Motion In A Plane

Motion in a Plane: A Summary

Motion in a plane is 2D motion. This means that the object can move both horizontally and vertically. Vectors provide a framework for quantifying and analyzing the motion, including its position, velocity, and acceleration

Key Concepts:

Position vector: Position vector defines the object’s position relative to a reference point.

Displacement vector: Displacement is the shortest distance between an object’s starting point and its ending point

Velocity vector: Velocity is the instantaneous change in position with respect to time

Acceleration vector: Acceleration is a vector quantity that describes how fast an object is speeding up, slowing down, or changing direction.

Scalar and Vector Quantities:

Scalar quantities: Have only magnitude (e.g., speed, mass, time).

Vector quantities: Have both magnitude and direction (e.g., velocity, acceleration, force).

Projectile Motion:

The motion of an object under the influence of gravity alone is called projectile motion.

The path of a projectile is parabolic.

The horizontal and vertical components of the projectile’s motion can be treated independently.

Uniform Circular Motion:

An object moving in a circle at a steady pace exhibits uniform circular motion.

Centripetal acceleration pulls an object towards the center of its circular path.

Relative Velocity:

The velocity of an object relative to another object is the difference between their individual velocities.

Important Formulas:

Velocity vector: v = dr/dt

Acceleration vector: a = dv/dt

Centripetal acceleration: a = v²/r

Projectile motion equations:

x = u_x t

y = u_y t – (1/2)gt²

Key Points:

The velocity vector of a particle in uniform circular motion is always tangent to the circle.

The direction of the acceleration vector in uniform circular motion is always towards the center of the circle, causing the object to change its direction continuously.

The time period of a particle in uniform circular motion is the time taken to complete one revolution.

The frequency of a particle in uniform circular motion is the number of revolutions per second.

By understanding these concepts and formulas, you can analyze and solve problems involving motion in a plane.

1. State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

Ans : Scalar vs. Vector Quantities

Scalar Quantities: Scalar quantities are described by a single number without specifying a direction

Examples include:

Volume

Mass

Speed

Density

Number of moles

Angular frequency

Vector Quantities: A vector is a quantity that requires both a numerical value and a specified direction to be fully defined.

Examples include:

Velocity

Acceleration

Displacement

Angular velocity

Quantities that can be completely described by a single numerical value are scalars. Quantities that require both a numerical value and a direction are vectors.

2. Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

Ans : The two scalar quantities in the given list are:

Work

Current

Scalar quantities have magnitude but no direction. While force, angular momentum, linear momentum, electric field, and magnetic moment are vector quantities with both magnitude and direction, work and current are scalar quantities.

3. 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.

Ans : The only vector quantity in the given list is impulse.

Impulse is the product of force and time, and it has both magnitude (the product of force and time) and direction (the same direction as the force). The other quantities listed are scalars, which have magnitude but no direction.

4. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful :

(a) adding any two scalars,

(b) adding a scalar to a vector of the same dimensions ,

(c) multiplying any vector by any scalar,

(d) multiplying any two scalars,

(e) adding any two vectors,

(f) adding a component of a vector to the same vector.

Ans : (a) No. Only quantities with the same units (dimensions) can be added.

(b) No. Scalars and vectors are fundamentally different and cannot be directly added.

(c) Yes. Multiplying a vector by a scalar results in a new vector with a magnitude scaled by the scalar factor. This is a common operation in physics, such as calculating force (F = ma).

(d) Yes. Multiplying two scalars results in a new scalar quantity. For instance, multiplying mass and temperature change gives the heat energy absorbed by a substance.

(e) No. Only vectors of the same dimensions can be added using vector addition rules.

(f) Yes. Two vectors of the same dimension can be added using vector addition rules to produce a resultant vector.

5 Read each statement below carefully and state with reasons, if it is true or false : (a) The magnitude of a vector is always a scalar,

(b) each component of a vector is always a scalar,

(c) the total path length is always equal to the magnitude of the displacement vector of a particle.

(d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time,

(e) Three vectors not lying in a plane can never add up to give a null vector.

Ans :

(a) True. The magnitude of velocity, also known as speed, is equal to the rate at which an object covers distance. In straight-line motion, the magnitude of velocity and speed are identical.

(b) False. Each component of a vector is a scalar quantity, not a vector.

(c) False. The magnitude of the displacement vector represents the shortest distance between the initial and final points. The total path length can be longer than this shortest distance.

(d) True. The equality holds only when the motion is in a straight line without any change in direction.

(e) True. The resultant of two vectors lies in the plane defined by those two vectors. If a third vector is not in this plane, it cannot be canceled out by the resultant of the first two, preventing the net vector from being zero.

6 Establish the following vector inequalities geometrically or otherwise :

(a) |a+b| < |a| + |b|

(b) |a+b| > ||a| −|b||

(c) |a−b| < |a| + |b|

(d) |a−b| > ||a| − |b||

When does the equality sign above apply?

Ans :

7 Given a + b + c + d = 0, which of the following statements are correct :

(a) a, b, c, and d must each be a null vector,

(b) The magnitude of (a + c) equals the magnitude of ( b + d),

(c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d,

(d) 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 ?

Ans : (a) Incorrect. Each vector does not necessarily need to be a null vector for their sum to be zero. For example, if a =− b and c =− d , then a + b +c + d = 0 .

(b) Correct. If ∣ a + c ∣=∣ b + d ∣ and a + c =−( b + d ), then the vectors a + c and b + d are equal in magnitude but opposite in direction, resulting in a null vector.

(c) Correct. If ∣ a ∣>∣ b +c + d ∣, then the vector sum a + b + c + d cannot be zero. Even if b , c , and d form a triangle and their sum is zero, adding a to this sum will result in a non-zero vector.

(d) Correct. If b + c does not lie in the plane of a + d , then the vectors ( a + b ) and ( c + d ) will have different magnitudes and directions. Therefore, their sum cannot be zero.

8 Three girls skating on a circular ice ground of radius 200 m 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. 3.19. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ?

Ans : Analyzing the Girls’ Paths

Displacement: The displacement vector is the shortest straight-line distance between the starting point (P) and the ending point (Q). Since all three girls start at P and end at Q, which is diametrically opposite, the magnitude of the displacement vector is the same for all three girls.

Path Length: The actual length of the path skated depends on the specific path taken.

Girl A: Takes a curved path, longer than the straight-line distance between P and Q.

Girl B: Takes a straight-line path directly from P to Q.

Girl C: Takes a curved path, longer than the straight-line distance but shorter than Girl

A’s path.

Therefore:

Magnitude of displacement vector: Same for all three girls (equal to the diameter of the circle, which is 400 m)

Girl B skated a path length that is equal to the magnitude of her displacement vector. This is because she took the shortest possible path between P and Q.

In summary, all three girls have the same magnitude of displacement vector (400 m), but only Girl B’s path length is equal to this displacement

i.e

The displacement of each girl is equal to the length of the line segment PQ.

Since PQ is the diameter of the circular ice ground, the magnitude of the displacement for each girl is 400 meters (2 × 200 meters).

9. A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the

(a) net displacement,

(b) average velocity, and

(c) average speed of the cyclist ?

Ans : (a) As the initial and final positions are the same, the net displacement is zero.

(b) Since the net displacement is zero, the average velocity, which is the ratio of net displacement to total time, is also zero.

(c) To calculate the average speed, we need to find the total distance covered and the total time taken.

Total Distance:

Distance traveled on the straight paths (OP and QO)

= 1 km + 1 km

= 2 km

Distance traveled along the circular path (PQ) = (1/4) * 2π * 1 km

= π/2 km

Total distance = 2 km + π/2 km ≈ 3.57 km

Total Time:

Time taken = 10 minutes = 10/60 hours = 1/6 hours

Average Speed:

Average speed = Total distance / Total time

= 3.57 km / (1/6) hours ≈ 21.43 km/h

10. On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. 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.

Ans : Understanding the Problem:

A motorist follows a circular path that forms a hexagon. We are asked to calculate the displacement, total path length, and the ratio of displacement to path length for different scenarios.

Scenario (i): After Three Turns

Displacement:

The motorist starts at point O and reaches point C after three turns.

The displacement is the straight-line distance between O and C.

Using geometry and trigonometry, the displacement is calculated as:

Displacement = OC = √((OB + FB)² + (BC)²) = 1000 m = 1 km

Total Path Length:

The motorist travels 500 m in each turn.

After three turns, the total path length is 3 × 500 m = 1500 m = 1.5 km.

Ratio of Displacement to Path Length:

The ratio is calculated as:

Ratio = Displacement / Path Length

= 1 km / 1.5 km = ⅔

= 0.67

Scenario (ii): After Six Turns

Displacement:

The motorist returns to the starting point O after six turns.

The net displacement is 0 because the object returns to its starting point

Total Path Length:

The total path length is 6 × 500 m = 3000 m = 3 km.

Ratio of Displacement to Path Length:

The ratio is zero since the displacement is zero.

Scenario (iii): After Eight Turns

Displacement:

The motorist reaches point B after eight turns.

The displacement is the shortest distance between the starting point O and the final point B.

Using trigonometry, the displacement is calculated as:

Displacement = 2 × 500 × cos(30°) = 500√3 m

≈ 0.866 km

Total Path Length:

The total path length is 8 × 500 m = 4000 m = 4 km.

Ratio of Displacement to Path Length:

The ratio is calculated as:

Ratio = Displacement / Path Length = 0.866 km / 4 km

≈ 0.22

In summary, the displacement, total path length, and their ratio vary depending on the number of turns the motorist makes.

11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km 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 ?

Ans : Understanding the Problem:

The actual distance traveled by the cab is 23 km.

The displacement is 10 km (straight-line distance from station to hotel).

The time taken is 28 minutes.

(a) Average Speed

Average speed = Total distance / Time

Average speed = 23 km / (28/60) hours ≈ 49.3 km/h

(b) Magnitude of Average Velocity

Average velocity = Displacement / Time

Average velocity = 10 km / (28/60) hours ≈ 21.4 km/h

Comparing Average Speed and Average Velocity:

The average speed (49.3 km/h) is greater than the magnitude of the average velocity (21.4 km/h). This is because the cab took a circuitous path, resulting in a longer total distance traveled compared to the straight-line displacement.

Therefore:

The average speed of the taxi is approximately 49.3 km/h.

The magnitude of the average velocity is approximately 21.4 km/h.

The average speed is not equal to the magnitude of the average velocity due to the circuitous path taken by the cab.

12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall ?

Ans : Given:

* Maximum height, h = 25 m

* v stands for Velocity of projection, = 40 m/s

* g stands for Acceleration due to gravity, = 9.8 m/s²

To find:

* Horizontal range, R

Solution:

We know that the maximum height reached by a projectile is given by:

h = (v² sin²θ) / (2g)

Substituting the given values:

25 = (40² sin²θ) / (2 * 9.8)

Solving for sinθ:

sin²θ = 0.30625

sinθ = 0.5534

θ = sin⁻¹(0.5534) ≈ 33.6°

Given,

horizontal range of a projectile :

R = (v² sin2θ) / g

Substituting the values:

R = (40² sin(2 * 33.6°)) / 9.8

R ≈ 150.5 m

Therefore, the maximum horizontal distance the ball can travel without hitting the ceiling is approximately 150.5 meters.

13 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ?

Ans : Given:

Maximum range (R_max) = 100 m

Calculation:

Finding the initial velocity (u):

given,

maximum range of a projectile

R_max = (v^2) / g

Substituting the given value of R_max:

100 = (v^2) / g

Therefore, the initial velocity (u) is equal to the velocity (v) at the maximum height.

Finding the maximum height (s):

Using the equation of motion:

v^2 = u^2 + 2as

Here,

Final velocity (v) = 0 (at the maximum height)

Initial velocity (u) = v (as calculated above)

Acceleration (a) = -g (acceleration due to gravity)

Distance (s) = maximum height

Substituting these values:

0^2 = v^2 + 2(-g)s

Rearranging the equation:

s = (v^2) / (2g)

Substituting the value of (v^2) from the first step:

s = (1/2) * 100 = 50 m

Conclusion:

The maximum height reached by the projectile is 50 meters.

14 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?

Ans : Here,

r = 80 cm = 0.8 m

v = 14/25 rev/s

Angular velocity, ω = 2πv = 2π × (14/25) rad/s

= 88/25 rad/s

The centripetal acceleration, a = ω²r = (88/25)² × 0.80

= 9.90 m/s²

15. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Ans : To compare the centripetal acceleration of the aircraft with the acceleration due to gravity, we need to calculate both values.

1. Centripetal acceleration:

The formula for centripetal acceleration is:

a = v²/r

where a is the centripetal acceleration, v is the linear speed, and r is the radius of the loop.

Converting the speed to m/s:

900 km/h = 900 * 1000 m/km / 3600 s = 250 m/s

Substituting the values:

a = (250 m/s)² / (1000 m)

a = 62.5 m/s²

2. Acceleration due to gravity:

Gravity’s acceleration is about 9.81 m/s²

Comparing the two accelerations:

a/g = 62.5 m/s² / 9.81 m/s² ≈ 6.37

This means that the centripetal acceleration of the aircraft is approximately 6.37 times the acceleration due to gravity.

Therefore, the centripetal acceleration of the aircraft is significantly greater than the acceleration due to gravity.

16 Read each statement below carefully and state, with reasons, if it is true or false:

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

Ans :

(a) False. The net acceleration of a particle in circular motion is always directed towards the center of the circle, regardless of whether the motion is uniform or non-uniform.

(b) True. When a particle leaves a circular path, it follows a tangent to the circle at the point of departure. This is due to inertia, which causes the particle to continue moving in a straight line unless acted upon by an external force.

(c) True. In uniform circular motion, the acceleration vector is always directed towards the center of the circle, but its direction constantly changes to maintain this inward orientation. As a result, when we add up all these acceleration vectors over one complete revolution, they cancel each other out, leading to a net acceleration of zero over the entire cycle.

17 The position of a particle is given by

where t is in seconds and the coefficients have the proper units for r to be in metres.

(a) Find the v and a of the particle?

(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?

Ans : Given:

Position vector: r = 3.0t i – 2.0t² j + 4.0 k m

(a) Finding Velocity (v) and Acceleration (a)

Velocity:

Velocity is the time derivative of position.

v = dr/dt = d(3.0t i – 2.0t² j + 4.0 k)/dt

v = 3.0 i – 4.0t j m/s

Acceleration:

Acceleration is the time derivative of velocity.

a = dv/dt = d(3.0 i – 4.0t j)/dt

a = -4.0 j m/s²

(b) Finding the magnitude and direction of velocity at t = 2.0 s

Evaluating v(t) at t = 2.0 s

v(2) = 3.0 i – 4.0(2.0) j = 3.0 i – 8.0 j m/s

The length of the velocity vector is..

|v(2)| = √(3.0² + (-8.0)²) = √73 ≈ 8.54 m/s

The angle between the velocity vector and the positive x-axis can be determined using the arctangent function

θ = arctan(-8.0 / 3.0) ≈ -69.45°

Therefore:

The velocity of the particle at t = 2.0 s is 8.54 m/s, directed at an angle of approximately -69.45° (below the positive x-axis).

18 A particle starts from the origin at t = 0 s with a velocity of 10.0 jɵ m/s and moves in the x-y plane with a constant acceleration of (8.0 2.0 ) ɵ ɵ i j + m s-2.

(a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

(b) What is the speed of the particle at the time ?

Ans : Given:

Initial position: r(0) = 0

Initial velocity: v(0) = 10.0 j m/s

Constant acceleration: a = 8.0 i + 2.0 j m/s²

Equations of Motion:

x(t) = x(0) + v_x(0)t + (1/2)a_xt²

y(t) = y(0) + v_y(0)t + (1/2)a_yt²

where:

x(t) and y(t) are the x and y coordinates at time t

v_x(0) and v_y(0) are the initial x and y components of velocity

a_x represents the acceleration in the x-direction, and a_y represents the acceleration in the y-direction.

Substituting the given values:

x(t) = 0 + 0*t + (1/2)(8.0)t² = 4t²

y(t) = 0 + 10.0t + (1/2)(2.0)t² = 10t + t²

(a) Finding the time when x-coordinate is 16 m:

Set x(t) = 16 and solve for t:

16 = 4t²

t² = 4

t = ± 2 seconds

Since time cannot be negative, t = 2 seconds.

To find the y-coordinate at this time, substitute t = 2 into the y(t) equation:

y(2) = 10(2) + (2)² = 24 m

(b) Finding the speed at t = 2 seconds:

The velocity of the object at any given moment can be described by the vector v(t).

v(t) = v(0) + at

v(t) = 10.0 j + (8.0 i + 2.0 j)t

At t = 2 seconds:

v(2) = 10.0 j + (8.0 i + 2.0 j)(2) = 16.0 i + 14.0 j m/s

The speed of an object is the rate at which it is moving, regardless of its direction.

Speed = |v(2)| = √(16.0² + 14.0²) ≈ 21.26 m/s

Therefore:

The x-coordinate of the particle is 16 m at t = 2 seconds.

The y-coordinate of the particle at t = 2 seconds is 24 m.

The speed of the particle at t = 2 seconds is approximately 21.26 m/s.

19. ɵ i and ɵ j are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors ɵ ɵ i j + , and ɵ ɵ i j − ? What are the components of a vector A= 2 ɵ ɵ i j + 3 along the directions of ɵ ɵ i j + and ɵ ɵ i j − ? [You may use graphical method]

Ans : (i)|i + j| = √(1² + 1² + 2*1*1*cos 90°) = √2 = 1.414 units

tan θ = 1/1 = 1 => θ

= 45°

So, the vector i + j makes an angle of 45° with the x-axis.

(ii) |i – j| = √(1² + (-1)² – 2*1*1*cos 90°)

= √2 = 1.414 units

The vector i – j makes an angle of -45° with the x-axis.

(iii) Let’s determine the component of A = 2i + 3j in the direction of i + j.

Let B = i + j

A.B =

(A cos θ)B = AB cos θ

So, the component of A in the direction of B is A.B/B

= [(2i + 3j) . (i + j)] / √(1² + 1²)

= (2i.i + 2i.j + 3j.i + 3j.j) / √2

= 5/√2 units

(iv) Component of A in the direction of i – j = [(2i + 3j) . (i – j)] / √2 = (2*1 + 3*(-1)) / √2 = -1/√2 units

20. For any arbitrary motion in space, which of the following relations are true :

(a) vaverage = (1/2) (v (t 1 ) + v (t 2 ))

(b) v average = [r(t 2 ) – r(t 1 ) ] /(t 2 – t 1 )

(c) v (t) = v (0) + a t

(d) r (t) = r (0) + v (0) t + (1/2) a t 2

(e) a average =[ v (t 2 ) – v (t 1 )] /( t 2 – t 1 )

(The ‘average’ stands for average of the quantity over the time interval t 1 to t 2 )

Ans : For any arbitrary motion in space, the following relations are true:

(b) v_average = [r(t2) – r(t1)] / (t2 – t1)

(e) a_average = [v(t2) – v(t1)] / (t2 – t1)

These equations are valid for both constant and non-constant acceleration.

The other options are not always true for arbitrary motion:

(a) v_average = (1/2)(v(t1) + v(t2)) is only true for uniformly accelerated motion.

(d) r(t) = r(0) + v(0)t + (1/2)at^2 is only true for uniformly accelerated motion.

In summary, the average velocity and average acceleration can be calculated using the equations (b) and (e) for any arbitrary motion, while the other equations are only valid for specific types of motion, such as uniformly accelerated motion.

21 Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that

(a) is conserved in a process

(b) can never take negative values

(c) must be dimensionless

(d) does not vary from one point to another in space

(e) has the same value for observers with different orientations of axes.

Ans :

(a) False. Kinetic energy, being a scalar quantity, only possesses magnitude.

(b) False. Gravitational potential energy can be negative, especially when the object is below the reference level (e.g., an object in a well).

(c) False. Physical quantities like mass, length, time, speed, and work have dimensions, which are combinations of fundamental units like mass, length, and time.

(d) False. Quantities like speed and energy can vary depending on the location in space. For example, the potential energy of an object changes with its height.

(e) True. Scalar quantities are independent of the orientation of the coordinate system. They have magnitude only and no direction. Therefore, their values remain the same for observers with different orientations.

22 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, wat is the speed of the aircraft ?

Ans : Here, O represents the observation point situated on the ground. A and B are the positions of the aircraft, with ∠AOB = 30°.

Drawing a perpendicular OC to AB, we form right-angled triangles AOC and BOC.