## BSEB Bihar Board Class 9 Science Solutions Chapter 9 Force and Laws of Motion

Bihar Board Class 9 Science Solutions Chapter 9 Force and Laws of Motion Textbook Questions and Answers, Additional Important Questions, Notes.

### Bihar Board Class 9 Science Chapter 9 Force and Laws of Motion InText Questions and Answers

Page 118

Question 1.

Which of the following has more inertia :

(а) a rubber ball and a stone of the same size?

(b) a bicycle and a train?

(c) a five rupee coin and a one rupee coin?

Answer:

(a) A stone

(b) A train

(c) A five rupee coin.

Question 2.

In the following example, try to identify the number of times the velocity of ball changes :

“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team.” Also identify’ the agent supplying the force in each case.

Answer:

Question 3.

Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer:

When we vigorously shake the branch of a tree the branch is set in motion. Due to inertia of rest, the leaves tend to remain at rest and get detached.

Question 4.

Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer:

A passenger sitting in a running bus is also in motion in the same direction. When the running bus stops suddenly, the lower part of the passenger’s body comes to rest; while the upper portion continues to remain in motion. As a result, the passenger is thrown in the forward direction.

Page 126

Question 1.

If action is always equal to the reaction, explain how a horse can pull a cart.

Answer:

The force on the cart determines whether the cart will move or not. The force exerted by the cart on the horse affects only the horse. Hence, if the horse is able to apply enough force to overcome the frictional force, the cart will move. Thus, in order to make the cart move, the horse bends forward and pushes the ground with its feet. When the forward reaction to the backward push of the horse is greater than the opposing frictional force of the wheels, the cart moves.

Question 2.

Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer:

This is because the hose pipe has a tendency to go backward. Ejection of water at a high velocity is the\ action which causes the hose pipe to move backward (the reaction). This is in accordance to Newton’s third law of motion.

Question 3.

From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s-1 calculate the initial recoil velocity of the rifle.

Answer:

Mass of the bullet, m_{1}

= 50 g = \(\frac {50}{1000}\) kg = 0.05 kg

Mass of the rifle, m_{2} = 4 kg

Initial velocity of the bullet, u_{1} = 0 ms^{-1}

Initial velocity of the rifle, u_{2} = 0 ms^{-1}

Final velocity of bullet, υ_{1} = 35 ms^{-1}

Final velocity of rifle = υ_{2}

According to law of conservation of momentum, Total momentum before the fire

= Total momentum after the fire

m_{1}u_{1} + m_{2}u_{2} = m_{1}υ_{1} + m_{2}υ_{2}

⇒ 0.05 kg x. 0 ms^{-1} + 4 kg x 0 ms^{-1} = 0.05 kg x 35 ms^{-1} + 4 kg x υ_{2}

⇒ 0 = 1.75 kgm s^{-1} + 4 kg υ_{2}

υ_{2} = \(\frac{-1.75 \mathrm{kgms}^{-1}}{4 \mathrm{~kg}}\) = 0.4375 ms^{-1}

Question 4.

Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms^{-1} and 1 ms^{-1} respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^{-1}. Determine the velocity of ; the second object.

Answer:

Mass of the first object,

m_{1} = 100 g = \(\frac {100}{1000}\) kg = 0.1 kg

Mass of the second object,

m_{2} = 200 g = \(\frac {200}{1000}\)kg = 0.2 kg

Initial velocity of the first object,

u_{1} = 2 ms^{-1}

Initial velocity of the second object, u_{2} = 1 ms^{-1}

Final velocity of the first object, υ_{2} = 1.67 ms^{-1}

Final velocity of the second object = υ_{2}

According to the law of conservation momentum, Total momentum before collision = Total momentum after collision

⇒ m_{1}u_{1} + m_{2}u_{2} = m_{1}υ_{1} + m_{2} υ_{2
}⇒ 0.1 kg x 2 ms^{-1} + 0.2 kg x 1 ms^{-1}

= 0.1 kg x 1.67 ms^{-1} + 0.2 kg x υ_{2}

⇒ 0.4 kgms^{-1}

= 0.167 kgms^{-1} + 0.2 kg υ_{2}

= 0.233 kgm s^{-1} = 0.2 kg υ_{2}

υ_{2} = \(\frac{0.233}{0.2}\) = 1.165 ms^{-1}

### Bihar Board Class 9 Science Chapter 9 Force and Laws of Motion Textbook Questions and Answers

Question 1.

An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of velocity. If no, provide a reason.

Answer:

No, the object will not travel with non-zero velocity. This is because by Newton’s first law of motion an object at rest will remain at rest unless an unbalanced force acts upon it.

Question 2.

When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer:

When the carpet is shaken, or beaten with a stick, the carpet is set in motion. Due to inertia of rest, the dust particles tend to remain at rest and fall off.

Question 3.

Why is it advised to tie and luggage kept on the roof of a bus with a rope?

Answer:

When the luggage is placed on the bus roof, it is not fixed to the bus. When the bus starts suddenly, the luggage due to inertia of rest tends to remain in the state of rest gtnd as a result, some of the luggage might, fall towardsHhe backside of the bus. When the moving bus suddenly stops, due to the inertia of motion, the luggage , might slide past the bus and fall in the front. To avoid it, luggage; is tied up with a rope, so that the luggage becomes^ a part pf the bus.

Question 4.

A batsman hits a cricket ball which them rolls on a level ground. After covering a short distance, the ball comes! to rest. The ball slows to a stop because

(а) the batsman did not hit the ball hard enough.

(b) velocity is proportional to the force exerted on the ball.

(c) there is a force on the ball opposing the motion. ‘

(d) there is no unbalanced force on the ball, so s the ball would want to come to rest.

Answer:

(c)

Question 5.

A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of v 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes.

[Hint: 1 metric tonne = 1000 kg]

Answer:

Initial velocity, u = 0 ms^{-1}

Distance, s = 400 m

Time, t = 20 s

Mass, m = 7 metric tonnes = 7000 kg

By second equation of motion,

s = ut + \(\frac {1}{2}\) at^{2}

⇒ 400 m = 0 ms^{-1} x 20 s + \(\frac {1}{2}\) x a x (20 s)^{2}

⇒ 400 m = \(\frac {1}{2}\) x a x 400 s^{2}

a = \(\frac{400 \mathrm{~m}}{200 \mathrm{~s}^{2}}\) = 2 ms^{-2}

By Newton’s second law of motion

F = m x a

F = 7000 kg x 2 ms^{-2}

= 14,000 N

Question 6.

A stone of 1 kg is thrown with a velocity of 20 ms-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Answer:

Mass, m = 1 kg

Initial velocity, u = 20 ms^{-1}

Final velocity, υ = 0 ms^{-1}

Distance, s = 50 m

By third equation of motion,

υ^{2} – u^{2} = 2as

⇒ (0 ms^{-1})^{2} – (20 ms^{-1})^{2}= 2 x a x 50 m

⇒ 400 m^{2}s^{-2} = 100a m

⇒ \(\frac{-400 \mathrm{~m}^{2} \mathrm{~s}^{-2}}{100 \mathrm{~m}}\) = a

⇒ – 4 ms^{-2} = a

By Newton’s second law of motion,

F = m x a

⇒ F = 1 kg x (- 4 ms^{-2}) = – 4 N

Question 7.

A 8000 kg engine pulls a train of 5 wagons, 1 each of 2000 kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a friction force of 5000 N, then calculate :

(a) the net accelerating force;

(6) the acceleration of the train; and

(c) the force of wagon 1 on wagon 2.

Answer:

Mass of engine, m_{1} = 8000 kg

Mass of each wagon, m_{2} = 2000 kg

Number of wagons, n = 5

Total mass of 5 wagons, m

= 2000 kg x 5 = 10,000 kg

Force exerted by engine, F_{1} = 40,000 N

Friction force of the track, F_{2} = 5000 N

(a) Net accelerating force, F = F_{1} – F_{2}

= (40,000 – 5000) N = 35,000 N

(b) Acceleration of the train, a = \(\frac {F}{m}\)

a = \(\frac{35,000 \mathrm{~N}}{10,000 \mathrm{~kg}}\) = 3.5 ms^{-2}

Question 8.

An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms-2 ?

Answer:

Mass of vehicle, m = 1500 kg

Acceleration, a = -1.7 ms^{-2}

By second law of motion,

F = m x a

F = 1500 x -1.7 = -2550 N

Question 9.

What is the momentum of an object of mass m, moving with a velocity v ?

(a) (mυ)^{-2}

(b) mυ^{2}

(c) \(\frac {1}{2}\) mυ^{2}

(d) mυ

Answer:

(d)

Question 10.

Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction a force that will be exerted on the cabinet?

Answer:

The friction force will also be 200 N.

Question 11.

Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms^{-1} before the collision during which they stick together. What will be the velocity of the combined objects after collision?

Answer:

Mass of the first object, m_{1} = 1.5 kg

Initial velocity of the first object, u _{1} = 2.5 ms^{-1}.

Mass of the second object, m_{2} = 1.5 kg^{-1}

Initial velocity of the second object, u_{2} = – 2.5 ms^{-1}

Let velocity of the combined object be = υ ms^{-1}

By law of conservation of momentum, Momentum before collision = Momentum after collision

⇒ m_{1}u_{1} + m_{2}u_{2}= (m_{1} + m_{2}) υ

⇒ 1.5 kg x 2.5 ms^{-1} + 1.5 kg x (- 2.5 ms^{-1}) = (1.5 kg + 1.5 kg) x υ ⇒ 0 = 3 υ

υ = 0 ms^{-1}

Question 12.

According to the third law of motion when i we push on an object, the object pushes back on us with an equal and opposite force. If the object is a A massive truck parked along the roadside, it will | probably not move. A student justifies this by answer-

ing that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Answer:

A massive truck has a high mass. As a result, it has a high inertia. (Mass is a measure of its inertia.) Due to high inertia, it remains at rest. By first law of motion, an object at rest remains at rest unless an imbalanced force acts upon it. The force created by our push is not strong enough to overcome the inertia of the truck.

Question 13.

A hockey ball of mass 200 g travelling at 10 ms-1 is struck by a hockey stick so as to return it • along its original path with a velocity at 5 ms-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer:

Mass of hockey ball, m = 200 g = \(\frac {200}{1000}\) = 0.2 kg

Initial velocity of the ball, u = 10 ms^{-1}

Final velocity of the ball, υ = -5 ms^{-1}

Initial momentum of the ball = m x u = 0.2 kg x 10 ms^{-1} = 2 kgms^{-1}

Final momentum of the ball = m x υ

= 0.2 kg x (-5 ms^{-1} )

= -1 kgms^{-1}

Change of momentum = mυ – mu

= (-1 – 2) kgm s^{-1}

v = -3 kgm s^{-1}

Question 14.

A bullet of mass 10 g travelling horizontal iy with, a velocity of 150 ms^{-1} strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the for :e exerted by the wooden block on the bullet.

Answer:

Mass, m = 10 g = \(\frac {200}{1000}\)kg = 0.01 kg

Initial velocity, u = 150 ms^{-1}

Final velocity, v = 0 ms^{-1}

Time, t = 0.03 s

Acceleration, a = \(\frac {υ – u}{t}\)

a = \(\frac{(0-150) \mathrm{ms}^{-1}}{0.03 \mathrm{~s}}\)

a = – 5000 ms^{-2}

By second equation of motion,

s = ut + \(\frac {1}{2}\) at^{2}

s = 150 ms^{-1} x 0.03 s + \(\frac {1}{2}\) x (- 5000 ms^{-2}) x (0.03 s)^{2}

s = 4.50 m – 2500 ms^{-2}

s = 4.50 m – 2.25 m

s = 2.25 m

By second law of motion,

F = m x a

F = 0.01 kg x (-5000 ms^{-2})

F = -50 N.

Question 15.

An object of mass 1 kg travelling in a straight line with a velocity of 10 ms-1 collides with and sticks to ‘a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also calculate the velocity of the combined object.

Answer:

MasSof the object, m_{1} = 1 kg

Initial velocity, u = 10 ms^{-1}

Mass of stationary wooden block,

m_{2} = 5 kg

Let velocity of the combined object = υ

Total momentum just before impact = m_{1} x υ

= 1 kg x 10 ms^{-1} = 10 kgm s^{-1}

By law of conservation of momentum,

Total momentum just before impact = Total momentum just after impact Total momentum just after impact = 10 kgm s^{-1}

Total momentum of the combined object , = (m_{1} + m_{2}) x υ

= (1 kg + 5 kg) x υ

10 = 6 υ

or υ = \(\frac {10}{6}\)= \(\frac {5}{3}\) ms^{-1}

Question 16.

An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms^{-1} to 8 ms^{-1} in 6 s. Calculate the initial and final momentum of the object. Also find the magnitude of the force exerted on the object.

Answer:

Mass of the object, m = 100 kg

Initial velocity, u = 5 ms^{-1}

Final velocity, υ = 8 ms^{-1}

Time, t = 6 s

Initial momentum = m x u = 100 kg x 5 ms^{-1}

= 500 kgm s^{-1}

Final momentum = m x υ

= 100 kg x 8 ms^{-1} = 800 kgms

F = 50 N

Question 17.

Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar.) Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. As a result, the insect died. Rahul, while putting an ehtirely new explanation, said that both the motorcar and insect experienced the same force and a change in their momentum. Comment on these suggestions.

Answer:

Rahul gave the correct explanation. This is because due to the law of conservation of momentum, during a collision, momentum of the system is conserved. Both the bodies suffer the same momentum change. However, the big having smaller mass will suffer greater change in velocity. The bus having much larger mass does not suffer any noticeable change in velocity

Question 18.

How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms^{-2}.

Answer:

Mass, m = 10 kg

Height or Distance, s = 80 cm

= \(\frac {80}{100}\) = 0.8 m

Acceleration, a = 10 ms^{-2}

Initial velocity, u = 0 ms^{-2}

By third equation of motion, v^{2} – u^{2} = 2as .

⇒ υ^{2} – (0 ms^{-1})2 = 2 x 10 ms^{-2} x 0.8 m

⇒ υ^{2} = 16 ms^{-1}

⇒ υ = 4 ms^{-1}

Momentum transferred to the floor = m x υ

= 10 kg x 4 ms^{-1} =40 kgm s^{-1}

### Bihar Board Class 9 Science Chapter 9 Force and Laws of Motion Textbook Activities

Based on Activity 9.1

Question 1.

Make a pile of similar carrom coins on a table. Then remove the lower coin without touching the other coins. With your fingers, you may give a sharp horizontal hit at the bottom of the pile using another carom coin or striker.

(i) What will happen if the hit is strong enough?

(ii) What will happen to the other coins once the lower coin is removed?

Answer:

(i) If the hit is strong enough, then the bottom coin moves quickly so that any horizontal force between it and the above will not move the rest of the pile *n horizontal direction.

(ii) Once the lower coin is removed, the inertia of the bther coins makes them fall vertically on the table.

Based on Activity 9.2

Question 2.

Take an empty glass. Place a square cardboard on the glass. Place a 50 paisa coin on the card [Fig. 9.11 (a)]. Now snap the card. We will find that the card moves on, but the coin falls into the glass [Fig. 9.11 (6)].

Answer:

Explanation. The coin was at rest on the card. When the card is snapped, it flies away. The coin, however, remains at rest due to the inertia of rest, and falls into the

glass.

Based on Activity 9.3

Question 3.

(i) Place a water-filled tumbler on a tray. Hold the tray and turn around as fast as you can. What will you observe and why?

(ii) Why is a groove provided in a saucer to place the cup?

Answer:

(i) We will observe that the tumbler topples down and water spills. This is because when we turn around fast the tumbler comes into motion but the water continues to remain at rest due to inertia.

(ii) The groove prevents the cup from toppling over due to sudden jerks.

Activity 9.4

Question 4.

Request two children to stand an two separate carts as shown in the figure. Give them a bag full of sand or some other heavier object. Ask them to play a game of catch with the bag.

(i) Does each of them receive an instantaneous reaction as a result of throwing the sand bag (action)? You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other.

(ii) Now, place two children on one cart and one on another cart. Ask them to repeat the activity. What will you observe?

(iii) Why are skateboards not effective in the above activity?

Answer:

(i) Yes, each of them will receive a reaction as a result of throwing the sand bag. Due to reaction, the cart will move backwards.

(ii) In this case, the cart will move further back because of the greater mass. Second law of motion can be seen, as this arrangement would show different accelerations for the same force.

(iii) Skateboards are not as effective because it is difficult to maintain straight line motion using a skateboard.

Based on Activity 9.5

Question 5.

Take a big size rubber balloon and inflate it fully. Tie its neck using a thread. Also using adhesive tape, fix a straw on the surface of this balloon. Pass a thread through the straw and hold one end of the thread in jmur hand. Ask your friend to hold the other end of the thread at some distance.

(i) What will happen if we remove the thread tied on the neck of the balloon?

(ii) In which direction will the cart move?

Answer:

(i) If we remove the thread tied on the neck of the balloon, the air will escape from the mouth of the balloon.

(ii) Since the air moves down, the cart will move up.

Based on Activity 9.6

Question 6.

Take a dry test tube and put a small amount of water in it. Place a stop cork on the mouth of it. Now suspend the test tube horizontally by two strings or wires as shown in the figure. Heat the test tube with a burner until water vapourises and the cork blows out.

(i) What happens to the test tube when the cork blows out?

(ii) How will you calculate the velocity of the cork?

Answer:

(i) When the cork blows out, the test tube recoils in the direction opposite to the direction of cork blow.

(ii) Let mass of the cork = m_{1}

Velocity of the cork = υ_{1}

Mass of test tube = m_{2}

Velocity of test tube = υ_{2} By law of conservation of momentum,

m_{1}υ_{1} = m_{1}υ_{1}

or υ_{1} = \(\frac{m_{2} v_{2}}{m_{1}}\)

Since the mass of the test tube is greater than the mass of the cork, the velocity of the cork will be higher.

### Bihar Board Class 9 Science Chapter 9 Force and Laws of Motion Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.

A ball on the ground when given a small hit does not move forever. What does this observation suggest ?

Answer:

This observation suggests that rest is the natural state of an object.

Question 2.

Name two scientists who developed an entirely different approach to understand motions.

Answer:

Galileo Galili (1564-1642) and Isaac Newton (1642-1727).

Question 3.

What can we do to an object to change its state of motion ?

Answer:

We can push or hit or pull an object to change its state of motion.

Short Answer Type Questions

Question 1.

What is force ? What is the SI unit of force ?

Answer:

(i) Forces something that changes or tends to change the state of rest, or the state of uniform motion of a body. ,

(ii) The SI unit of force is Newton (N).

Question 2.

Write the effects of force.

Answer:

Force can produce three effects :

(i) It can change the magnitude of velocity of an object. (i.e., to make the object move faster or slower).

(ii) It can change the direction of motion of the object.

(iii) It can change the shape or size of the object.

Question 3.

Give examples to show the effects of force.

Answer:

Illustrations :

(i) Place a ball on the ground. Kick it with your foot. The ball starts moving. The ball moves because of the force applied to it.

(ii) If a ball is coming towards you, you can kick it in any direction. The direction of motion of the ball changes because of the force applied to it.

(iii) Place a rubber ball on the ground. Press it with your foot. It is found that the ball is no longer round but takes the shape of an egg, i.e., it is oblong. The shape of the ball has been changed because of the force applied on the ball.

Question 4.

Name some non-SI units of force. Define them.

Answer:

Some commonly used non-SI units of force are :

(a) Dyne. The dyne is the CGS unit of force. One dyne is that much force which produces an acceleration of 1 cm/ s^{2} in a body of mass of 1 g, i.e.,

1 dyne =.l g x 1 cm/s^{2}

(b) Gram-weight. It is the gravitational unit of force in CGS system. This is denoted as g – wt.

One gram-weight is that much force which produced an acceleration of 981 cm/s^{2} in a body of mass 1 g. Thus,

1 g-wt = 1 g x Acc. due to gravity (g)

or, 1 g-wt = 1 g x 981 cm/s^{2} = 981 dynes [∵ 1 g x 1 cm/s^{2} = 1 dyne]

(c) Kilogram-weight. It is the gravitational unit of force in SI system. This unit is denoted as kg-wt. One kg- wt is that much force which produces an acceleration of5 9.81 m/s^{2} in a body of mass 1 kg. Thus,

1 kg-wt = 1 kg x 9.81 m/s^{2}

Question 5.

Convert :

(i) 1 newton into dynes

(ii) 1 g-wt into dynes

(iii) 1 kg-wt into newton

(iv) 1 kg-wt into dynes.

Answer:

(i) By definition,

1 N = 1 kg x 1 m/s^{2}

= 1000 g x 100 cm/s^{2}

= 105 x g x cm/s^{2}

= 105 dynes [ ∵ lgxl cm/s^{2} = 1 dyne]

(ii) By definition,

1 g-wt = 1 g x 1 g

1 g-wt = 1 g x 981 cm/s^{2}

= 981 dynes [ ∵ 1 g x 1 cm/s^{2} = 1 dyne]

(iii) By definition,

1 kg-wt = 1 kg x 1 g

1 kg-wt = 1 kg x 9.81 m/s^{2}

= 9.81 N [∵ 1 kg x 1 m/s^{2} = 1 N]

(iv) l kg-wt = 9.81 N

= 9.81 x 10^{5} dynes [∵ 1 N = l°r’ dynes]