Chapter 5 Laws of motion Solutions
Question - 21 : - Astone of mass 0.25 kg tied to the end of a string is whirled round in a circleof radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is thetension in the string? What is the maximum speed with which the stone can bewhirled around if the string can withstand a maximum tension of 200 N?
Answer - 21 : -
Mass of the stone, m = 0.25kg
Radius of the circle, r =1.5 m
Numberof revolution per second, 
Angularvelocity, ω = 
The centripetal force for the stone isprovided by the tension T, in the string, i.e.,
Maximum tension in the string, Tmax =200 N
Therefore,the maximum speed of the stone is 34.64 m/s.
Question - 22 : - If,in Exercise 5.21, the speed of the stone is increased beyond the maximumpermissible value, and the string breaks suddenly, which of the followingcorrectly describes the trajectory of the stone after the string breaks:
(a)the stone moves radially outwards,
(b)the stone flies off tangentially from the instant the string breaks,
(c)the stone flies off at an angle with the tangent whose magnitude depends on thespeed of the particle ?
Answer - 22 : -
Answer: (b)
Whenthe string breaks, the stone will move in the direction of the velocity at thatinstant. According to the first law of motion, the direction of velocity vectoris tangential to the path of the stone at that instant. Hence, the stone willfly off tangentially from the instant the string breaks.
Question - 23 : - Explainwhy
(a)a horse cannot pull a cart and run in empty space,
(b)passengers are thrown forward from their seats when a speeding bus stopssuddenly,
(c)it is easier to pull a lawn mower than to push it,
(d)a cricketer moves his hands backwards while holding a catch.
Answer - 23 : -
(a) In order to pull a cart, a horse pushes theground backward with some force. The ground in turn exerts an equal andopposite reaction force upon the feet of the horse. This reaction force causesthe horse to move forward.
Anempty space is devoid of any such reaction force. Therefore, a horse cannotpull a cart and run in empty space.
(b) When a speeding bus stops suddenly, thelower portion of a passenger’s body, which is in contact with the seat,suddenly comes to rest. However, the upper portion tends to remain in motion(as per the first law of motion). As a result, the passenger’s upper body isthrown forward in the direction in which the bus was moving.
(c) While pulling a lawn mower, a force at anangle θ is applied on it, as shown in the following figure.
Thevertical component of this applied force acts upward. This reduces theeffective weight of the mower.
On the other hand, while pushing a lawnmower, a force at an angle θ is applied on it, as shown in thefollowing figure.
Inthis case, the vertical component of the applied force acts in the direction ofthe weight of the mower. This increases the effective weight of the mower.
Sincethe effective weight of the lawn mower is lesser in the first case, pulling thelawn mower is easier than pushing it.
(d) According to Newton’s second law of motion,we have the equation of motion:
Where,
F = Stopping force experienced by thecricketer as he catches the ball
m = Mass of the ball
Δt = Time of impact of the ballwith the hand
It can be inferred from equation (i)that the impact force is inversely proportional to the impact time, i.e.,
Equation (ii) shows that the forceexperienced by the cricketer decreases if the time of impact increases and viceversa.
Whiletaking a catch, a cricketer moves his hand backward so as to increase the timeof impact (Δt). This is turn results in the decrease in the stoppingforce, thereby preventing the hands of the cricketer from getting hurt.
Question - 24 : - Figure5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest asuitable physical context for this motion. What is the time between twoconsecutive impulses received by the body? What is the magnitude of eachimpulse?
Answer - 24 : -
A ball rebounding between two walls locatedbetween at x = 0 and x = 2 cm; after every 2s, the ball receives an impulse of magnitude 0.08 × 10–2 kg m/sfrom the walls
The given graph shows that a body changesits direction of motion after every 2 s. Physically, this situation can bevisualized as a ball rebounding to and fro between two stationary wallssituated between positions x = 0 and x = 2cm. Since the slope of the x–t graph reverses afterevery 2 s, the ball collides with a wall after every 2 s. Therefore, ballreceives an impulse after every 2 s.
Mass of the ball, m = 0.04kg
The slope of the graph gives the velocity ofthe ball. Using the graph, we can calculate initial velocity (u) as:
.
Velocity of the ball before collision, u =10–2 m/s
Velocity of the ball after collision, v =–10–2 m/s
(Here,the negative sign arises as the ball reverses its direction of motion.)
Magnitudeof impulse = Change in momentum
=0.08 × 10–2 kg m/s
Question - 25 : - Figure 5.18 shows a man standing stationary withrespect to a horizontal conveyor belt that is accelerating with 1 m s–2.What is the net force on the man? If the coefficient of static friction betweenthe man’s shoes and the belt is 0.2, up to what acceleration of the belt canthe man continue to be stationary relative to the belt? (Mass of the man = 65kg.)
Figure5.18
Answer - 25 : -
Mass of the man, m = 65 kg
Acceleration of the belt, a =1 m/s2
Coefficient of static friction, μ =0.2
The net force F, acting on theman is given by Newton’s second law of motion as:
=65 × 1 = 65 N
The man will continue to be stationary withrespect to the conveyor belt until the net force on the man is less than orequal to the frictional force fs, exerted by the belt, i.e.,
∴a‘ = 0.2 × 10 = 2 m/s2
Therefore,the maximum acceleration of the belt up to which the man can stand stationaryis 2 m/s2.
Question - 26 : - A stone of mass m tied tothe end of a string revolves in a vertical circle of radius R. Thenet forces at the lowest and highest points of the circle directed verticallydownwards are: [Choose the correct alternative]
| Lowest Point | Highest Point |
| (a) | mg – T1 | mg + T2 |
| (b) | mg + T1 | mg – T2 |
| (c) | mg + T1 – /R | mg – T2 + /R |
| (d) | mg – T1 – /R | mg + T2 + /R |
T1 and v1 denotethe tension and speed at the lowest point. T2 and v2 denotecorresponding values at the highest point.
Answer - 26 : -
(a)The free body diagram of the stone at thelowest point is shown in the following figure.
Accordingto Newton’s second law of motion, the net force acting on the stone at thispoint is equal to the centripetal force, i.e.,
Where, v1 =Velocity at the lowest point
Thefree body diagram of the stone at the highest point is shown in the followingfigure.
UsingNewton’s second law of motion, we have:
… (ii)
Where, v2 =Velocity at the highest point
Itis clear from equations (i) and (ii) that the net force acting atthe lowest and the highest points are respectively (T – mg)and (T + mg).
Question - 27 : - A helicopter of mass 1000 kg rises with avertical acceleration of 15 m s–2. The crew and the passengers weigh300 kg. Give the magnitude and direction of the
(a)force on the floor by the crew and passengers,
(b)action of the rotor of the helicopter on the surrounding air,
(c)force on the helicopter due to the surrounding air.
Answer - 27 : -
(a) Mass of the helicopter, mh =1000 kg
Mass of the crew and passengers, mp =300 kg
Total mass of the system, m =1300 kg
Acceleration of the helicopter, a =15 m/s2
Using Newton’s second law of motion, thereaction force R, on the system by the floor can be calculated as:
R – mpg =ma
= mp(g + a)
=300 (10 + 15) = 300 × 25
=7500 N
Sincethe helicopter is accelerating vertically upward, the reaction force will alsobe directed upward. Therefore, as per Newton’s third law of motion, the forceon the floor by the crew and passengers is 7500 N, directed downward.
(b) Using Newton’s second law of motion, thereaction force R’, experienced by the helicopter can be calculatedas:
= m(g + a)
=1300 (10 + 15) = 1300 × 25
=32500 N
Thereaction force experienced by the helicopter from the surrounding air is actingupward. Hence, as per Newton’s third law of motion, the action of the rotor onthe surrounding air will be 32500 N, directed downward.
(c) The force on thehelicopter due to the surrounding air is 32500 N, directed upward.
Question - 28 : - A stream of water flowing horizontally witha speed of 15 m s–1 gushes out of a tube of cross-sectionalarea 10–2 m2, and hits a vertical wall nearby. Whatis the force exerted on the wall by the impact of water, assuming it does notrebound?
Answer - 28 : -
Speed of the water stream, v =15 m/s
Cross-sectional area of the tube, A =10–2 m2
Volumeof water coming out from the pipe per second,
V = Av = 15 × 10–2 m3/s
Density of water, ρ = 103 kg/m3
Mass of water flowing out through the pipeper second = ρ × V = 150 kg/s
Thewater strikes the wall and does not rebound. Therefore, the force exerted bythe water on the wall is given by Newton’s second law of motion as:
F =Rate of change of momentum 
Question - 29 : - Ten one-rupee coins are put on top of eachother on a table. Each coin has a mass m. Give the magnitude anddirection of
(a) the force on the 7th coin(counted from the bottom) due to all the coins on its top,
(b) the force on the 7th coinby the eighth coin,
(c) the reaction of the 6th coin on the 7th coin.
Answer - 29 : -
(a) Force on the seventh coin is exerted by theweight of the three coins on its top.
Weight of one coin = mg
Weight of three coins = 3mg
Hence, the force exerted on the 7th coinby the three coins on its top is 3mg. This force acts verticallydownward.
(b) Force on the seventh coin by the eighth coinis because of the weight of the eighth coin and the other two coins (ninth andtenth) on its top.
Weight of the eighth coin = mg
Weight of the ninth coin = mg
Weight of the tenth coin = mg
Total weight of these three coins = 3mg
Hence, the force exerted on the 7th coinby the eighth coin is 3mg. This force acts vertically downward.
(c) The 6th coin experiences adownward force because of the weight of the four coins (7th, 8th,9th, and 10th) on its top.
Therefore, the total downward forceexperienced by the 6th coin is 4mg.
Asper Newton’s third law of motion, the 6th coin will produce anequal reaction force on the 7th coin, but in the oppositedirection. Hence, the reaction force of the 6th coin on the 7th coinis of magnitude 4mg. This force acts in the upward direction.
Question - 30 : - Anaircraft executes a horizontal loop at a speed of 720 km/h with its wingsbanked at 15°. What is the radius of the loop?
Answer - 30 : - Speed of the aircraft, v = 720km/h 
Acceleration due to gravity, g = 10 m/s2
Angle of banking, θ = 15°
For radius r, of the loop, wehave the relation:
=14925.37 m
=14.92 km