Case Study Questions Class 11 Physics Motion in a Plane

Case Study Questions Class 11 Physics Chapter 4 Motion in a Plane

CBSE Class 11 Case Study Questions Physics Motion in a plane. Important Case Study Questions for Class 11 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Motion in a plane.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 11 Physics Motion in a Plane

 

Case Study – 1

1.)

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

1) Force is example of

a) Scalar

b) Vector

c) Tensor

d) None of these

2) Mass of an object is

a) Scalar

b) Vector

c) Tensor

d) None of these

3) Define scalar quantity and vector quantity

4) Can we add vectors like ordinary algebra? If not then how vectors are added?

5) Differentiate between scalar and vectors

 

Answer key-1

1) b

2) a

3) A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit are: the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened.

4) No. We can’t add vectors like ordinary algebra rules. Then question naturally arises that how can we add them so we can add then with the help of triangle law of vector addition and parallelogram law of vector addition.

5) Difference between scalar and vector is given below

Sr No.

Scalar quantity

Vector quantity

1 A scalar quantity has only magnitude, but no direction. Vector quantity has both magnitude and direction.
2 It changes with the change in their magnitude It changes with the change in their direction or magnitude or both.
3 Every scalar quantity is one-dimensional. Vector quantity can be one, two or three-dimensional.
4 Scalar quantity cannot be resolved as it has exactly the same value regardless of direction. Vector quantity can be resolved in any direction using the sine or cosine of the adjacent angle.
5 Examples are; between two points, mass of an object Examples are; displacement, velocity, acceleration and force

 

Case Study – 2

2.) 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 :

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).

Answer the following-

1) Two vectors A and B are said to be equal if

a) they have the same magnitude

b) they have the same direction

c) they have the same magnitude and the same direction

d) None of these

2) Multiplying a vector A with a positive number will impact

a) Change in magnitude

b) Change in direction

c) Change in both magnitude and the same direction

d) None of these

3) What is null vector?

4) How we can perform subtraction of two vectors?

5) Enlist any 4 properties of vectors.

 

Answer key-2

1) c

2) a

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

0 A = 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 :

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)

 

Case Study – 3

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.

Ax = A cos(θ)

Ay = A sin(θ)

If A and θ are given, Ax and Ay can be obtained using If Ax and Ay are given, A and θ can be obtained as follows –

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.

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(θ)= vx/vy

 

1) If A is vector given by A = Ax i + Ay j .if the magnitude of vector is  A and the angle θ it makes with the x-axis Ax can be given by

a) Ax = A cos(q)

b) Ax = A sin(q)

c) Ax = A tan(q)

d) None of the above

2) If A is vector given by A = Ax i + Ay j .if the magnitude of vector is  A and the angle θ it makes with the x-axis Ay can be given by

a) Ax = A cos(q)

b) Ax = A sin(q)

c) Ax = A tan(q)

d) None of the above

3) Write a note on position vector and displacement of object

4) Write a note on average velocity

5) 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.

 

Answer key-3

1) a

2) b

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.

 

If A and θ are given, Ax and Ay can be obtained using If Ax and Ay are given, A and θ can be obtained as follows

Case Study – 4

When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word “uniform” refers to the speed, which is uniform (constant) throughout the motion. Suppose an object is moving with uniform speed v in a circle of radius R Since the velocity of the object is changing continuously in direction, the object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. Thus, the acceleration of an object moving with speed v in a circle of radius R has a magnitude V2/R and is always directed towards the centre. This is why this acceleration is called centripetal acceleration (a term proposed by Newton). A thorough analysis of centripetal acceleration was first published in 1673 by the Dutch scientist Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. “Centripetal” comes from a Greek term which means ‘centre-seeking’. Since v and R are constant, the magnitude of the centripetal acceleration is also constant. However, the direction changes pointing always towards the centre. Therefore, a centripetal acceleration is not a constant vector. We can express centripetal acceleration ac in terms of angular speed as

ac = ω2R

The time taken by an object to make one revolution is known as its time period T and the number of revolution made in one second is called its frequency v (=1/T). However, during this time the distance moved by the object is s = 2πR. Therefore, v = 2πR/T =2πRv In terms of frequency n, we have

ω = 2πv

v = 2πRv

ac = 4π2v2R

1) SI unit of angular velocity is

a) Rev/sec

b) m/s

c) m/s2

d) None of these

2) A centripetal acceleration is not a constant vector. True or false?

a) True

b) False

3) Define Uniform circular motion

4) What is meaning of word centripetal?

5) What is centripetal acceleration? Give its relation with angular velocity

 

Answer key-4

1) a

2) a

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

ac = ω2R

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

ω = 2πv

v = 2πRv

ac = 4π2v2R

 

Case Study – 5

we consider the motion of a projectile. An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity. It was Galileo who first stated this independency of the horizontal and the vertical components of projectile motion in his Dialogue on the great world systems.

Horizontal range of a projectile: The horizontal distance travelled by a projectile from its initial position (x = y = 0) to the position where it passes y = 0 during its fall is called the horizontal range, R. It is the distance travelled during the time of flight Tf . Therefore, the range

R is R = (v0 cos θ0) (Tf)

R = (v0 cos θ0) (2 v0 sin θ0)/g

R = (v02 sin 2 θ0)/g

This shows that for a given projection velocity, R is maximum when sin 2θ0 is maximum, i.e., when θ0 = 450. The maximum horizontal range is, therefore

R= v02/g

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

Hm = (v0 sin θ0)2/2g

 

1) Range in projectile motion is maximum when θ0

a) 450

b) 00

c) 900

d) None of these

2) Who was first stated this independency of the horizontal and the vertical components of projectile motion in his Dialogue on the great world system?

a) Galileo

b) Newton

c) Einstein

d) None of these

3) What is projectile motion?

4) What is horizontal range of projectile? Give its formula

5) What is maximum height of projectile? Give its formula

 

Answer key-5

1) a

2) a

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 Tf . Therefore, the range R is

R = (v0 cos θ0) (Tf)

R = (v0 cos θ0) (2 v0 sin θ0)/g

R = (v02 sin 2 θ0)/g

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

Hm = (v0 sin θ0)2/2g

Updated: March 23, 2022 — 10:30 am

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