Motion in a Plane Class 11 Important Questions

Motion in a Plane Class 11 Important Questions and Answers are covered in this article. The chapter has a weightage of about 4 to 5 marks in the annual examinations. Motion in a plane is also known as motion which happens in two dimensions. For example, projectile motion, circular motion, etc. For analysing such kinds of motion, the reference point that is taken is of an origin and the two coordinate axes Y and X.

Motion in a plane is the point where we consider the motion to be happening in just two dimensions as a plane is made of only two dimensions. Here, understanding the above situation, we generally take two axes into consideration, the X-axis and Y-axis. Motion in a Plane Class 11 Important Questions are as given below.

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Very Short Answer Questions (1 Mark Questions)

Ques. Under what condition |a + b| = |a| + |b| holds good?

Ans. |a + b| = |a| + |b| holds good when a and b both act in the same direction, which means, when 0 = 0 between them.

Ques. The sum and difference of the two vectors are equal in magnitude, i. e. |a + b| = |a – b|. What conclusion do you draw from this?

Ans. When the sum and difference of two vectors are equal, then we can conclude that the two vectors have equal magnitude and are found to be perpendicular to each other.

Ques. What is the minimum number of coplanar vectors of different magnitudes which can give zero resultant?

Ans. The minimum number of coplanar vectors is found to be 3. As in, If three vectors are shown completely by a triangle’s three sides that are taken in the same order, then their resultant is found to be zero.

Ques. When a – b = a + b condition holds good, then what can you say about b?

Ans. If we want a – b = a + b condition to hold good, we must have b as a null vector.

Ques. What is the effect on the dimensions of a vector if it is multiplied by a non-dimensional scalar?

Ans. Upon multiplying a vector with a dimensionless scalar, the vector’s dimensions do not go through any change. 

Read More: Difference Between Scalar and Vector Quantities

Ques. Is finite rotation a vector quantity? Why?

Ans. No, a finite rotation cannot be considered a vector quantity as the addition of two finite rotations never obeys commutative law.

Ques. Is infinitesimally small rotation a vector quantity? Why?

Ans. Yes, an infinitesimally small rotation is considered a vector quantity because the two infinitesimally small rotations, in the addition, will obey commutative law.

Ques. A quantity has both magnitude and direction. Is it necessarily a vector? Why? Give an example.

Ans. No quantity cannot be considered a vector just because of magnitude and direction, it must obey the laws of vector addition as well to be that. An example is an Electric current.

Read More:

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Short Answer Questions (2 Marks Questions)

Ques. Name two quantities that are the largest when the maximum height attained by the projectile is the largest.

Ans. The two quantities that are found to be the largest when a maximum height is attained by a projectile are Time of flight and the vertical component of velocity.

Ques. Can a particle accelerate when its speed is constant? Explain.

Ans. Yes. A particle can surely go through acceleration even though its speed is constant. A particle that has a uniform circular motion has a constant speed, however, its direction of motion continuously changes. This creates a change in velocity and the particle now moves with variable velocity. This is how the particle is accelerating.

Read More: Differences between Acceleration and Velocity

Ques. (a) Is circular motion possible at a constant speed or constant velocity? Explain.

(b) Define frequency and time.

Ans. (a) Circular motion can happen even at a constant speed. This is because, in a circular motion, speed remains constant, however, the motion’s direction continuously keeps changing.

(b) Frequency is known as the number of rotations completed by a body in one second and the time is known as the time taken to complete one rotation by an object.

Ques. When vector A’s component along the direction of vector B’s component is zero, what can you understand about both of these vectors?

Ans. The two vectors that are A and B are found to be perpendicular to each other.

This can be explained as follows: Let θ = angle between the two vectors A and B component of vector A along the direction of B is obtained by resolving A i.e. A cos θ.

As the statement says

A cosθ= 0

or

cos θ = 0 = cos 90°

θ = 90°

i.e. A ⊥ B

Hence proved.

Read More: Types of Vectors

Ques. Comment on the statement whether it is true or false “Displacement vector is fundamentally a position vector.’’ Why?

Ans. The statement given above is true. The displacement vector and position vector both give the position of a point. The only difference between both of these vectors is that the displacement vector gives the point’s position with reference to a point that is not only the origin, while the position vector provides a point’s position with the origin’s reference.

Ques. Does the nature of a vector experience any change when it is multiplied by a scalar?

Ans. A vector’s nature may or may change when it is multiplied. For example, when we multiply a vector by a pure number as 1, 2, or 3, etc., then the vector’s nature does not alter. However, when a vector is multiplied by a scalar physical quantity, then its nature changes.

For example, when acceleration (a vector quantity) is multiplied by a mass (a scalar quantity) of a particular body, then it provides force (vector quantity) and its nature is not the same as acceleration.

Ques. Can the walk of a man be an example of the resolution of vectors? Explain.

Ans. Yes, when the action of walking is done by a man, the ground is pushed with his foot. An opposite and equal reaction acts on his foot in return. The reaction can be resolved into two types of components: the first is horizontal and the second the vertical component. The reaction’s horizontal component helps the walking man to move in the forward direction whereas its vertical component does the balancing of the man’s weight.

Read More: Properties of Vector Addition

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Long Answer Questions (3 Marks Questions)

Ques. Derive an equation for the path of a projectile fired parallel to horizontal.

Ans. Let us assume a projectile that has an initial uniform horizontal velocity as u, that is under the gravity’s influence, then at any of the instant t at the position P, e have the horizontal and vertical.

For horizontal motion

s = ut + ½ at²

s = x, u = u, t = t, a = 0 

x = ut

⇒ t = x/u … (1)

For vertical motion,

s = ut + ½ at2

s = −y, t = t, u = 0, a = −g

−y = − ½ gt²

⇒ y = ½ gt² …… (2)

Using equations (1) and (2)

⇒ y = ½ g (x/u)² = ½ gx²/u², which is the path’s required equation.

Ques. a) Define time of flight and horizontal range.

  b) From a certain height above the ground stone A is dropped gently. Simultaneously another stone B is fired horizontally. Which of the stones between them will hit the ground earlier?

Ans. (a) The time of flight is the time taken by the projectile to complete its trajectory.

The horizontal range is the maximum horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground.

(b) In both of these cases, the initial vertical velocity is zero and the acceleration of their fall is the same, hence Both the stones will reach the ground simultaneously.

Ques. Discuss the problem of a swimmer who wants to cross the river in the shortest time.

Ans. Let vs and v_r be the velocities of swimmer and river respectively.

Let v = resultant velocity of v_r and v_s.

Let us assume that the swimmer starts swimming at an angle θ with the line OA where OA is perpendicular to the river flow.

If t is the time taken to cross the river, then

t = l/vcosθ …(i)

where l = breadth of the river

To make a minimum, we should make cos 0 should maximum.

i.e. cos θ = 1

This is possible if θ = 0

Thus, we now know that the swimmer must swim in a direction that is perpendicular to the direction of the river flow.

v = \(\sqrt v\)s² + vr²

where v is the resultant velocity of vs and vr.

tan θ = vr/vx = x/l

or

x = l vr/vs

If cos θ = 0 = θ = 1, so the shortest time will be

t = l/vs 

Read More: Relative Velocity

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Very Long Answer Questions (5 Marks Questions)

Ques. A cyclist rides at a speed of 27km/h. As he is approaching a circular turn that is on the road with a radius of 80m, he applied his brakes and reduced his speed at a constant rate of 0.5m/s each second. What will be the direction and magnitude and the net acceleration of the cyclist when he is on the circular turn?

Ans. Speed of the cyclist is given as, v = 27 km/h = 7.5 m/s

Radius of the turn that is circular, r = 80 m

Centripetal acceleration can be given as ac = v²/r

⇒ ac = (7.5)²/80 = 0.7m/s

Now, Let us suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the brakes and decelerates the speed of the bicycle by

0.5 m/s².

This acceleration is found to be along the tangent at Q and the direction of motion of the cyclist’s opposite direction.

Because the angle between ac and at is 90â, the resultant acceleration a is given by:

 a = \(\sqrt a\)_c² + a_t²

\(\implies\) a = \(\sqrt 0\).72 + 0.52

\(\implies\) a = √0.74 = 0.86 m/s²

\(\implies\)a= \(\sqrt 0.74 \)= 0.86 m/s²

tanθ= a_c/a_t

Where θ becomes the angle of the resultant with velocity’s direction.

tanθ= 0.7/0.5 =1.4

⇒ θ = tan−1(1.4)

⇒ θ = 54.460, which is the required angle of direction here.

Read More: Angle Between a Line and a Plane

Ques. Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of

(a) the force on the 7th coin (counted from the bottom) due to all the coins on its top,

(b) the force on the 7th coin by the 8th coin,

(c) the reaction of the sixth coin on the seventh coin.

Ans. (a) The Force on the 7th coin is exerted by the three coins’ weight on its top.

• Weight of 1 coin = mg
• Weight of 3 coins = 3mg

Hence, the force exerted on the 7th coin by the 3 coins on its top is 3mg. This force acts vertically downward.

(b) The Force on the 7th coin by the 8th coin is because of the weight of the 8th coin and the other 2 coins (9th and 10th) on its top.

• Weight of the 8th coin = mg
• Weight of the 9th coin = mg
• Weight of the 10th coin = mg
• Total weight of these 3 coins = 3mg

Hence, the total force exerted on the seventh coin by the eighth coin is 3mg. This force acts vertically downward.

(c) The 6th coin experiences a downward force because of the weight of the 4 coins (7th, 8th, 9th, and 10th) on its top.

Therefore, the downward force in total experienced by the sixth coin is 4mg.

In accordance with Newton"s third law of motion, the 6th coin will produce an equal reaction force on the 7th coin, but in the opposite direction. Therefore the reaction force of the 6th coin on the 7th coin is of magnitude 4mg. This force acts in the upward direction.

Read More: Force and Motion

Ques. A man of mass of 70 kg stands on a weighing scale in a lift which is moving

(a) upwards along with a uniform speed of 10m/s

(b) downwards along with a uniform acceleration of 5 m/s

(c) upwards with a uniform acceleration of 5m/s2.

What would be the readings on the scale in each case?

(d) Calculate what would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Ans. (a) The Mass of the given man is m = 70 kg

Acceleration that is a = 0

In accordance with Newton"s second law of motion, we shall write the equation of motion as follows:

R - mg = ma

Where ma is the net force that is acting on the man.

As the lift is found to be moving at a uniform speed, acceleration is a = 0

∴ R = mg

= 70 × 10 = 700 N

∴ The reading on the weighing scale will be 700/g = 700/10 = 70 kg

(b) Mass of the given man is m = 70 kg

Acceleration, a = 5 m/s2 downward

Using Newton"s second law of motion, we can write the equation of motion as:

R + mg = ma

= R = m (g - a)

= 70 (10 - 5) = 70 x 5

= 350 N

∴ The Reading on the weighing scale is found to be = 350/g = 350/10 = 35 kg

(c) Mass of the given man, m = 70 kg

The Acceleration, a = 5m/s² upward

By Using Newton"s second law of motion, we shall write the equation of motion as:

R - mg = ma

R = m (g + a)

= 70 (10 + 5) = 70 × 15

= 1050 N

∴ The Reading on the weighing scale is found to be = 1050/g = 1050/10 = 105 kg

(d) The acceleration is a = g, as the lift moves freely under gravity.

In accordance with Newton"s second law of motion, we shall write the equation of motion as:

R + mg = ma

R = m (g - a)

= m (g - g) = 0

∴ Reading on the weighing scale = 0/g = 0kg

The man will be found in a state of weightlessness.

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