projectile at the top of its trajectory is g = 9.8 ms^-1

Q9. Can two vectors of unequal magnitude add up to give the zero vector? Can three unequal vectors add up to give the zero vector?
Ans.

No. two vectors of unequal magnitude cannot add up to give zero vector.
Yes, three vectors of unequal magnitude can add up to give zero vector.

Short Question Answers

Q1. State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector?

Ans. Parallelogram law of vectors: if two vectors drawn from a point be represented adjacent sides of a parallelogram, then the diagonal drawn from the same point represents their resultant vector both in magnitude and direction.
As shown in the diagram $\vec{(O P)}$ and $\vec{(O Q)}$ drawn from a point be represented adjacent sides of the parallelogram and $\vec{(O M)}$ is diagonal of the parallelogram and their magnitudes are $A, B$ and $R$ respectively.
Magnitude: According to Pythagoras theorem,
From ∆ OMN
$(O M)^{2} = (O N)^{2} + (O Q)^{2}$
$R^{2} = (O P + P N)^{2} + (B s i n θ)^{2}$
$R^{2} = (O P)^{2} + (P N)^{2} + 2 (O P) (P N) + (B s i n θ)^{2}$
$R^{2} = A^{2} + (B c o s θ)^{2} + 2 (A) (B c o s θ) + (B s i n θ)^{2}$
$R^{2} = A^{2} + B^{2} + 2 A B c o s θ$
$R = \sqrt{A^{2} + B^{2} + 2 A B c o s θ}$
The above equation is represented magnitude of resultant vector.
Direction: From ∆ OMN
$t a n α = \frac{M N}{O N}$
$t a n α = \frac{M N}{(O P + P N)}$
$t a n α = \frac{B s i n θ}{A + B c o s θ}$
$α = t a n^{- 1} \frac{B s i n θ}{(A + B c o s θ}$
The above equation is represented direction of resultant vector.

Q2. What is relative motion? Explain it?

Ans. Relative motion: it is a motion of one body with respect to another body.

The relative velocity of body ‘A’ with respect to ‘B’ is given by $V_{A B} = | V_{A} - V_{B} | = \sqrt{{V_{A}}^{2} + {V_{B}}^{2} - 2 V_{A} V_{B} c o s θ}$
The relative velocity of body ‘B’ with respect to ‘A’ is given by $V_{B A} = | V_{B} - V_{A} | = \sqrt{{V_{A}}^{2} + {V_{B}}^{2} - 2 V_{A} V_{B} c o s θ}$
$V_{A B}$ and $V_{B A}$ are equal and in magnitude and opposite in direction
If ‘A’ and ‘B’ are opposite in direction(θ=180°) then $V_{r} = V_{A} - V_{B}$
If ‘A’ and ‘B’ are same in direction(θ=0°) then $V_{r} = V_{A} + V_{B}$
If ‘A’ and ‘B’ are perpendicular to each other in direction(θ=90°) then $V_{r} = \sqrt{{V_{A}}^{2} + {V_{B}}^{2}}$

Q3. Show that a boat must move at an angle of $90^{\circ}$ with respect to river water in order to cross the river in a minimum time?

Ans. River - Boat : As shown in the diagram, the width of the river is $d$ and the velocity of the river water is $V_{r w}$ . A boat velocity in the river water is $V_{b}$ and the angle between $V_{b}$ and $V_{r w}$ is $θ$ . Horizontal component $V_{b} c o s θ$ is acting in the direction of velocity $V_{r w}$ .Vertical component $V_{b} s i n θ$ is acting along the width $d$ . Therefore, time take by the boat to cross the river is $t = \frac{d}{(V_{b} s i n θ)}$ (t is minimum for θ=90°) .According to above equation, a boat move at an angle of 90° with respect to river water in order to cross the river in a minimum time.

Q4. Define unit vector, null vector and position vector?

Ans. Unit Vector : A vector whose magnitude is equal to one is known as a unit vector. $U n i t v e c t o r \hat{a} = \frac{\vec{a}}{| a |}$ Null Vector : A vector whose magnitude is equal to zero is known as a zero vector.
Position Vector: A vector which is drawn from the origin of a reference frame is known as position vector. This helps to locate the position of a body in space.
Position vector $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

Q5. If |a+b|=|a-b |, prove that the angle between a and b is 90°?
Ans.

$| a + b | = | a - b |$ $\sqrt{a^{2} + b^{2} + 2 a b c o s θ} = \sqrt{a^{2} + b^{2} - 2 a b c o s θ}$ $a^{2} + b^{2} + 2 a b c o s θ = a^{2} + b^{2} - 2 a b c o s θ$ $4 a b c o s θ = 0$ $c o s θ = 0$ $θ = 45^{\circ}$

Q6. Show that the maximum height and range of a projectile are $\frac{u ² s i n ² θ}{2 g}$ and $\frac{u ² s i n 2 θ}{g}$ respectively where the terms have their regular meanings?

Ans. i) At the maximum height of a trajectory $y = h_{m}, t = \frac{(V_{o} s i n θ)}{g}, a_{y} = - g, V_{o} y = V_{o} s i n θ$ . $h_{m} = V_{o} s i n θ (\frac{(V_{o} s i n θ)}{g}) + 1 / 2 (- g) (\frac{V_{o} s i n θ}{g})^{2}$ $h_{m} = \frac{(V_{o} s i n θ)^{2}}{2 g}$ ii) if projectile reaches the horizontal $x = R, T = (2 V_{o} s i n θ) / g, V_{o} x = V_{o} c o s θ, a_{x} = 0$ $x = V_{o} x t + 1 / 2 a_{x} t^{2}$ $R = V_{o} c o s θ (\frac{2 V_{o} s i n θ}{g})$ $R = \frac{V_{o}^{2} s i n 2 θ}{g}$

Q7. Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola?
Ans.

Let a body projected with an initial velocity

V_{o}

at an angle

θ

(other than

90 °

)with the horizontal axis , from the Origin

O

. The path of the body is called trajectory.

The resolutions of initial velocity are

V_{o x} = V_{o} c o s θ, V_{o y} = V_{o} s i n θ

. The horizontal component

(V_{o x})

is constant throughout the motion of the body. The vertical component

(V_{o y})

changes both in magnitude and direction due to gravity.

V_{o x} = V_{o} c o s θ

Let the projectile is at a point

P (x, y)

after a time interval

t

Displacement along the

X - a x i s

x = V_{o x} t + \frac{1}{2} a_{x} t^{2}

where

a_{x} = 0

x = V_{o x} t

x = (V_{o} c o s θ) t

t = \frac{x}{V_{o} c o s θ}

Displacement along the

Y - a x i s

y = V_{o y} t + \frac{1}{2} a_{y} t^{2}

where

a_{y} = - g

y = V_{o y} t - \frac{1}{2} g t^{2}

y = (V_{o} s i n θ) (\frac{x}{V_{o} c o s θ}) - \frac{g}{2} {(\frac{x}{V_{o} c o s θ})}^{2}

y = (t a n θ) x - (\frac{g}{2 V_{o}^{2} c o s^{2} θ}) x^{2}

y = A x - B x^{2}

Where

A = t a n θ

and

B = (\frac{g}{2 V_{o}^{2} c o s^{2} θ})

both are constants . The above equation represent a parabola . So, the trajectory of an object thrown at certain angle with the horizontal is a parabola.

Q8. Explain the terms the average velocity and instantaneous velocity. When are they equal?

Ans. Average velocity: The ratio between total displacement to the total time is called average velocity.

Average velocity $\overset{―}{V} = \frac{T o t a l D i s p l a c e m e n t}{T o t a l T i m e}$ $\overset{―}{V} = \frac{Δ x}{Δ t}$

Instantaneous velocity:The velocity of particle at any instant of time is called instantaneous velocity.

instantaneous velocity

V = lim_{t \to 0} \frac{Δ x}{Δ t}

= \frac{d x}{d t}

Average velocity and Instantaneous velocity are equal in uniform motion
Average velocity and Instantaneous velocity will differ in non-uniform motion

Q9. If the trajector of a body is parabolic in one reference frame, can it be parabolic in another reference frame that moves at constant velocity with respect to the first reference frame? If the trajectory can be other than parabolic, what else it can be?

Ans. No, In another frame of reference, the trajectory is a vertical straight line.

MOTION IN A PLANE

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