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6 months ago

Class 11th Work, Energy and Power

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?

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Payal Gupta

Contributor-Level 10

6.18 The length of the pendulum, l = 1.5 m

The potential energy of the bob at horizontal position = mgl

Since it dissipates 5% of its kinetic energy to come to the horizontal position, from the law of conservation of energy we get,

$(\frac{1}{2}) m v^{2}$ = (0.95) $* m g l$

$v^{2}$ = 2 $* 0.95 * 9.81 * 1.5$

v = 5.287 m/s

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6 months ago

Class 11th Work, Energy and Power

6.17 The bob A of a pendulum released from 30 $°$ to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic.

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Payal Gupta

Contributor-Level 10

6.17 In an elastic collision, when the ball A will hit the ball B, A comes to rest immediately and the ball B acquires the velocity of ball A. The momentum thus gets transferred from a moving body to a stationary body.

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6 months ago

Class 11th Work, Energy and Power

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 (Fig. 6.14) is a possible result after collision ?

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Payal Gupta

Contributor-Level 10

6.16 The mass of the ball bearing = m

Before the collision, the total K.E. of the system = K.E. of the stationary ball bearing + K.E. of the striking ball bearing = 0 + $(\frac{1}{2}) {m v}^{2}$

After the collision, the K.E. of the total system is

(a) Case (i) = 0 + $(\frac{1}{2}) (2 m) {(\frac{v}{2})}^{2}$ = $(\frac{1}{4}) {m v}^{2}$

(b) Case (ii) = 0 + $(\frac{1}{2}) m v^{2}$

(c) Case (iii) = $(\frac{1}{2}) (3 m) {(\frac{v}{3}) /}^{2}$ = $(\frac{1}{6}) m v^{2}$

Case (ii) is possible since K.E. is conserved in this case.

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6 months ago

Class 11th Work, Energy and Power

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?

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Payal Gupta

Contributor-Level 10

6.15 Given, the volume of the tank, V = 30 $m^{3}$

Time required to fill the tank, t = 15 min = 900 s

Height of the tank above the ground, h = 40 m

The efficiency of the pump? = 30%

The density of water, $?$ = $10^{3}$ kg/ $m^{3}$

Now, the mass of the water pumped, m = $? V$ = 30 $* 10^{3}$ kg = 3 $* 10^{4}$ kg

Power consumed = W/t = mgh/t = 3 $* 10^{4} * 9.8 *$ 40 / 900 = 13066 W

P input = Power consumed /? = 43.6 kW

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6 months ago

Class 11th Work, Energy and Power

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 ?

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Payal Gupta

Contributor-Level 10

6.14 Momentum is always conserved for an elastic or inelastic collision.

The molecule's initial velocity, u = final velocity v = 200 m/s

Initial kinetic energy = (1/2)m $u^{2}$

Final kinetic energy = (1/2)m $v^{2}$ = (1/2)m $u^{2}$

Therefore, kinetic energy is also conserved

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6 months ago

Class 11th Work, Energy and Power

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 ?

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Payal Gupta

Contributor-Level 10

6.13 The radius of the rain drop = 2 mm = 2 $* 10^{- 3}$ m

The height of drop, s = 500 m

Density of water, $?$ = $10^{3}$ kg/ $m^{3}$

Mass of the rain drop = volume $* d e n s i t y$ = (4/3) $? r^{3}$ $* ?$ = 3.35 $* 10^{- 5}$ kg

The gravitational force on the raindrop, F = mg = 3.28 $* 10^{- 4}$ N

Work done by the gravity on the drop is = mgs where s = 250 m

Work done = 0.082 J

The work done during the second half will remain same.

The total energy of the raindrop will be conserved during the motion.

Total energy at the top

E1 = mgh where h = 500 m, E1 = 0.164 J

Due to resistive force, the energy of the drop on reaching the ground

E2 = (1/2)mv² where

...more

6.13 The radius of the rain drop = 2 mm = 2 $* 10^{- 3}$ m

The height of drop, s = 500 m

Density of water, $?$ = $10^{3}$ kg/ $m^{3}$

Mass of the rain drop = volume $* d e n s i t y$ = (4/3) $? r^{3}$ $* ?$ = 3.35 $* 10^{- 5}$ kg

The gravitational force on the raindrop, F = mg = 3.28 $* 10^{- 4}$ N

Work done by the gravity on the drop is = mgs where s = 250 m

Work done = 0.082 J

The work done during the second half will remain same.

The total energy of the raindrop will be conserved during the motion.

Total energy at the top

E1 = mgh where h = 500 m, E1 = 0.164 J

Due to resistive force, the energy of the drop on reaching the ground

E2 = (1/2)mv² where v = 10 m/s, E2 = 0.001675 J

Work done by the resistive force, W = E1 – E2 = 0.1623 J

less

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6 months ago

Class 11th Work, Energy and Power

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).

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Payal Gupta

Contributor-Level 10

6.12 Electron mass, me = 9.11*10-31 kg

Proton mass, madhya pradesh = 1.67*10–27 kg

Electron's kinetic energy = 10 keV = 10 $*$ $10^{3}$ $*$ 1.60 *10–19 J = 1.60 *10–15 J

Proton's kinetic energy = 100 keV = 100 $* 10^{3}$ $*$ 1.60 *10–19 J = 1.60 *10–14 J

The electron kinetic energy is given by Eke = (1/2)m $v_{e}^{2}$ where $v_{e}$ is the velocity of electron

$v_{e}$ = $?$ { (2 $*$ Eke )/m} = 5.92 $* 10^{7}$ m/s

The velocity of proton $v_{p}$ = $?$ { (2 $*$ Pke )/m} = 4.37 $* 10^{6}$ m/s

The speed ratio = $v_{e}$ / $v_{p}$ = 13.5

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6 months ago

Class 11th Work, Energy and Power

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

F = − $\overset{?}{i} + 2 \overset{?}{j}$ +3 $\overset{?}{k}$ N

where $\overset{?}{i}$ , $\overset{?}{j}$ , $\overset{?}{k}$ 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 ?

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Payal Gupta

Contributor-Level 10

6.11: Force exerted on the body, F =? $\overset{?}{i} + 2 \overset{?}{j}$ +3 $\overset{?}{k}$ N

Displacement, s = 4 km

Work done, W = F.s

= (? $\overset{?}{i} + 2 \overset{?}{j}$ +3 $\overset{?}{k}) . (4 \overset{?}{k})$

= 0+0-3 $* 4$

= 12 J

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6 months ago

Class 11th Work, Energy and Power

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) t ^1/2

(ii) t

(iii) t ^3/2

(iv) t ²

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Payal Gupta

Contributor-Level 10

6.10 Power is given by the relation

P = Fv = mav = mv $\frac{d v}{d t}$ = constant ( say, k)

vdv = $\frac{k}{m} d t$

$\frac{v^{2}}{2} = \frac{k}{m} t$

v = $\sqrt{\frac{2 k t}{m}}$

For displacement x of the body, we have:

v = $\frac{d x}{d t}$ = $\sqrt{\frac{2 k}{m}} t^{1 / 2}$

dx = k' $t^{1 / 2}$ dt where k' = $\sqrt{\frac{2 k}{3}}$ = constant

On integrating both sides, we get

x = $\frac{2}{3} k' t^{3 / 2}$

Therefore x $? t^{3 / 2}$

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6 months ago

Class 11th Work, Energy and Power

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) t ^1/2 (ii) t (iii) t ^3/2 (iv) t ²

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Payal Gupta

Contributor-Level 10

6.9 Let us assume

Body mass = m

Acceleration = a

According to Newton's 2nd law F = MA (constant)

We know a = dv/dt = constant. Hence dv = dt $*$ constant

On integrating, v = t + constant

The relation of power is given by P = F $* v$

We have $v = u + a t$

Hence, $P = F * (u + a t)$

= $m a (u + a t)$

= $m a u + m a^{2} t$

Therefore P $? t$

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