$|\lambda\text{A}|=\lambda|\text{A}|,\ \text{if}$
- A$\lambda>0$
- B$\lambda<0$
- C$\lambda=0$
- D$\lambda\neq0$
Question types
Motion in a Plane question types
379 questions across 8 question groups — pick any mix to generate a Physics paper with step-by-step answer keys.
379
Questions
8
Question groups
5
Question types
01
M.C.Q (1 Marks)
57 Q→02Fill In The Blanks[1 Marks ]
13 Q→03True False[1 Marks ]
17 Q→041 Marks Question
23 Q→052 Marks Questions
64 Q→063 Marks Question
120 Q→074 Marks Question
9 Q→085 Marks Questions
76 Q→Sample Questions
Motion in a Plane questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
A boy aims a gun at a target from a point, at a horizontal distance of 100m. If the gun can impart a horizontal velocity of 500ms-1 to the bullet, the height above the target where he must aim his gun, in order to hit it is (Take g = 10ms-2)
- A20cm
- B10cm
- C50cm
- D100cm
The angle between $\vec{\text{A}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{B}}=\hat{\text{i}}-\hat{\text{j}}$ is
- A45°
- B90°
- C–45°
- D180°
A plane is inclined at an angle of 30° with horizontal. The magnitude of component of a vector $\vec{\text{A}}=-10\hat{\text{k}}$ perpendicular to this plane is (here z-direction is vertically upwards):
- A$5\sqrt{2}$
- B$5\sqrt{3}$
- C$5$
- D$2.5$
Given, |A + B| = P, |A - B| = Q. The value of P2 + Q2 is:
- A2(A2 + B2)
- BA2 - B2
- CA2 + B2
- D2(A2 - B2)
$\vec{\text{C}}=\vec{\text{A}}\times\vec{\text{B}},$ direction of $\vec{\text{C}}$ is _______.
View full solution →Value of $\vec{\text{A}}\times\vec{\text{A}}$ is ______.
View full solution →$\vec{\text{A}}\times\vec{\text{B}}=\vec{\text{A}}.\vec{\text{B}},$ the angle between $\vec{\text{A}}$ and $\vec{\text{B}}$ is $\theta=$______.
View full solution →The necessary condition for a physical quantity to be a vector is that it must obey ________.
Resultant of two same vector physical quantity perpendicular to each other ________.
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that
Does not vary from one point to another in space.
Read each statement below carefully and state with reasons, if it is true or false:
The total path length is always equal to the magnitude of the displacement vector of a particle.
The total path length is always equal to the magnitude of the displacement vector of a particle.
For any arbitrary motion in space, which of the following relations are true:
$\text{v}(\text{t})=\text{v}(0)+\text{a t}$
(The ‘average’ stands for average of the quantity over the time interval t1 to t2)
View full solution →$\text{v}(\text{t})=\text{v}(0)+\text{a t}$
(The ‘average’ stands for average of the quantity over the time interval t1 to t2)
For any arbitrary motion in space, which of the following relations are true:
$\text{v}_{\text{average}}=\frac{[\text{r}(\text{t}_2)-\text{r}(\text{t}_1)]}{(\text{t}_2-\text{t}_1)}$
(The ‘average’ stands for average of the quantity over the time interval t1 to t2)
View full solution →$\text{v}_{\text{average}}=\frac{[\text{r}(\text{t}_2)-\text{r}(\text{t}_1)]}{(\text{t}_2-\text{t}_1)}$
(The ‘average’ stands for average of the quantity over the time interval t1 to t2)
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that
Has the same value for observers with different orientations of axes.
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that
Has the same value for observers with different orientations of axes.
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that
Does not vary from one point to another in space.
Read each statement below carefully and state, with reasons, if it is true or false:
The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
A passenger arriving in a new town wishes to go from the station to a hotel located 10km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?
Given a + b + c + d = 0, which of the following statements are correct:
b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?
Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
An insect trapped in a circular groove of radius $12 cm$ moves along the groove steadily and completes 7 revolutions in $100 s$. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector? What is its magnitude?
View full solution →A cricket ball is thrown at a speed of $28 m s ^{-1}$ in a direction 30 above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.
View full solution →The position of a particle is given by
$
r =3.0 t \hat{ i }+2.0 t^2 \hat{ j }+5.0 \hat{ k }
$
where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres. (a) Find $v (t)$ and $a(t)$ of the particle. (b) Find the magnitude and direction of $v (t)$ at $t=1.0 s$.
View full solution →$
r =3.0 t \hat{ i }+2.0 t^2 \hat{ j }+5.0 \hat{ k }
$
where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres. (a) Find $v (t)$ and $a(t)$ of the particle. (b) Find the magnitude and direction of $v (t)$ at $t=1.0 s$.
Given a + b + c + d = 0, which of the following statements are correct:
a, b, c, and d must each be a null vector.
a, b, c, and d must each be a null vector.
Three girls skating on a circular ice ground of radius 200m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?
Given a + b + c + d = 0, which of the following statements are correct:
The magnitude of (a + c) equals the magnitude of ( b + d).
The magnitude of (a + c) equals the magnitude of ( b + d).
Shows that the projection angle $\theta_0$ for a projectile launched from the origin is given by,
$\theta_0=\tan^{-1}\Big(\frac{4\text{h}_\text{m}}{\text{R}}\Big)$
where the symbols have their usual meaning.
View full solution →$\theta_0=\tan^{-1}\Big(\frac{4\text{h}_\text{m}}{\text{R}}\Big)$
where the symbols have their usual meaning.
Given a + b + c + d = 0, which of the following statements are correct:
The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d.
The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d.
Read the passage given below and answer the following questions from 1 to 5.
View full solution →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).
- Two vectors A and B are said to be equal if:
- they have the same magnitude
- they have the same direction
- they have the same magnitude and the same direction
- None of these
- Multiplying a vector A with a positive number will impact:
- Change in magnitude
- Change in direction
- Change in both magnitude and the same direction
- None of these
- What is null vector?
- How we can perform subtraction of two vectors?
- Enlist any 4 properties of vectors.
Read the passage given below and answer the following questions from 1 to 5.
In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are: the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided just as the ordinary numbers. A vector quantity is a quantity that has both a magnitude and a direction and obeys the triangle law of addition or equivalently the parallelogram law of addition. So, a vector is specified by giving its magnitude by a number and its direction. Some physical quantities that are represented by vectors are displacement, velocity, acceleration and force. Answer the following
- Force is example of:
- Scalar
- Vector
- Tensor
- None of these
- Mass of an object is:
- Scalar
- Vector
- Tensor
- None of these
- Define scalar quantity and vector quantity
- Can we add vectors like ordinary algebra? If not then how vectors are added?
- Differentiate between scalar and vectors
Read the passage given below and answer the following questions from 1 to 5.
If A is vector given by A = Ax i + Ay j where
The quantities Ax and Ay are called x, and y- components of the vector A. Note that Ax is itself not a vector, but Ax i is a vector, and so is Ay j. Using simple trigonometry, we can express Ax and Ay in terms of the magnitude of A and the angle θ it makes with the x-axis.
$\text{Ax} = \text{A} \cos(\theta)$
$\text{Ay} = \text{A} \text{ sin}(\theta)$
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 –
$\text{A}^2_\text{x}+\text{A}^2_\text{y}=(\text{A}\cos\theta)^2+(\text{A}\sin\theta)^2$
$\text{A}^2_\text{x}+\text{A}^2_\text{y}=\text{A}^2\cos^2\theta+\text{A}^2\sin^2\theta$
$\Rightarrow\text{A}^2_\text{x}+\text{A}^2_\text{y}=\text{A}^2(\cos^2\theta+\sin^2\theta)$
${A}^2_\text{x}+\text{A}^2_\text{y}=\text{A}^2(\because\sin^2\theta+\cos^2\theta=1)$
$\text{A}^2=\text{A}^2_\text{y}+\text{A}^2_\text{y}$
$\Rightarrow\text{A}=\sqrt{\text{A}^2_\text{x}+\text{A}^2_\text{y }}...$
$\text{Dividing}\text{ A}_\text{y}\text{ by} \text{ A}_\text{y},\text{we get}$
$\frac{\text{A}_\text{y}}{\text{A}_\text{x}}=\frac{\text{A}\sin\theta}{\text{A}\cos\theta}$
$\Rightarrow\frac{\text{A}_\text{y}}{\text{A}_\text{x}}=\tan\theta$
$\tan\theta=\frac{\text{A}_\text{y}}{\text{A}_\text{x}}$
$\theta=\tan^{-1}\Big[\frac{\text{A}_\text{y}}{\text{A}_\text{x}}\Big]$
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 or simply they are the coordinates of the object. Suppose a particle moves along the Then, 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.
The average velocity (v) of an object is the ratio of the displacement and the corresponding time Interval.
$\text{V}=\frac{\triangle\text{r}}{\triangle\text{t}}$
$=\frac{\text{i}\triangle\text{x}-\text{j}\triangle\text{y}}{\triangle\text{t}}$
$=\text{i}\times\frac{\triangle\text{x}}{\triangle\text{t}}+\text{j}\times\frac{\triangle\text{y}}{\triangle\text{t}}$
$=\text{V}_\text{x}\text{i}+\text{V}_\text{y}\text{j}$
So, if the expressions for the coordinates x and y are known as functions of time, we can use these equations to find vx and vy. The magnitude of v is then
V = ( vx2 + vy2) and the direction of v is given by the angle q and given by $\tan(\theta)=\frac{\text{vx}}{\text{vy}}$
View full solution →If A is vector given by A = Ax i + Ay j where
The quantities Ax and Ay are called x, and y- components of the vector A. Note that Ax is itself not a vector, but Ax i is a vector, and so is Ay j. Using simple trigonometry, we can express Ax and Ay in terms of the magnitude of A and the angle θ it makes with the x-axis.
$\text{Ax} = \text{A} \cos(\theta)$
$\text{Ay} = \text{A} \text{ sin}(\theta)$
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 –
$\text{A}^2_\text{x}+\text{A}^2_\text{y}=(\text{A}\cos\theta)^2+(\text{A}\sin\theta)^2$
$\text{A}^2_\text{x}+\text{A}^2_\text{y}=\text{A}^2\cos^2\theta+\text{A}^2\sin^2\theta$
$\Rightarrow\text{A}^2_\text{x}+\text{A}^2_\text{y}=\text{A}^2(\cos^2\theta+\sin^2\theta)$
${A}^2_\text{x}+\text{A}^2_\text{y}=\text{A}^2(\because\sin^2\theta+\cos^2\theta=1)$
$\text{A}^2=\text{A}^2_\text{y}+\text{A}^2_\text{y}$
$\Rightarrow\text{A}=\sqrt{\text{A}^2_\text{x}+\text{A}^2_\text{y }}...$
$\text{Dividing}\text{ A}_\text{y}\text{ by} \text{ A}_\text{y},\text{we get}$
$\frac{\text{A}_\text{y}}{\text{A}_\text{x}}=\frac{\text{A}\sin\theta}{\text{A}\cos\theta}$
$\Rightarrow\frac{\text{A}_\text{y}}{\text{A}_\text{x}}=\tan\theta$
$\tan\theta=\frac{\text{A}_\text{y}}{\text{A}_\text{x}}$
$\theta=\tan^{-1}\Big[\frac{\text{A}_\text{y}}{\text{A}_\text{x}}\Big]$
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 or simply they are the coordinates of the object. Suppose a particle moves along the Then, 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.
The average velocity (v) of an object is the ratio of the displacement and the corresponding time Interval.
$\text{V}=\frac{\triangle\text{r}}{\triangle\text{t}}$
$=\frac{\text{i}\triangle\text{x}-\text{j}\triangle\text{y}}{\triangle\text{t}}$
$=\text{i}\times\frac{\triangle\text{x}}{\triangle\text{t}}+\text{j}\times\frac{\triangle\text{y}}{\triangle\text{t}}$
$=\text{V}_\text{x}\text{i}+\text{V}_\text{y}\text{j}$
So, if the expressions for the coordinates x and y are known as functions of time, we can use these equations to find vx and vy. The magnitude of v is then
V = ( vx2 + vy2) and the direction of v is given by the angle q and given by $\tan(\theta)=\frac{\text{vx}}{\text{vy}}$
- If A is vector given by A = Ax i + Ay j .if the magnitude of vector is A and the angle $\theta$ it makes with the x-axis Ax can be given by:
- Ax = A cos(q)
- Ax = A sin(q)
- Ax = A tan(q)
- None of the above
- If A is vector given by A = Ax i + Ay j .if the magnitude of vector is A and the angle $\theta$ it makes with the x-axis Ay can be given by:
- Ax = A cos(q)
- Ax = A sin(q)
- Ax = A tan(q)
- None of the above
- Write a note on position vector and displacement of object:
- Write a note on average velocity:
- If A is vector given by A = Ax i + Ay j where obtain expression for resultant amplitude of vector and its angle with x axis:
Read the passage given below and answer the following questions from 1 to 5.
Projectile motion is a form of motion in which an object or particle is thrown with some initial velocity near the earth’s surface and it moves along a curved path under the action of gravity alone. The path followed by a projectile is called its trajectory, which is shown below. When a projectile is projected obliquely, then its trajectory is as shown in the figure below.
Here velocity u is resolved into two components, we get (a) u cosθ along OX and (b) u sinθ along OY.
Projectile motion is a form of motion in which an object or particle is thrown with some initial velocity near the earth’s surface and it moves along a curved path under the action of gravity alone. The path followed by a projectile is called its trajectory, which is shown below. When a projectile is projected obliquely, then its trajectory is as shown in the figure below.
Here velocity u is resolved into two components, we get (a) u cosθ along OX and (b) u sinθ along OY.
- The example of such type of motion is:
- Motion of car on a banked road.
- Motion of boat in sea.
- A javelin thrown by an athlete.
- Motion of ball thrown vertically upward.
- The acceleration of the object in horizontal direction is:
- Constant
- Decreasing
- Increasing
- Zero
- The vertical component of velocity at point H is:
- Maximum
- Zero
- Double to that at O
- Equal to horizontal component
- A cricket ball is thrown at a speed of 28m/s in a direction 30° with the horizontal. The time taken by the ball to return to the same level will be:
- 2.0s
- 3.0s
- 4.0s
- 2.9s
- In above case, the distance from the thrower to the point where the ball returns to the same level will be:
- 39m
- 69m
- 68m
- 72m
Read the passage given below and answer the following questions from (i) to (v).
Relative Velocity
Every motion is relative as it has to be observed with respect to an observer. Relative velocity is a measurement of velocity of an object with respect to other observer. It is defined as the time rate of change of relative position of one object with respect to another.
For example, if rain is falling vertically with a velocity v, and a man is moving horizontally with vm, the man can protect himself from the rain if he holds his umbrella in the direction of relative velocity of rain w.r.t. man.
View full solution →Relative Velocity
Every motion is relative as it has to be observed with respect to an observer. Relative velocity is a measurement of velocity of an object with respect to other observer. It is defined as the time rate of change of relative position of one object with respect to another.
For example, if rain is falling vertically with a velocity v, and a man is moving horizontally with vm, the man can protect himself from the rain if he holds his umbrella in the direction of relative velocity of rain w.r.t. man.
- Two bodies are held separated by 9.8m vertically one above the other. They are released simultaneously to fall freely under gravity. After 2s, the relative distance between them is:
- 4.9m
- 19.6m
- 9.8m
- 39.2m
- If two objects P andQ move along parallel straight lines in opposite direction with velocities vP and vQ respectively, then relative velocity of P w.r.t.Q,
- vPQ = vP = v0
- vP - v0
- vP + v0
- v0 - vp
- A train is moving towards East and a car is along North, both with same speed. The observed direction of car to the passenger in the train is:
- East - North direction
- West - North direction
- South - East direction
- None of the above
- Buses A and B are moving in the same direction with velocities $20\hat{\text{i}}\text{ms}^{-1}$ and $15\hat{\text{i}}\text{ms}^{-1},$ respectively. Then, relative velocity of A w.r.t. B is:
- $35\hat{\text{i}}\text{ms}^{-1}$
- $5\hat{\text{i}}\text{ms}^{-1}$
- $5\hat{\text{j}}\text{ms}^{-1}$
- $35\hat{\text{j}}\text{ms}^{-1}$
- A girl riding a bicycle with a speed of 5 ms -1 towards east direction sees raindrops falling vertically downwards. On increasing the speed to 15ms-1, rain appears to fall making an angle of 45° of the vertical. Find the magnitude of velocity of rain.
- $5\text{ms}^{-1}$
- $5\sqrt5\text{ms}^{-1}$
- $25\text{ms}^{-1}$
- $10\text{ms}^{-1}$
A particle starts from the origin at t = 0s with a velocity of $10.0\hat{\text{j}}\text{m/s}$ and moves in the x - y plane with a constant acceleration of $(8.0\hat{\text{i}}+2.0\hat{\text{j}})\text{ms}^{-2}.$ (a) At what time is the x- coordinate of the particle 16m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time?
View full solution →On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
In a harbour, wind is blowing at the speed of 72km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51km/h to the north, what is the direction of the flag on the mast of the boat?
A fighter plane flying horizontally at an altitude of 1.5km with speed 720km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600ms-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10ms-2).
A cyclist is riding with a speed of 27km/h. As he approaches a circular turn on the road of radius 80m, he applies brakes and reduces his speed at the constant rate of 0.50m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
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