Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 6.9:

A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to

(i)   (ii) t (iii)   (iv) 

Answer:

Answer: (ii) t

Mass of the body = m

Acceleration of the body = a

Using Newton’s second law of motion, the force experienced by the body is given by the equation:

F = ma

Both m and a are constants. Hence, force F will also be a constant.

F = ma = Constant … (i)

For velocity v, acceleration is given as,

Power is given by the relation:

P = F.v

Using equations (i) and (iii), we have:

P ∝ t

Hence, power is directly proportional to time.

Question 6.10:

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to

(i)   (ii) t (iii)   (iv) 

Answer:

Answer: (iii)

Power is given by the relation:

P = Fv

Integrating both sides:

On integrating both sides, we get:

Question 6.11:

A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by 

Where  are unit vectors along the x-, y– and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?

Answer:

Force exerted on the body, 

Displacement, s =  m

Work done, W = F.s

Hence, 12 J of work is done by the force on the body.

Question 6.12:

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass = 9.11 × 10^–31 kg, proton mass = 1.67 × 10^–27 kg, 1 eV = 1.60 × 10^–19 J).

Answer:

Electron is faster; Ratio of speeds is 13.54 : 1

Mass of the electron, m_e = 9.11 × 10^–31 kg

Mass of the proton, m_p = 1.67 × 10^{– 27} kg

Kinetic energy of the electron, E_Ke = 10 keV = 10⁴ eV

= 10⁴ × 1.60 × 10^–19

= 1.60 × 10^–15 J

Kinetic energy of the proton, E_Kp = 100 keV = 10⁵ eV = 1.60 × 10^–14 J

Hence, the electron is moving faster than the proton.

The ratio of their speeds:

Question 6.13:

A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s^–1?

Answer:

Radius of the rain drop, r = 2 mm = 2 × 10^–3 m

Volume of the rain drop, 

Density of water, ρ = 10³ kg m^–3

Mass of the rain drop, m = ρV

= 

Gravitational force, F = mg

= 

The work done by the gravitational force on the drop in the first half of its journey:

W_I = Fs

=

 × 250

= 0.082 J

This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey, i.e., W_II, = 0.082 J

As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same.

∴Total energy at the top:

E_T = mgh + 0

=

 × 500 × 10^–5

= 0.164 J

Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s.

∴Total energy at the ground:

∴Resistive force = E_G – E_T = –0.162 J

Question 6.14:

A molecule in a gas container hits a horizontal wall with speed 200 m s^–1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Answer:

Yes; Collision is elastic

The momentum of the gas molecule remains conserved whether the collision is elastic or inelastic.

The gas molecule moves with a velocity of 200 m/s and strikes the stationary wall of the container, rebounding with the same speed.

It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.

Question 6.15:

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Answer:

Volume of the tank, V = 30 m³

Time of operation, t = 15 min = 15 × 60 = 900 s

Height of the tank, h = 40 m

Efficiency of the pump,  = 30%

Density of water, ρ = 10³ kg/m³

Mass of water, m = ρV = 30 × 10³ kg

Output power can be obtained as:

For input power Pi,, efficiency   is given by the relation:

Question 6.16:

Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following figure is a possible result after collision?

Answer:

Answer: Case (ii)

It can be observed that the total momentum before and after collision in each case is constant.

For an elastic collision, the total kinetic energy of a system remains conserved before and after collision.

For mass of each ball bearing m, we can write:

Total kinetic energy of the system before collision:

Case (i)

Total kinetic energy of the system after collision:

Hence, the kinetic energy of the system is not conserved in case (i).

Case (ii)

Total kinetic energy of the system after collision:

Hence, the kinetic energy of the system is conserved in case (ii).

Case (iii)

Total kinetic energy of the system after collision:

Hence, the kinetic energy of the system is not conserved in case (iii).