NCERT Solutions for Work, Energy and Power

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Link to NCERT Solutions for Class 11 Physics, Chapter 6, Work, Energy and Power: Question 6.1 to 6.15

Question 6. 16. Two identical ball bearings in contact with each other and resting on a friction-less 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 (Fig.) is a possible result after collision?NCERT Solutions for Work, Energy and Power 1

Let m be the mass of each ball bearing.
K.E. of the system before collision is =\frac{1}{2}mv^2
After collision, K.E. of the system is
Case I, E_1 = 2 \times \frac{1}{2}m (\frac{v}{2})^2= \frac{1}{4}mv^2
Case II, E_2 = \frac{1}{2}mv^2
Case III, E_3 = 3\times \frac{1}{2}m (\frac{v}{3})^2 = \frac{1}{6}mv^2
As the collision is elastic, total kinetic energy is conserved. Case II is the possible result.

Question 6.17. The bob A of a pendulum released from 30^o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.NCERT Solutions for Work, Energy and Power 2

Since the collision is completely elastic in nature, A will come to rest and and B will start to move with the velocity of A.
The bob A does not rise at all.

Question 6. 18. The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Let the mass of the bob be m kg
Initial Total Energy = Potential energy = mgh= 15m Joule [g=10m/s^2]
As the 5% of the initial energy is dissipated while at the bottom most point,
Final Total Energy = Kinetic energy = 0.95 \times Initial Potential Energy= 14.25m Joule
\frac{1}{2}mv_f^2 = 14.25 m
v_f = 5.34 m/sec

Question 6. 19. A trolley of mass 300 kg carrying a sandbag 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^-1. What is the speed of the trolley after the entire sand bag is empty?

There shall be no change in the speed of the trolley as the trolley and the sand remains a conserved system.

Question 6. 20. A particle of mass 0.5 kg travels in a straight line with velocity u = a x^\frac{3}{2}, where a = 5 m^{-1/2}s^{-1}. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Work done= Change in K.E=\frac{1}{2}mv_f^2 -\frac{1}{2}mv_i^2
v=ax^{3/2} and a=5 m^{-1/2}s^{-1}
Initial velocity at x = 0, v_i = 0
Final velocity at x = 2, v_f = a (2)^{3/2 }= 5 \times (2)^{3/2}
Work done = \frac{1}{2} m(v_f^2-v_i^2) = 50 Joule.

Question 6. 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 the air passing through it in time t? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m-3 . What is the electrical power produced?

(a) Volume of the air cylinder is Avt where A is the cross sectional area, v is the velocity and t is the time period.
If the density of air is \rho then the mass of air cylinder is = Av \rho t
(b) Kinetic energy of the air cylinder will be \frac{1}{2}mv^2 = \rho A v^3 \frac{t}{2}
(c) Power generated = \frac{1}{2} \times 30 \times 10 \times 1.2 \times 100 = 18kW
Now the efficiency of the wind mill is 25%
Thus electrical power generated = 0.25 \times 18 KW= 4.5 KW

Question 6. 22. A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated, (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 x 10^7J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Mass m = 10 kg, height h = 0.5 m, Number of repeat n = 1000
(a) Work done against the gravitational force in each repeat
W = mgh = 10 \times 10\times 0.5 = 50 J
Total work done in 1000 repeats = 1000 \times 50 = 50,000 Joule

(b) As mentioned, Fat supplies 3.8 \times 10^7 Joule of energy per kg.
However, mechanical energy converts to 20% of total energy
Thus 1 kg of fat supplies 0.2 \times 3.8 \times 10^7 = 0.76 \times 10^7 Joule of mechanical energy
Thus the person loses \frac{50000}{0.76 \times 10^7} = 6.58 \times 10^{-3} Kg = 6.58 gm

Question 6. 23. A family uses 8 kW of power, (a) 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? (b) Compare this area to that of the roof of a typical house.

(a) Power converted from each sq meter= 200 \times 0.2 watt= 40 watt
Thus the area required to generate 8000 watt is 8000/40= 200 sq-m

Question 6. 24. A bullet of mass 0.012 kg and horizontal speed 70 ms^{-1} 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 thin wire. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.NCERT Solutions for Work, Energy and Power 3

Mass of the bullet m_b = 0.012 kg and initial velocity before touching the wood block is u_b = 70 m/s
Mass of the wood block m_w = 0.4 kg an initial velocity u_w = 0
Let the initial velocity of the wood block + bullet just after collision be V
From the conservation of total linear momentum
0.012 \times 70 =(0.012+0.4)V or V = 2.04 m/sec
Now this composite system starts rising and let h be the maximum height. At this height total initial kinetic energy has converted into potential energy.
\frac{1}{2}mV^2 = mgh or h = \frac{V^2}{2 g} = 0.208 m [g = 10m/s^2]
To calculate the heat produced, we will calculate loss in kinetic energy before and after of collision.
\Delta E = \frac{1}{2} \times 0.012 \times 70^2 -\frac{1}{2} \times 0.412 \times 2.04^2 =28.54 Joule
Thus Heat produced H =\Delta E/J=28.54/4.2= 6.8 Cal

Question 6. 25. Two inclined friction-less tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given theta_1=30^o, \theta_2=60^o and h = 10 m, what are the speeds and times taken by the two stones?NCERT Solutions for Work, Energy and Power 5

At BC plane total potential energy will be converted into kinetic energy. If V_B and V_C be the velocity at point B and C respectively the
mgh = \frac{1}{2}mV_B^2 = \frac{1}{2}mV_c^2 Thus V_B=V_C= \sqrt{2gh}= 14.14 m/s
They will reach the base plane with same speed.
To calculate the time required, Sin \theta =h/l or l= h/Sin\theta
Component of the gravitational force along the slanted plane is g Sin \theta
Thus l=\frac{h}{Sin \theta}=\frac{1}{2}g Sin \theta t^2 or t=\frac{1}{Sin \theta} \sqrt{\frac{2h}{g}}
t_B= 2 \sqrt{2}s
t_C= \frac{2\sqrt{2}}{\sqrt{3}} s

Question 6. 26. A 1 kg block situated on a rough incline is connected to a spring with spring constant 100 Nm-1 as shown in Figure. The block is released from rest with the spring in the unstretched 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 has negligible mass and the pulley is frictionless.NCERT Solutions for Work, Energy and Power 6

From the figure above
Reaction force R = mg cos \theta
Friction force F= \mu R = \mu mg cos \theta
Net force on the block anong the incline = mg sin \theta - \mu mg cos \theta
Distance travelled x = 10 cm = 0.1 m
In equilibrium, work done = Potential energy of stretched spring
or mg (sin \theta - \mu cos \theta) x = -\frac{1}{2}kx^2
\mu= 0.126

Question 6. 27. A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 ms^{-1}. It hits the floor of the elevator (length of elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary ?

Potential energy of the bolt = mgh = 0.3 \times 10 \times 3 = 9 J [g=10 m/s^2]
The bolt does not rebound thus the total energy is converted into heat.
W = JH or H = W/J= 2.14 Cal
Since the value of acceleration due to gravity is the same in all inertial system, therefore the answer will not change even if the elevator is stationary.

Question 6. 28. A trolley of mass 200 kg moves with a uniform speed of 36 km h^{-1} on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 ms^{-1} relative to 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 moved from the time the child begins to run?

Let there be an observer travelling parallel to the trolley with the same speed. He will observe the initial momentum of the trolley of mass M and child of mass m as zero. When the child jumps in opposite direction, he will observe the increase in the velocity of the trolley by \Delta v.
Let u be the velocity of the child. He will observe child landing at velocity (u - \Delta u) Therefore, initial momentum = 0
Final momentum = M \Delta v - m (u - \Delta v)
Hence, M \Delta v - m (u - \Delta v) = 0
Whence \Delta v= \frac{m u}{M+m}
Putting values \Delta v = \frac{4 \times 20}{20+220} ms^{-1}
Final speed of trolley is 10.36 ms^-1.
The child take 2.5 s to run on the trolley.
Therefore, the trolley moves a distance = 2.5 \times 10.36 m = 25.9 m.

Question 6.29. Which of the following potential energy curves in Fig. cannot possibly describe the elastic collision of two billiard balls? Here r is distance between centers of the balls.NCERT Solutions for Work, Energy and Power 7

The potential energy of a system of two masses varies inversely as the distance (r) between 1
them i.e., V (r) \propto \frac{1}{r}. When the two billiard balls touch each other, P.E. becomes zero i.e., at r = R + R = 2 R; V (r) = 0. Out of the given graphs, curve (v) only satisfies these two conditions. Therefore, all other curves cannot possibly describe the elastic collision of two billiard balls.


Link to Work, Energy and Power: Question 6.1 to 6.15