Question.15. Two bodies of masses 10 kg and 20 kg kept on a smooth horizontal surface are tied to the end of a light string. A horizontal force F = 600 N is applied to (i) 10 kg and (ii) 20 kg along the direction of string . What is the tension in the string in each case ?
Solution: Given, F = 600 N, m1 = 10 kg and m2 = 20 kg
If T = tension of the string, a = acceleration of the system along the direction of the applied force
$a = \frac{F}{m_1 + m_2}$
$a = \frac{600}{10 + 20}$
a = 20 ms-2
(i)When force is applied on lighter block,
T = m2a
T = 20 × 20 N = 400 N
(ii)When force is applied of heavier block,
T = m1 a
T = 10 × 20 N = 200 N
Question.16. Two masses 8 kg and 12 kg are connected at the two ends of light inextensible string that passes over a frictionless pulley. Find the acceleration of the masses and the tension in the string, when the masses are released.
Solution: Given, m1 = 8 kg , m2 = 12 kg
For m2 block ,
m2 g – T = m2 a …(i)
For m1 block ,
T – m1 g = m1 a …(ii)
Adding eqn (i) & (ii)
( m2 – m1) g = ( m1 + m2 )a
$\displaystyle a = \frac{(m_2 – m_1)g}{m_1 + m_2}$
$\displaystyle a = \frac{(12 – 8)9.8}{12 + 8}$
a = 1.96 ms-2
$\displaystyle T = \frac{2 m_1 m_2 g }{m_1 + m_2} $
$\displaystyle T = \frac{2 \times 8 \times 12 \times 9.8 }{8 + 12} $
and, T = 94.1 N
Question.18. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6ms-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other.
Solution: Given, Initial momentum of ball A = 0.05(6) = 0.3 kg ms-1,
When ball is reversed after collision, speed would be reversed,
final momentum = 0.05(-6) = – 0.3 kg ms-1,
Impulse imparted to to ball A = change in momentum of ball A
= final momentum – initial momentum
= (-0.3 -0.3) = – 0.6 ms-1
Question.19. A shell of mass 0.02 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80ms-1 what is the recoil speed of the gun?
Solution: Given, Mass of shell, m = 0.02 kg .
Mass of gun, M = 100 kg and Muzzle speed of shell = 80 ms-1
According to the principle of conservation of linear momentum,
mV + Mv = 0
v = -mv/M
= -0.02 × 80/100
v = 0.016 ms-1
Question.20. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse to the ball? Mass of the ball is 0.15 kg .
Solution: Impulse on the ball = mu cosθ – (-mu cosθ)
= 2 mu cosθ
= 2 × 0.15 ×15 cos 22.5°
= 4.16 kgms-1.
Question.21. A stone of mass 0.25 kg tied to the end of string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min, in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can with stand a maximum tension of 200 N ?
Solution: Given, m = 0.25 kg, r = 1.5 m, v = 40 rpm = 40/60 rps
T = m rω2
T = m r(2πν)2
T = 4π2 m rν2
T = 4 ×(22/7)2 × (0.25) × 1.5 × (2/3)2
T = 6.6 N
If Tmax = 200 N , then
$\displaystyle T_{max} = \frac{m v_{max}^2}{r}$
$\displaystyle v_{max} = \sqrt{\frac{T_{max} r }{m}}$
$\displaystyle v_{max} = \sqrt{\frac{200 \times 1.5 }{0.25}}$
vmax = √(1200) = 34.6 m/s
Question.23. Explain why
(a) a horse cannot pull a cart and run in the empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn roller than to push it ?
(d) a cricketer moves his hands backwards when holding ?
Solution: (a) When a cart is pulled, the horse pushes the ground with a force at an angle. The ground offers an equal reaction in opposite direction, through the feet of the horse. The horizontal component of this reaction helps the cart to move in forward direction. In empty space, reaction force would be zero, so horse cannot pull the cart.
(b)If the speeding but stops suddenly, the lower part of the passengers, in contact with bus, would be at rest. However, upper part of the body would still be in motion. Therefore, passengers would move forward. This occurs due to inertia of motion.
(c) When a lawn roller is pulled, the force acts along the handle, which has two components. The upwardly directed vertical component reduced the weight of roller , however, horizontal component of the force helps to push forward the mover. Whereas, when a lawn roller is pushed, force is applied downwards and thereby, vertical component, directed downward, increase the weight of roller , which creates difficulties in pushing the roller . Therefore, pushing of roller is difficult than pulling the same.
(d) When a cricketer holds a catch the impulse received at hand = F × t = change in linear momentum of the ball = constant. By moving the hand backward, the cricketer increases the time of impact and reduces the force, so as the reaction, thereby reduces the changes of hurting severely.
Question: 27. A helicopter of mass 1000 kg rises with vertical acceleration of 15 ms-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of
(a) force on the floor by the crew and passengers
(b) action of the rotor of the helicopter on surrounding air,
(c) force on the helicopter due to the surrounding air. Take g = 9.8 ms-2.
Solution: Given, Mass of the helicopter, m1 = 1000 kg,
Mass of the crew and passengers, m2 = 300 kg, upwards acceleration a = 15 ms-2 and g = 9.8 ms-2
(a) Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers
= m2 (g + a)
= 300(10 + 15) = 7500 N
(b) Action of the rotor of helicopter is downwards when helicopter rising up,
therefore, force of action = (m2 + m1)(g + a)
= (1000 + 300)(9.8 + 15)
= 32500 N (vertically downwards)
(c) Force on the helicopter due to surrounding air is the reaction . Since action and reaction are equal & opposite , therefore, force of reaction = 32500 N (vertically upwards)
Question: 28. A steam of water flowing horizontally with a speed of 15 ms-1 pushes out of tube of cross sectional area 10-2 m² and hits at a vertical wall nearby. What is the force exerted on the wall by the impact of water assuming that it does not rebound ?
Solution: Given, v = 15 m/s;
Area of cross section, a = 10-2 m2 and
mass of the water per sec = m = (volume/sec) × (density of water)
= (0.15) × (103) = 150 kg-1
force exerted on the wall = mv/t = 150 × 15/1 = 2250 N
Question: 29. Ten one rupee coins are put on top of one another on a table. Each coin has a mass m kg. Give the magnitude and direction of (a) the force on the 7th coin (counted from the bottom) due to all coins above it, (b) the force on the 7th coin by the 8th coin and (c) the reaction of the 6th coin on the 7th coin.
Solution: (a) Force on the 7th coin = weight of the 3 coins lying above 7th coin,
F = 3 mg N (acts vertically)
(b) Eight coin is under the weight of 2 coins above it and its own weight,
force on 7th coin due to 8th coin = sum of the 3 coins = 2 mg + mg = 3 mg N
(c) Sixth coin is under the weight of 4 coins above it,
therefore, reaction = -F = -4mg N (negative sign shows the reaction is vertically upwards)
Question: 30. An aircraft executes a horizontal loop at speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ?
Solution: Given, α = 15°, v = 720 km/h = 200 m/s, g = 9.8 ms-2.
tanα = v2/rg ,
r = v2/g tanα
= (200)2/ (9.8 × tan 15°)
= 15.23 m
Question: 31. A train rounds an unbanked circular bend of radius 30 m at a speed of 54km/h. The mass of the train is 106kg. What provides the centripetal force required for this purpose ? The engine or the rails? The outer of the inner rails? Which rail will wear out faster, the outer or the inner rail? What is the angle of banking required to prevent wearing out the rails?
Solution: Here, lateral thrust exerted by the rails to wheel is proving the necessary centripetal force and the train would exert an equal an opposite thrust (by Newton’s 3rd law) on the rails causing its wear and tear.
Therefore, outer rail’s wear and tear would faster due to the larger force applied on it by train.
tanα = v²/rg,
Given, r = 30 m, v = 54 km/h = 15 m/s, g = 9.8 ms-2
α = tan-1 (v²/rg ) = tan-1{ (15)² / (30 × 9.8)} = 37.4°
Q:32 . A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in fig . What is the action on the floor by the man in two cases ? If the floor yields to a normal force of 700 N , which mode should the man adopt to lift the block without the floor yielding ?
Sol: Force applied to lift the block
F = m g = 25 × 9.8 = 245 N
Weight of man = 50 × 9.8 = 490 N
case (i) : In this case force is applied by man in upward direction . This increases the apparent weight of man .Hence action on the floor , W’ = 490 + 245 = 735 N
case (ii) In this case force is applied by the man in downward direction . This decreases the apparent weight of man . Hence action on the floor W” = 490 – 245 = 245 N
As the floor yields to a normal force of 700 N , the mode (ii) has to be adopted by the man to lift the block