Physics 4 Marks Question

1. (c) they have the same magnitude and the same direction
2. (a) Change in magnitude
3. Null vector is defined as the vector having zero magnitude and any direction. Consider two vectors A and –A. Their sum is A + (–A). Since the magnitudes of the two vectors are the same, but the directions are opposite, the resultant vector has zero magnitude and is represented by 0 called a null vector or a zero vector. . Properties of 0 are

A + 0 = A

λ 0 = 0

4. Subtraction of vectors can be defined in terms of addition of vectors. We define the difference of two vectors A and B as the sum of two vectors A and –B:

 A – B = A + (–B)

5. Following are properties of vectors

a) Two vectors A and B are said to be equal if, and only if, they have the same magnitude and the same direction.

b) Multiplying a vector A with a positive number λ gives a vector whose magnitude is changed by the factor λ but the direction is the same as that of A :

$|\ \lambda\text{ A }|=\lambda\text{ A }|$

c) The null vector also results when we multiply a vector A by the number zero. Properties of 0 are

A + 0 = A

λ 0 = 0

0 A = 0

d) Subtraction of vectors can be defined in terms of addition of vectors. We define the difference of two vectors A and B as the sum of two vectors A and –B :

A – B = A + (–B)

1. (a) Ax = A cos(q)
2. (b) Ax = A sin(q)
3. Position vector- The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame is given by r = x i + y j where x and y are components of r along x-, and y- axes. The displacement is: Δr = r₂-r₁. We can write this in a component form:

Δr = (x’ i + y’ j) – ( x i + y j)

= iΔx – jΔy

Where Δx = x’ – x,  Δy = y’ – y.

4. The average velocity (v) of an object is the ratio of the displacement and the corresponding time Interval.

$\text{V}=\frac{\triangle\text{r}}{\triangle\text{t}}$

$=\frac{\text{i}\triangle\text{x}-\text{j}\triangle\text{y}}{\triangle\text{t}}$

$=\text{i}\times\frac{\triangle\text{x}}{\triangle\text{t}}+\text{j}\times\frac{\triangle\text{y}}{\triangle\text{t}}$

$=\text{V}_\text{x}\text{i}+\text{V}_\text{y}\text{j}$

5. If A and $\theta$ are given, Ax and Ay can be obtained using If Ax and Ay are given, A and $\theta$ can be obtained as follows

$\text{A}^2_\text{x}+\text{A}^2_\text{y}=(\text{A}\cos\theta)^2+(\text{A}\sin\theta)^2$

$\text{A}^2_\text{x}+\text{A}^2_\text{y}=\text{A}^2\cos^2\theta+\text{A}^2\sin^2\theta$

$\Rightarrow\text{A}^2_\text{x}+\text{A}^2_\text{y}=\text{A}^2(\cos^2\theta+\sin^2\theta)$

${A}^2_\text{x}+\text{A}^2_\text{y}=\text{A}^2(\because\sin^2\theta+\cos^2\theta=1)$

$\text{A}^2=\text{A}^2_\text{y}+\text{A}^2_\text{y}$

$\Rightarrow\text{A}=\sqrt{\text{A}^2_\text{x}+\text{A}^2_\text{y }}...$

$\text{Dividing}\text{ A}_\text{y}\text{ by} \text{ A}_\text{y},\text{we get}$

$\frac{\text{A}_\text{y}}{\text{A}_\text{x}}=\frac{\text{A}\sin\theta}{\text{A}\cos\theta}$

$\Rightarrow\frac{\text{A}_\text{y}}{\text{A}_\text{x}}=\tan\theta$

$\tan\theta=\frac{\text{A}_\text{y}}{\text{A}_\text{x}}$

$\theta=\tan^{-1}\Big[\frac{\text{A}_\text{y}}{\text{A}_\text{x}}\Big]$

1. (c) 9.8m

Explanation:

Since, they are following freely, so both the bodies will fall same distance in same time
interval.

So, the relative separation between them will remain unchanged.

2. (c) v_P + v₀

Explanation:

Relative velocity of P w.r.t. Q is given by

v_PQ = v_p - (-v_Q) = v_p + v_Q

3. (b) West - North direction

Explanation:

Velocity of car w.r.t. train, v_d = v_c - v_t

⇒ va = vc + (-v_t)

Velocity of car w.r.t. train (v_d) ct is towards West-North.

4. (b) $5\hat{\text{i}}\text{ms}^{-1}$

Explanation:

Given, $\text{v}_{\text{A}}=20\hat{\text{i}}\text{ms}^{-1}$

$\text{v}_{\text{B}}=15\hat{\text{i}}\text{ms}^{-1}$

Relative velocity of A w.r.t. B,

$\text{v}_{\text{AB}}=\text{v}_\text{A}-\text{v}_\text{B}=20\hat{\text{i}}-15\hat{\text{i}}=5\hat{\text{i}}\text{ms}^{-1}$

5. (b) $5\sqrt5\text{ms}^{-1}$

Explanation:

Given, velocity of girl, $\text{v}_{\text{g}}=5\hat{\text{i}}\text{ms}^{-1}$

Let velocity of rain, $\text{v}_{\text{r}}=\text{v}_{\text{x}}\hat{\text{i}}+\text{v}_{\text{y}}\hat{\text{j}}\text{ms}^{-1}$

Relative velocity of rain

$=\text{v}_{\text{r}}-\text{v}_{\text{g}}=(\text{v}_{\text{x}}-5)\hat{\text{i}}+\text{v}_{\text{y}}\hat{\text{j}}$

Now, it is vertical, so 

$\tan\theta=\frac{\text{v}_{\text{x}}-5}{\text{v}_{\text{y}}}=0$

⇒ v_x - 5 = 0

⇒ v_x = 5 ....(i)

On increasing the speed of the girl, relative velocity becomes 

$(\text{v}_{\text{x}}-5)\hat{\text{i}}+\text{v}_{\text{y}}\hat{\text{j}}$

$\tan\theta=\tan45^\circ=\frac{\text{v}_{\text{x}}-15}{\text{v}_{\text{y}}}=1$

⇒ v_x - 15 = v_y

⇒ v_y = -10 [using Eq. (i)]

$\therefore\text{Velocity of rain}=(5\hat{\text{i}}-10\hat{\text{j}})\text{ms}^{-1}$

⇒ Magnitude of velocity of rain

$=\sqrt{(5)^2+(10)^2}$

$=\sqrt{125}=5\sqrt{5}\text{ms}^{-1}$

1. (c) Temperature

Explanation:

Temperature is not a vector quantity because it has magnitude only.

However, force, acceleration and velocity have both a magnitude and a direction.

So, these are vectors in nature.

2. (c) A and B

Explanation:

Two vectors are said to be equal, if and only if they have the same magnitude and direction.

Among the given vectors A and B are equal vectors as they have same magnitude (length) and direction.

However, P and Q are not equal even though they are of same magnitude because their directions are different.

3. (c) $\lambda>0$

Explanation:

$\mid\lambda\text{A}\mid=\lambda\mid\text{A}\mid,$ if $\lambda>0$ as multiplication of vector A with a positive number $\lambda$ gives a

vector whose magnitude is changed by the factor $\lambda$ but the direction is same as that of A.

4. (b) $\lambda0=\lambda$

Explanation:

Null vector 0 is a vector, whose magnitude iszero and its direction cannot be specified.

So, it means, $\mid0\mid=0.$

Thus, $\lambda0=0.$

Hence, property given in option (b) is incorrect.

5. (c) 60° and 8

Explanation:

Let P be as shown in the figure, then according to the given information 

Px = 5, Py = 6

$\therefore\mid\text{P}\mid=\sqrt{\text{P}_\text{x}^2+\text{P}_\text{y}^2}$

$=\sqrt{25+336}$

$\Rightarrow\mid\text{P}\mid=\sqrt{61}$ and $\tan\theta=\frac{\text{P}_\text{y}}{\text{P}_\text{x}}=\frac{6}{5}$

$\Rightarrow\theta=\tan^{-1}\big(\frac{6}{5}\big)$

1. (a) Rev/ sec
2. (a) True
3. When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
4. “Centripetal” comes from a Greek term which means ‘centre-seeking’ i.e. always directed towards centre of a circle.
5. Acceleration of particle performing uniform circular motion which is always directed towards centre of a circle is called centripetal acceleration. We can express centripetal acceleration ac in terms of angular speed as.

$\omega=2\pi\text{v}$

However, during this time the distance moved by the object is s = 2πR. In terms of frequency V, we have

$\omega=2\pi\text{v}$

$\text{V}=2\pi\text{ RV}$

$\text{ac}=4\pi^2\text{v}^2\text{R}$

1. (d) Both (b) and (c)

Explanation:

Circular motion is an example of two-dimensional motion with radius vector as

$\text{r}=\text{a}\cos\omega\text{t}\hat{\text{i}}+\text{a}\sin\omega\text{t}\hat{\text{j}}$

Both the components $\text{r}=\text{a}\cos\omega\text{t}\hat{\text{i}}$ and $\text{a}\sin\omega\text{t}\hat{\text{j}}$ are perpendicular to each other.

2. (c) Direction of acceleration keeps changing as particle moves.

Explanation:

If a particle is performing uniform circular motion, then its

1. Speed will be constant throughout the motion.
2. Velocity will be tangential in the direction of motion at a particular point.

2. Acceleration, $\text{a}=\frac{\text{v}^2}{\text{r}}$ will always be towards centre of the circular path.
3. Angular momentum (mvr) is constant in magnitude but direction keeps on changing.

3. (b) $\frac{\text{r}_{\text{A}}}{\text{r}_\text{B}}$

Explanation:

Angular velocity $\omega$ is constant.

$\text{v}=\text{r}\omega$

$\therefore\text{v}\propto\text{r}$ or $\frac{\text{v}_\text{A}}{\text{v}_\text{B}}=\frac{\text{r}_{\text{A}}}{\text{r}_{\text{B}}}$

4. (b) 0,10ms^-1

Explanation:

On a circular path in completing one turn,the distance travelled is $2\pi\text{r},$ while displacement
is zero. Hence, average velocity

$=\frac{\text{displacement}}{\text{time interval}}=\frac{0}{\text{t}}=0$

Average speed $=\frac{\text{Distance}}{\text{Time interval}}$

$=\frac{2\pi\text{r}}{\text{t}}=\frac{2\times3.14\times100}{62.8}=10\text{ms}^{-1}$

5. (b) 4740ms^-2

Explanation:

^{Given, V = 1200 rpm $\frac{1200}{60}\text{rps}$}

^{$\text{r}=30\text{cm}=\frac{30}{100}\text{m}$}

^{Acceleration of the particle = Centripetal acceleration}

$=\omega^2\text{r}=(2\pi\text{v})^2\text{r}$

$=\Big(2\times\frac{22}{7}\times\frac{1200}{60}\Big)^2\times\frac{30}{100}\approx4740\text{ms}^{-2}$

1. (a) 45⁰
2. (a) Galileo
3. The motion of object under only gravity force in the air is called projectile motion.
4. The horizontal distance travelled by a projectile from its initial position to the  position where it passes same horizontal position during its fall is called the horizontal range, R. It is the distance travelled during the time of flight T_f . Therefore, the range R is.

$\text{R is R} =(\text{v}_o\cos\theta_o(\text{T}_\text{f})$

$\text{R}=\frac{(\text{v}_\text{o}\cos\theta_\text{o})(2\text{v}_\text{o}\sin\theta_\text{o})}{\text{g}}$

$\text{R}=\frac{(\text{v}_\text{o}^{2}\sin\theta_\text{o})}{g}$

5. Maximum height of a projectile: Maximum height that can be achieved during projectile and it is given by

$\text{H}_\text{m}=\frac{\text{(v}_\text{o}\sin\theta)^2}{2g}$