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  $( 𝒚)\,=1 \,$

$angle\:\:\theta=tan^{-1}\left(\:\frac{y}{x}\right)$

$\theta=tan^{-1}\left(\:1\right)$

$\theta=45^{\circ}$

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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.

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

Ans. $P = 3 \,\,Units , \:\:Q =  5\:\: and\:\: θ = 60^{\circ}$

$Magnitude \,\,  R=\sqrt{P^{2}+Q^{2}+2PQ\,cos\theta}$

$R=\sqrt{3^{2}+5^{2}+2\times 3\times 5 \times cos60^{\circ }}$

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

$R=\sqrt{49}$

$\therefore R= 7$

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

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

$angle\:\:\theta=tan^{-1}\left(\:\frac{y}{x}\right)$

$\theta=tan^{-1}\left(\:1\right)$

$\theta=45^{\circ}$

 

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

Ans. $P = 7 \,\,Units , \,\,Q =  24\,\, and \,\,θ = 90^{\circ}$

^{$Magnitude \,\,  R=\sqrt{P^{2}+Q^{2}+2PQ\,cos\theta}$}

^{$R=\sqrt{7^{2}+24^{2}+2\times 7\times 24 \times cos90^{\circ} }$}

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

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

^{$\therefore R= 25$}

 

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

Ans. \begin{gathered}

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

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

$\therefore\left|\,\vec{P}+\vec{Q}\right|=\sqrt{6^{2}+8^{2}+24^{2}}$

$ =\sqrt{676}$

$\left|\,\vec{P}+\vec{Q}\right|=26$

\end{gathered}

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

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

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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

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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.

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

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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{(OP)} $and  $\vec{(OQ)} $ drawn from a point be represented adjacent sides of the parallelogram and $\vec{(OM)} $ is diagonal of the parallelogram and their magnitudes are$ A, B$ and $R$ respectively.
    Magnitude: According to Pythagoras theorem,
From ∆ OMN
$(OM)^{2}=(ON)^{2}+(OQ)^{2}$
$R^{2}=(OP+PN)^{2}+(B sin⁡θ)^{2} $
$R^{2}=(OP)^{2} +(PN)^{2}+2(OP)(PN)+(B sin⁡θ)^{2} $
$R^{2}=A^{2} +(B cos⁡θ)^{2}+2(A)(B  cos⁡θ)+(B sin⁡θ)^{2} $
$R^{2}=A^{2}+B^{2} +2AB  cos⁡θ $
$R=\sqrt{A^{2}+B^{2}  +2AB cos⁡θ } $
The above equation is represented magnitude of resultant vector.
Direction: From ∆ OMN
$tan⁡α=\frac{MN}{ON}$
$tan⁡α=\frac{MN}{(OP+PN)}$
$tan⁡α=\frac{B sin⁡θ}{A+B cos⁡θ }$
$α=tan^{-1} \frac{⁡B sin⁡θ}{(A+B cos⁡θ }$
The above equation is represented direction of resultant vector.

 

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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 $V_{AB}=|V_{A}-V_{B} |=\sqrt{{ V_{A}}^2+{V_{B}}^2-2V_{A} V_{B}  cos⁡θ }$
•  The  relative velocity of body ‘B’ with respect to ‘A’ is given by  $V_{BA}=|V_{B}-V_{A} |=\sqrt{{ V_{A}}^2+{V_{B}}^2-2V_{A} V_{B}  cos⁡θ }$
•  $V_{AB}$ and $V_{BA}$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 $V_{r}=\sqrt{ {V_{A}}^2+{V_{B}}^2 }$
   

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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$V_{rw}$. A boat velocity in the river water is $ V_{b}$  and  the angle between $V_{b}$ and  $V_{rw}$ is $θ$. Horizontal component  $V_{b}  cos⁡θ$ is acting in the direction  of  velocity$V_{rw}$.Vertical component $ V_{b}  sin⁡θ$ is acting along the width $d$. Therefore, time take by the boat to cross the river is $$t= \frac{d}{(V_{b}  sin⁡θ )} $$ (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. 

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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.$$ Unit \,\:vector \,\, \hat{a}=\frac{\vec{a}} {|a|} $$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 $\vec{r} =x\hat{i}+y\hat{j}+z\hat{k}$

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

Ans.

$$|a+b|= |a-b| $$ $$ \sqrt{a^{2}+b^{2}  +2ab cos⁡θ }=\sqrt{a^{2}+b^{2}  -2ab cos⁡θ }$$ $$  a^{2}+b^{2}  +2ab cos⁡θ = a^{2}+b^{2}  -2ab cos⁡θ $$ $$   4ab cos⁡θ = 0 $$ $$  cos⁡θ = 0 $$ $$ θ = 45^{\circ}$$

 

 

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Q6. Show that the maximum height and range of a projectile are\( {u²sin²\theta \over 2g} \) and \( {u²sin2\theta \over g} \) respectively where the terms have their regular meanings?

Ans. i)  At the  maximum height of a trajectory \( y=h_m, t={(V_o  sin⁡θ )\over g} , a_y=-g, V_oy =V_o  sin⁡θ \).$$ h_m=V_o  sin⁡θ  ({(V_o  sin⁡θ)\over g})+1/2(-g)({V_o  sin⁡θ \over g})^2 $$$$ h_m={ (V_o  sin⁡θ  )^2\over 2g} $$ ii) if projectile reaches the horizontal  \(x=R ,     T=(2 V_o  sin⁡θ)/g, V_ox=V_o  cos⁡θ, a_x=0 \)$$ x=V_ox t+1/2 a_x t^2$$$$ R=V_o  cos⁡θ( {2 V_o  sin⁡θ \over g})$$ $$ R={ V_o^{2}  sin2⁡θ \over g} $$

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

Ans.

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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$$ \overline{V} =\frac{Total Displacement}{Total Time}$$$$ \overline{V} =\frac{Δx}{Δt}$$

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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.

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Problems Question Answers