Q:11. A body constrained to move along the z – axis of a coordinate system is subjected to a constant force $\displaystyle \vec{F} = \hat{i} + 2\hat{j} + 3\hat{k} $ N. What is the work done by this force in moving the body over a distance of 4m along the z – axis ?

Sol.

$ \displaystyle \vec{F} = \hat{i} + 2\hat{j} + 3\hat{k} $ N

and $ \displaystyle \vec{S} = 4\hat{k} $

$\displaystyle W = \vec{F}. \vec{S} $

$ \displaystyle W = (\hat{i} + 2\hat{j} + 3\hat{k}) .(4\hat{k}) $

= 12 J

Q: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 proton? Obtain the ratio of their speeds. Take mass of electron = 9.11×10^{-31} kg. mass of proton = 1.67 × 10^{-27} kg. and 1 keV = 1.6 × 10^{-16}J.

Sol.

E_{1} = (1/2) m_{e} v_{e}^{2} = 10 × 1.6 × 10^{-16} J

E_{2} = (1/2) m_{P} v_{P}^{2} = 100 × 1.6 × 10^{-16} J

On Dividing,

$\displaystyle \frac{m_e v_e^2}{m_p v_p^2} = \frac{1}{10} $

$ \displaystyle \frac{v_e}{v_p} = \sqrt{\frac{1}{10}.\frac{m_p}{m_e}} $

$ \displaystyle \frac{v_e}{v_p} = \sqrt{\frac{1}{10}.\frac{1.67\times 10^{-27}}{9.1\times 10^{-31}}} $

v_{e} / v_{P}= 13.53

v_{e} = 13.53 v_{P}

Hence electron is faster than the proton.

Q: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 air until half of 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 half and second half of the journey ?

[Density of H_{2}O = 10^{3} kg/m3]. What is the work done by the resistance force in the entire journey if its speed on reaching the ground is 10 m/s ?

Sol. r = 2 mm = 2 × 10^{-3} m, ρ = 10^{3} kg/m^{3}

Distance moved in half of the journey = s = 500/2 = 250 m.

Mass of the drop = vol. × density $ \displaystyle = \frac{4}{3}\pi r^3 \rho $

$ \displaystyle = \frac{4}{3}\times \frac{22}{7}(2\times 10^{-3})^3 \times 10^3 $

= 3.35 × 10^{-5} kg

Work done = W = F × s

= 3.35 × 10^{-5} × 9.8 × 250 = 0.082 J

Work done by the gravitational force remains same whether the drop falls under decreasing acceleration or with uniform speed. So, the work done in second half of the journey is also 0.082 J.

Energy of the drop in reaching the ground = E = mgh

= 3.35 × 10^{-5} × 9.8 × 500 = 0.164 J

(considering no resistive forces).

Kinetic energy $\displaystyle = \frac{1}{2}mv^2 = \frac{1}{2}(3.35\times 10^{-5})\times 10^2 $

= 1.675 × 10^{-3}

= 0.001675

Work done by the resistive force = 0.164 – 0.001675 = 0.1623 J

Q: 14. A molecule in a gas container hits the speed 200 m/s at an angle 30° with the normal and rebound with the same speed. Is momentum conserved in the collision? Is collision elastic or inelastic?

Sol. Momentum is conserved in all types of collisions.

Let m → mass of the molecule, M → mass of the wall

K.E. after collision $ \displaystyle = \frac{1}{2}m(200)^2 + \frac{1}{2}M(0^2) $

= 2 × 10^{4} mJ.

K.E. before collision = $ \displaystyle = \frac{1}{2}m(200)^2 $

= 2 × 10^{4} m J.

As there is No loss of Kinetic Energy , hence the collision is elastic.

Q:15. A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 minutes. 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?

Sol. Volume of water = 30m^{3}, t = 15 min = 15 × 60 = 900 s.

h = 40m, efficiency = 30% = (Out put water)/(In put power) ,

ρ_{H2O} = 10^{3} kg/m^{3}

Mass of water pumped = Vol × density = 30 × 10^{3} kg

Output power = W/t = mgh/t

$ \displaystyle = \frac{30\times 10^3 \times 9.8\times 40}{900} $

= 13070 W

Input power = (Output power)/(Efficiency )

$ \displaystyle = \frac{13070}{0.30} $

= 43567 W = 43.567 kW

Q: 16. Two identical balls in contact with 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, with of the following is a possible result after collision?

Sol. Let m → mass of each ball bearing,

Before collision, total KE of the system

= (1/2)mv^{2} + 0 = (1/2)mv^{2}

After collision, KE of the system in case I is, E1 = (1/2)(2m) (v/2)^{2} = 1/4 mv^{2}

In case II is, E2 = 1/2 mv^{2}

In case III is, E3 = (1/2)(3m) (v/3)^{2} = (1/6) mv^{2}

KE is conserved in case II. The only possibility is case II.

Q: 17.The bob of a simple pendulum from 300 to the vertical hits anther bob B of the same mass at rest on a table. How high does the bob A rise after the collision?

Neglect the size of the bobs the assume the collision to be elastic.

Sol. When two bodies of same mass undergo elastic collision their velocities are interchanged. So after collision, A will not rise but it will come to rest and B will move with the velocity of A.

Q: 18. The bob of pendulum is released from its horizontal position A as shown in the figure. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowest point B, given that it pendulum is 1.5 m, what is the speed with the bob arrives at the lowest point B, given that it dissipates 5% of its initial energy air resistance?

Sol. h = 1.5 m, v = ?, Energy dissipated = 5% ]

∵ B is the lowest position of the bob, its P.E. at B is zero. At A, it is mgh.

While going from A to B, P.E. is converted to k.E. Energy = 95% (mgh).

If v the velocity at B, then

$\displaystyle KE = \frac{1}{2}mv^2 = \frac{95}{100}mgh $

$ \displaystyle v = \sqrt{\frac{95}{100}2gh}$

$ \displaystyle v = \sqrt{\frac{19}{20}(2\times 9.8\times 1.5)}$

= 5.258 m/s

Q: 19. A trolley of mass 300 kg carrying a sand bag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the trolley’s floor at the rate of 0.05 kg/s. what is the speed of the trolley after the entire sand bag is emptied ?

Sol. Since the trolley carrying the sand bag is moving uniformly, so the external force on the system is zero. When the sand leaks out, it doesn’t apply any external force on the trolley. So the speed of the trolley doesn’t change.

Q: 20. A particle of mass 0.5 kg travels in a straight line with velocity v = ax^{3/2} , where a = 5m^{-1/2}s^{-1}. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Sol. m = 0.5 kg, v = ax^{3/2}, a = 5m^{-1/2}s^{-1}

Initial velocity at x = 0 is v1 = a × 0 = 0

Final velocity at x = 2 is v_{2} = a.2^{3/2} = 5 × 2^{3/2}

Work done = Inerease in KE

$ \displaystyle = \frac{1}{2}m(v_2^2 – v_1^2) $

$\displaystyle = \frac{1}{2}\times 0.5((5\times2^{3/2})^2 – 0) $

= 50 J

Q:21.The blades of a windmill sweep out a circle of area A.

(a)If the wind flows at a velocity v perpendicular to the circle, what is the mass of air passing through it in time t ?

(b)What is KE of the air?

(c)Assume that the wind mill converts 25% of the wind’s energy into electrical energy and that A = 30 m^{2}, v = 36 km/h and the density of air is 1.2 kg/m3. What is the electrical power produced?

Sol. Vol. of wind flowing/s = Av, Mass of wind flowing/s = Avρ

Mass of wind flowing in t sec = Avρt ;

KE of air $ \displaystyle \frac{1}{2}mv^2 = \frac{1}{2}(Av\rho t)v^2 = \frac{1}{2}Av^3\rho t $

Electrical energy produced $\displaystyle = \frac{25}{100}\times KE \, of \, air $

$\displaystyle = \frac{1}{4}\times\frac{1}{2}Av^3\rho t $

Power = Energy/Time

$ \displaystyle = (\frac{1}{8}Av^3 \rho t )/t $

$ \displaystyle = \frac{1}{8}Av^3 \rho $

$\displaystyle = \frac{1}{8}\times 30\times 10^3 \times 1.2 $

= 4500 W = 4.5 kW

Q:22.A person trying to lose weight 10 kg mass, to a height 0.5, 1000 times. Assume that PE lost each time she lowers the mass the mass is dissipated

(a)How much work does she do against the gravitational force?

(b)Fat supplies 3.8 × 10^{7} J of energy /kg which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up ?

Sol. m = 10 kg, h = 0.5 m, n = 1000

Work done against the gravitational force = n (mgh)

= 1000 × 10 × 9.8 × 0.5 = 49000 J.

Mechanical energy supplied by 1 kg of fat = 3.8 × 10^{7} × 20/100

= 0.76 × 10^{7} J/kg

Fat used up by dieter = 1/(0.76 ×10^{7} ) × 49000

= 6.45 × 10^{-3} kg

Q:23.A family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

Sol. Let A = Area of the surface in square metre.

Total power = 200 A,

Useful electrical energy produced/s = 20/100(200A) = 8 kW

= 8000 W

A = 800/40 = 200 sq.m.

Q:24.A bullet of mass 0.012 kg and horizontal speed 70 m/s strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by a wire. Calculate the height to which the block rises. Also estimate the amount of heat produced in the block

Sol. m_{1} = 0.012 kg, u_{1} = 70 m/s, m_{2} = 0.4 kg, u_{2} = 0

As the bullet gets embedded in the block, the two behave as one body.

Let v = velocity of net combination.

Applying law of conservation of linear momentum,

(m_{1} + m_{2})v = mv_{1}u_{1} + m_{2}u_{2} = m_{1}u_{1}

$\displaystyle v = \frac{m_1 u_1}{m_1+m_2} $

$ \displaystyle v = \frac{0.012\times 70}{0.012+0.4} =\frac{0.84}{0.412} $

v = 2.04 m/s

Let h = height to which the block rises.

$ \displaystyle h = \frac{v^2}{2g} = \frac{2.04\times 2.04}{2\times 9.8} $

= 0.212 m

Heat produced = energy lost

= initial KE of the bullet – final KE of the combination

$\displaystyle = \frac{1}{2}m_1u_1^2 – \frac{1}{2}(m_1+m_2)v^2 $

$\displaystyle \frac{1}{2}\times 0.012 \times (70)^2 – \frac{1}{2}(0.412)(2.04)^2 $

$ \displaystyle = 29.4 – 0.86 $

= 28.54 J

Q: 26. A 1kg block situated on a rough incline is connected to a spring of spring constant 100 N/s as shown in the fig. The block is released form rest with the spring in the unstreched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring is of negligible mass and the pulley is frictionless.

Sol. R = mgcosƟ and F = μR = μmg cosƟ

Net force on the block in the downward direction = mgsin Ɵ – F

= mgsinƟ – μmgcosƟ )

= mg (sinƟ – μcosƟ)

Distance moved = x = 10 cm = 0.1 m

In equilibrium, work done = PE of the spring

= mg (sinƟ – μcosƟ) x = 1/2kx^{2}

2mg(sinƟ – μcosƟ) = kx

2 × 1 × 10 (sin370 – μcos370) = 100 × 0.1

20(0.601 – μ× 0.798 )= 10

μ = 0.126

Q: 27. A bob of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7m/s. it hits the floor of the elevator (length of the elevator = 3m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

Sol. m = 0.3 kg, v = m/s, h = 3 m

∵ The relative velocity of the ball w.r.t. the elevator is zero.

Only PE of the ball is converted into heat energy by the impact.

Amount of heat produced = PE lost by the ball = mgh

= 0.3 × 9.8 × 3 = 8.82 J

The answer will not change if the elevator was stationary as the relative velocity of the ball w.r.t. the elevator would still be zero.

Q:28. A trolly of mass 200 kg moves with uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolly from one end to the other (10 m away) with a speed of 4 m/s relative of the trolley in a direction opposite to the trolley’s motion and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley’s motion and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run ?

Sol. Mass of the trolley = m_{1} = 200 kg, mass of the child = 20 kg,

Speed of trolley = v = 36 km/h = 10 m/s

Before the child starts running,

Momentum of the system = P1 = (m_{1} + m_{2}) v

= (200 + 20)10 = 2200 kgm/s.

When the child starts running in the opposite direction of the trolley suppose v’ is the final speed of the trolley (w.r.t. earth)

Speed of the child relative to earth = v’ – 4

Momentum of the system when the child is running

= P2 = 200 v’ + 20 (v’ – 4)

= 220 v’ – 80

P_{1} = P_{2}

2200 = 220v’ – 80

2280 = 220v’

v` = 2280/220 = 10.36 m/s.

Time taken by the child to run a distance of 10m over the trolley = 10/4 = 2.5 s

Distance moved by the trolley in that time = 10.36 × 2.5 = 25.9 m.