# Laws of Motion

## NCERT Solution

### Part 3

Question 21: A stone of mass 0.25 kg tied to the end of a 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 withstand a maximum tension of 200 N?

**Answer:** Given, m = 0.25 kg, r = 1.5 m, v = 40 rev/min = `(40)/(60)=2/3` rps

Tension T `=(mv^2)/r=mrω`

`=mr(2πn)^2`

`=0,25xx1.5(2xx3.14xx2/3)^2`

`=(0.25xx(2/3)^2)/(1.5)`

`=0.25xx1.5xx4xx3.14xx3.14xx4/9`

`=(59.1576)/9=6.57` N

Now, T_{max} = 200 N

T_{max} `=(m(v_(ma\x))^2)/r`

Or, `200=(0.25(v_(ma\x))^2)/(1.5)`

Or, v_{max} `=(200xx1.5)/(0.25)`

Or, `v=20sqrt3=34.64` m/s

Question 22: If in question 21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:

- The stone moves readily outwards
- The stone flies off tangentially from the instant the string breaks
- The stone falls off at an angle with the tangent whose magnitude depends on the speed of the particle.

**Answer:** (b) The stone flies off tangentially from the instant the string breaks

Question 23: Explain why

(a) A horse cannot pull a cart and run in empty space

**Answer:** When a horse tries to pull a cart, it applies force on the ground. The ground, in turn, provides the reactionary force facilitating forward movement of the cart. The reactionary force will be absent in empty space. Hence, a horse cannot pull a cart and run in empty space.

(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly

**Answer:** When a passenger is sitting in a moving bus, his body is in state of motion. When the bus suddenly stops, the lower body of passenger comes to rest. But the upper body, being more flexible tries to remain in motion. Due to this, a passenger is thrown forward from the seat when a speeding bus suddenly stops.

(c) It is easier to pull a lawn mower than to push it

**Answer:** When a person pulls a lawn mower the force is applied at a certain angle. The vertical component of this force works upwards and helps in reducing the apparent weight of the lawn mower.

When a person pushes a lawn mower the vertical component (of the applied force) acts downwards. This increases the apparent weight of the lawn mower.

So, it is easier to pull a lawn mower than to push it.

(d) A cricketer moves his hands backwards while holding a catch.

**Answer:** We know force is given by following equation: `F=ma`

Or, `F=m(Δv)/(Δt)`

This equation shows that the force is inversely proportional to time. When the cricketer moves his hands backwards he tries to increase the time for acceleration to come to zero. This helps in reducing the force and thereby impact of the ball. It minimizes the chances of injury from catching the ball.

#### Additional Exercise

Question 24: This figure shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

**Answer:** This graph can be showing the motion of a ball which is bouncing back after colliding with two opposite walls. The time between two consecutive impulses is 2 second.

Velocity = Displacement ÷ Time

`=(0.02)/2=0.01` m/s

Impulse = Change in momentum

Initial momentum = mv = 0.04 × 0.01 = 0.0004 kg m s^{-1}

Final momentum = mv = 0.04 × -0.01 = -0.0004 kg m s^{-1}

Change in momentum = 0.0004 – ( - 0.0004) = 0.0008 kg m s^{-1}

Question 25: A man is standing stationary with respect 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 between the man’s shoes and the belt is 0.2, up to what acceleration of the belt, can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg)

**Answer:** Given, a = 1 m s^{-2}, m = 65 kg, μ = 0.2

As the man is stationary with respect to the conveyor belt so he is experiencing the same force as applied by him.

F = ma

= 65 × 1 = 65 N

The man will continue to be stationary until the net force on him will be less than or equal to the force of friction

So, F_{net} = F_{s}

Or, ma' = μmg

Or, a' = μg = 0.2 × 10 = 2 m s^{-2}

Question 26: A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points to the circle directed vertically downwards are:

Lowest Point | Highest Point |
---|---|

mg – T_{1} | mg + T_{2} |

mg + T_{1} | mg – T_{2} |

mg + T_{1} - (mv_{1}^{2})/R | mg – T_{2} - (mv_{1}^{2})/R |

mg – T_{1} - (mv_{1}^{2})/R | mg + T_{2} + (mv_{1}^{2})/R |

T_{1} and v_{1} denote the tension and speed at the lowest point. T_{2} and v_{2} denote corresponding values at the highest point.

**Answer:**(a) mg – T_{1} and mg + T_{1}

Question 27: A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s^{-2}. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the

(a) Force on the floor by the crew and passengers

**Answer:** Mass of helicopter m_{h} = 1000 kg, mass of passengers m_{p} = 300 kg, a = 15 m s^{-2}

The reaction force R can be calculated as follows:

R – m_{p}g = ma

Or, R = ma + m_{p}g = m(a + g)

`=300xx(15+10)=7500` N

Direction of this force is downwards

(b) Action of the rotor of the helicopter on the surrounding air

**Answer:** R – mg = ma

Or, R = m(a + g)

`=1300(15+10)=32500` N

Downwards

(c) Force on the helicopter due to the surrounding air

**Answer:** The force on helicopter due to surrounding air is 32500 N and is acting upwards.

Question 28: A stream of water flowing horizontally with a speed of 15 m s^{-1} gushes out of a tube of cross-sectional area 10^{-2} m^{2}, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

**Answer:** Speed of water v = 15 m/s, Area of cross section A = 10^{-2} m^{2}

Volume of water coming out per second

V = vA = 15 × 10^{-2} m^{3} s^{-1}

Density of water ρ = 10^{3} kg m^{-3}

Mass of water flowing per second = ρ × V

= 10^{3} kg m^{-3} × 15 × 10^{-2} m^{3} s^{-1}

= 150 kg s^{-1}

As the water does not rebound after striking the wall, the force can be calculated as follows:

F = rate of change of momentum

`=(ΔP)/(Δt)`

`=(mv)/t=150xx15=2250` N

Question 29: Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of

(a) The force on the 7th coin (counted from the bottom) due to all the coins on its top.

**Answer:** Mass of 3 coins = 3m

F = 3mg

Because there are three coins on top of the 7th coin

(b) The force on the 7th coin by the 8th coin.

**Answer:** F = 3mg

Because two more coins above the 8th coin are exerting force on the 7th coin

(c) The reaction of the 6th coin on the 7th coin.

**Answer:** 4 mg

Because four coins are imparting force on the 6th coin

Question 30: An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?

**Answer:** Speed of aircraft v = 720 km/h `=720xx5/(18)=200` m/s, g = 10 m s^{-2}, angle of banking θ = 15°

Radius can be calculated as follows:

Tan θ = `(v^2)/(rg)`

Or, `r=(v^2)/(g tan\n\θ)`

`=(200^2)/(10xx\ta\n\15)`

`=(4000)/(0.26)=14925.37` m

= 14.92 km