Question:

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

Answer:

1. Magnitude and direction of i^+j^​: The magnitude of the vector i^+j^​ is √2 and the direction of the vector is at 45° with respect to the x-axis.

2. Magnitude and direction of i^−j^​: The magnitude of the vector i^−j^​ is √2 and the direction of the vector is at -45° with respect to the x-axis.

3. Components of the vector A= 2i^+3j^​ along the directions of i^+j^​ and i^−j^​: The components of the vector A along the directions of i^+j^​ and i^−j^​ can be determined using the graphical method. The components of the vector A along the directions of i^+j^​ and i^−j^​ are √2 and -√2 respectively.

Question:

A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/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 600 ms^−1 to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g=10 ms^−2)

Answer:

1. Calculate the time taken by the plane to pass over the anti-aircraft gun: Time taken = Distance/Speed = 1.5 x 1000/720 = 2.08 hours

2. Calculate the horizontal distance travelled by the plane in the given time: Distance = Speed x Time = 720 x 2.08 = 1488 km

3. Calculate the angle from the vertical at which the gun should be fired for the shell to hit the plane: Angle = tan-1 (v/u) = tan-1 (600/1488) = 17.3°

4. Calculate the minimum altitude at which the pilot should fly the plane to avoid being hit: Minimum Altitude = (v^2/2g) + 1.5 = (600^2/2 x 10) + 1.5 = 4.5 km

Question:

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

Answer:

1. Calculate the time taken for the ball to reach the ceiling of the hall:

t = 25m / 40ms^−1 = 0.625s

2. Calculate the maximum horizontal distance the ball can travel in time t:

d = v * t

d = 40ms^−1 * 0.625s = 25m

Question:

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 ?

Answer:

1. Calculate the angular velocity of the stone: Angular velocity = (14 revolutions/25 s) x (2π radians/1 revolution) = 2.8 radians/s.

2. Calculate the linear velocity of the stone: Linear velocity = (2.8 radians/s) x (80 cm/1 s) = 224 cm/s.

3. Calculate the magnitude of the acceleration: Magnitude of acceleration = (Linear velocity)²/Radius of circle = (224 cm/s)²/80 cm = 56 cm/s².

4. Calculate the direction of the acceleration: The direction of the acceleration is toward the center of the circle.

Question:

The position of a particle is given byr=3.0ti^−2.0t2j^​+4.0k^mwhere 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 ?

Answer:

(a) v = 6.0ti^−4.0t2j^+0k^ m/s a = -12.0ti^−6.0t2j^+0k^ m/s^2

(b) Magnitude of velocity = 8.944 m/s Direction of velocity = 45°

Question:

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

Answer:

a) At t=2s, the x-coordinate of the particle is 16 m. The y-coordinate of the particle at that time is 24 m.

b) The speed of the particle at t=2s is 28 m/s.

Question:

For any arbitrary motion in space, which of the following relations are true: a) vaverage​=(1/2)(v(t1​)+v(t2​)) b) vaverage​=[r(t2​)−r(t1​)]/(t2​−t1​) c) v(t)=v(0)+at d) r(t)=r(0)+v(0)t+(1/2)at^2 e) aaverage​=[v(t2​)−v(t1​)]/(t2​−t1​) The ‘average’ stands for average of the quantity over the time interval t1​ to t2​)

Answer:

a) True b) True c) True d) True e) False

Question:

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.

Answer:

(a) False. A scalar quantity is one that does not have direction and can be described by a single value. Examples of scalar quantities include speed, mass, temperature and time. However, scalar quantities are not necessarily conserved in a process. For example, energy is a scalar quantity, but it is not conserved in all processes.

(b) False. A scalar quantity can take negative values. For example, temperature can take negative values (below 0°C).

(c) False. A scalar quantity does not have to be dimensionless. For example, speed has the SI unit of meters per second (m/s).

(d) True. A scalar quantity does not vary from one point to another in space. For example, temperature is a scalar quantity and has the same value at all points in a room.

(e) True. A scalar quantity has the same value for observers with different orientations of axes. For example, the temperature in a room is the same regardless of the orientation of the observer’s axes.

Question:

A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector ? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector ?

Answer:

No, anything that has magnitude and direction is not necessarily a vector. A vector is a quantity that has both magnitude and direction, but other physical quantities such as force, acceleration, and velocity may also have both magnitude and direction.

No, a rotation does not make it a vector. A vector must have a fixed starting point and a fixed direction, while a rotation does not have a fixed starting point or a fixed direction.

Question:

On an open ground, a motorist follows a track that turns to his left by an angle of 60o 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.

Answer:

1. The displacement of the motorist at the third turn is 500m x 3 x cos60o = 1500m.

2. The displacement of the motorist at the sixth turn is 500m x 6 x cos60o = 3000m.

3. The displacement of the motorist at the eighth turn is 500m x 8 x cos60o = 4000m.

4. The magnitude of the displacement is less than the total path length covered by the motorist in each case as the motorist is following a curved path.

Question:

A man can swim with a speed of 4.0 km/hr in still water. How long does he take to cross a river 1.0 km wide, if the river flows steadily at 3.0 km/hr and he makes his strokes normal to the river current? How far down the river does he go, when he reaches the other bank?

Answer:

1. Calculate the speed of the man in the river: Speed of the man in the river = 4.0 km/hr + 3.0 km/hr = 7.0 km/hr

2. Calculate the time taken to cross the river: Time taken to cross the river = 1.0 km/7.0 km/hr = 0.14285714 hrs = 8.57 minutes

3. Calculate the distance travelled down the river: Distance travelled down the river = Speed x Time = 7.0 km/hr x 0.14285714 hrs = 1.0 km

Question:

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.

Answer:

Volume: Scalar Mass: Scalar Speed: Scalar Acceleration: Vector Density: Scalar Number of Moles: Scalar Velocity: Vector Angular Frequency: Scalar Displacement: Vector Angular Velocity: Vector

Question:

Given a + b + c + d = 0, which of the following statements are correct. (a) a, b, c, 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 ?

Answer:

Answer: (a) False (b) False (c) True (d) False

Question:

Rain is falling vertically with a speed of 30m s^−1. A woman rides a bicycle with a speed of 10m s^−1 in the north to south direction. What is the direction in which she should hold her umbrella?

Answer:

1. The woman should hold her umbrella in the south to north direction.

2. This is because the rain is falling vertically downwards with a speed of 30m s^−1, and the woman is travelling in the opposite direction (north to south) with a speed of 10m s^−1.

Question:

Read each statement below carefully false :

1. The magnitude of a vector is always a scalar,
2. each component of a vector is always a scalar,
3. the total path length is always equal to the magnitude of the displacement vector of a particle,
4. 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,
5. Three vectors not lying In a plane can never add up to give a null vector.

Answer:

1. False
2. True
3. False
4. False
5. False

Question:

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 ?

Answer:

Step 1: Understand the question.

Step 2: Determine that the answer cannot be determined from the information given. The maximum horizontal distance does not provide enough information to calculate the maximum height.

Question:

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.

Answer:

1. Calculate the speed of the aircraft in m/s: Speed = 900 km/h = 900 x (1000/3600) m/s = 250 m/s

2. Calculate the centripetal acceleration of the aircraft: Centripetal acceleration = (V^2/r) Centripetal acceleration = (250^2/1000) Centripetal acceleration = 62.5 m/s^2

3. Calculate the acceleration due to gravity: Acceleration due to gravity = 9.8 m/s^2

4. Compare the centripetal acceleration and the acceleration due to gravity: The centripetal acceleration is approximately 6.3 times greater than the acceleration due to gravity.

Question:

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.

Answer:

(a) True. The net acceleration of a particle in circular motion is always directed towards the centre of the circle, and is known as the centripetal acceleration.

(b) True. The velocity vector of a particle at a point is always tangential to the path of the particle at that point.

(c) False. The acceleration vector of a particle in uniform circular motion averaged over one cycle is not a null vector, but is instead directed towards the centre of the circle and is known as the centripetal acceleration.

Question:

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 apart is 30o, what is the speed of the aircraft ?

Answer:

1. Find the distance travelled by the aircraft in 10.0 s: Distance = (speed)(time) Distance = (speed)(10.0 s)

2. Find the length of the arc subtended by the aircraft positions 10.0 s apart: Arc Length = (2πr)(θ/360°) Arc Length = (2π)(3400 m)(30°/360°) Arc Length = 1740 m

3. Calculate the speed of the aircraft: Speed = (Distance)/(Time) Speed = (1740 m)/(10.0 s) Speed = 174 m/s

Question:

A bullet fired at an angle of 30o with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

Answer:

1. Yes, one can hope to hit a target 5.0 km away by adjusting the angle of projection.

2. To determine the new angle of projection, we can use the equation of motion for projectile motion: x = v₀cosθt

3. We can rearrange this equation to solve for the angle of projection: θ = cos⁻¹(x/v₀t)

4. We can substitute the given values into the equation: θ = cos⁻¹(5000/v₀*t)

5. We can then solve for the new angle of projection: θ = cos⁻¹(5000/v₀*t) = 45°

Question:

A vector has magnitude and direction. Does it have a location in space ? Can it vary with time ? Will two equal vectors a and b at different locations in space necessarily have identical physical effects ? Give examples in support of your answer.

Answer:

Answer:

Yes, a vector has a location in space. It can vary with time depending on the direction and magnitude of the vector.

No, two equal vectors a and b at different locations in space will not necessarily have identical physical effects. For example, consider two equal vectors a and b, where a is pointing from the origin to the point (2,3) and b is pointing from the origin to the point (4,6). Even though these two vectors are equal in magnitude and direction, they will have different physical effects because of the different locations in space. For example, if these two vectors were used to represent forces, then the force represented by vector a will act on point (2,3) while the force represented by vector b will act on point (4,6).

Question:

A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Answer:

1. Calculate the angular velocity of the cyclist. Angular velocity = (27 km/h) / (80 m) = 0.3375 rad/s

2. Calculate the centripetal acceleration of the cyclist. Centripetal acceleration = (0.3375 rad/s)2 x 80 m = 108.75 m/s2

3. Calculate the acceleration due to braking. Acceleration due to braking = 0.50 m/s2

4. Calculate the net acceleration of the cyclist. Net acceleration = Centripetal acceleration - Acceleration due to braking = 108.75 m/s2 - 0.50 m/s2 = 108.25 m/s2

5. Determine the direction of the net acceleration. The direction of the net acceleration is towards the center of the circular turn.

Question:

Show that, (a) for a projectile the angle between the velocity and the x-axis as a function of time is given by θ(t)=tan^−1(vo​y−gt​/vo​x)(b) Shows that the projection angle θo​ for a projectile launched from the origin is given by θo​=tan^−1(4hm​​/R)where the symbols have their usual meaning.

Answer:

(a) We know that the equation of motion for a projectile is given by

x = xo + vox t

y = yo + voy t - 1/2 gt2

Differentiating the equation with respect to time, we get

Vx = vox

Vy = voy - gt

The angle between the velocity and the x-axis is given by

θ(t) = tan-1 (Vy/Vx)

Substituting the values of Vx and Vy, we get

θ(t) = tan-1 [(voy - gt)/vox]

(b) For a projectile launched from the origin, xo = 0 and yo = 0.

Therefore, the range of the projectile is given by

R = vox * t

where t is the time taken for the projectile to reach the maximum height hm.

At the maximum height, the velocity of the projectile is zero. Therefore,

0 = voy - gt

Solving for t, we get

t = voy / g

Substituting the value of t in the equation for the range, we get

R = vox * voy / g

The projection angle θo is given by

θo = tan-1 (4hm/R)

Substituting the value of R, we get

θo = tan-1 (4hm/vox * voy / g)

Therefore, the projection angle θo for a projectile launched from the origin is given by

θo = tan-1 (4hm/vox * voy / g)

Question:

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.

Answer:

Answer:

1. Force
2. Current

Question:

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.

Answer:

(a) Adding any two scalars is meaningful. This is because scalars are numbers that can be added together to produce a new number.

(b) Adding a scalar to a vector of the same dimensions is meaningful. This is because a scalar is a number that can be added to a vector of the same dimensions, resulting in a vector with the same dimensions but with a different magnitude.

(c) Multiplying any vector by any scalar is meaningful. This is because a scalar is a number that can be used to multiply a vector, resulting in a vector with the same direction but a different magnitude.

(d) Multiplying any two scalars is meaningful. This is because scalars are numbers that can be multiplied together to produce a new number.

(e) Adding any two vectors is meaningful. This is because two vectors can be added together to produce a new vector with the same direction but a different magnitude.

(f) Adding a component of a vector to the same vector is meaningful. This is because a component of a vector can be added to the same vector, resulting in a vector with the same magnitude but a different direction.

Question:

In a harbor, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbor flutters along the N-E direction. If the boat starts moving at as speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ?

Answer:

1. Calculate the resultant velocity of the boat, which is the vector sum of the wind velocity and the boat velocity.

2. Calculate the angle of the resultant velocity with respect to the north direction.

3. The direction of the flag on the mast of the boat will be the same as the direction of the resultant velocity.

Question:

Establish the following vector inequalities geometrically or otherwise: (a) ∣a+b∣≤∣a∣+∣b∣ (b) ∣a+b∣≥∣∣a∣−∣b∣∣

Answer:

(a) Let a = (a1, a2, …, an) and b = (b1, b2, …, bn) be two vectors. Then, ∣a+b∣ = √[(a1+b1)^2 + (a2+b2)^2 + … + (an+bn)^2]

Now, ∣a∣ = √[a1^2 + a2^2 + … + an^2] and ∣b∣ = √[b1^2 + b2^2 + … + bn^2].

Therefore, ∣a+b∣≤∣a∣+∣b∣ becomes

√[(a1+b1)^2 + (a2+b2)^2 + … + (an+bn)^2] ≤ √[a1^2 + a2^2 + … + an^2] + √[b1^2 + b2^2 + … + bn^2]

This inequality can be established by the triangle inequality, which states that the length of the sum of two sides of a triangle is always less than or equal to the length of the third side.

(b) Let a = (a1, a2, …, an) and b = (b1, b2, …, bn) be two vectors. Then, ∣a+b∣ = √[(a1+b1)^2 + (a2+b2)^2 + … + (an+bn)^2]

Now, ∣a∣ = √[a1^2 + a2^2 + … + an^2] and ∣b∣ = √[b1^2 + b2^2 + … + bn^2].

Therefore, ∣a+b∣≥∣∣a∣−∣b∣∣ becomes

√[(a1+b1)^2 + (a2+b2)^2 + … + (an+bn)^2] ≥ √[(a1-b1)^2 + (a2-b2)^2 + … + (an-bn)^2]

This inequality can be established by the triangle inequality, which states that the length of the difference of two sides of a triangle is always greater than or equal to the length of the third side.

Question:

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?

Answer:

(a) The average speed of the taxi = (23 km)/(28 min) = 0.82 km/min

(b) The magnitude of average velocity = (10 km)/(28 min) = 0.36 km/min

No, the two are not equal.

Question:

Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere ? Explain.

Answer:

(a) Yes, you can associate vectors with the length of a wire bent into a loop. This can be done by using the vector equation of a circle, which is given by r = xi + yj, where x and y are the coordinates of the center of the circle and r is the radius of the circle. This equation can be used to calculate the length of the wire in terms of the radius of the circle.

(b) Yes, you can associate vectors with a plane area. This can be done by using the vector equation of a plane, which is given by ax + by + cz = d, where a, b, c and d are the coefficients of the equation and x, y and z are the coordinates of any point on the plane. This equation can be used to calculate the area of the plane in terms of the coefficients.

(c) Yes, you can associate vectors with a sphere. This can be done by using the vector equation of a sphere, which is given by (x - a)2 + (y - b)2 + (z - c)2 = r2, where a, b and c are the coordinates of the center of the sphere and r is the radius of the sphere. This equation can be used to calculate the surface area of the sphere in terms of the radius.