*NCERT Solutions of Chapter: Force and Laws of Motion. NCERT Solutions along with worksheets and notes for Class 9.*

**Page No. 118 (Exercise)**

**(1)** **Which of the following has more inertia:**

**(a) a rubber ball and a stone?**

**(b) a bicycle and a train**

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

**Ans-** Inertia is related to mass of the body. Therefore, when mass of the object is greater the inertia of that object will be also greater.

**(a) **A rubber ball has less mass than the stone of same size. Therefore stone has more inertia than the ball.

**(b)** A bicycle has less mass than a train. Therefore train has more inertia than the bicycle.

**(c)** A five rupee coin has more mass than a one rupee coin. Therefore a five rupee coin has more inertia than a one rupee coin.

**(2)** **In the following example, try to identify the number of times the velocity of the ball changes;”A football player kicks a football to another player of his team who kicks the ball 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.**

**Ans-** The ball of above example has changed its velocity 4 times.

**Change 1-**

Ball is at rest,ie, velocity =0

Player 1 kicked the ball towards player 2.

Ball changed its velocity and direction here.

**Change 2-**

Player 2 kicked the ball towards goal.

Ball again changed its velocity and direction.

**Change 3-**

In goal ball again come to rest,ie, velocity become zero.

**Change 4-**

The goalkeeper kicked the ball towards the player 3.

The ball changes its velocity from zero to certain level and its direction also.

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

**Ans-** The leaves of tree fall down if tree is shaken forcibly. It happens because of inertia. When we started shaken tree vigorously branches of tree also started shaken but leaves of tree due to inertia wants to keep their position of rest and do not started shaken. As a result of it leaves fall down from the tree.

**(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?**

**Ans-** When a moving bus brakes to a stop our lower body or feet stops motion. But due to inertia our upper body wants to keep its position of motion. Therefore, we fall in the forward direction.

When a bus accelerates from the rest, our lower body or feet starts motion with the bus. But due to inertia our upper body wants to keep its position of rest. Therefore, we fall in the backward direction.

**Page No. 126 (Exercise)**

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

**Ans-**When a horse cart moves forward many forces applied on it instead of equal and opposite action and reaction forces. But net resulting force allows horse cart to move.

Let

Force applied on the cart by the horse = F_{HC}

Force applied by the cart on the horse = F_{CH}

These forces are equal and opposite. Besides these forces ground also exerts push force on horse and friction force on the cart.

Let

Friction force exerts by ground on the horse = F_{GH}

Friction force exerts by ground on the cart = F_{GC}

When, horse and cart are at rest.

Acceleration is zero.[Velocity is zero,ie,at rest]

Forces acted on cart are equal and opposite.

Froce F_{HC} applied by horse = Friction of ground on cart F_{GC}

F_{HC} = – F_{GC} [equal and opposite]

Forces acted on horse are equal and opposite.

Force F_{CH} applied by the cart on horse = Friction of ground on horse F_{GH}

F_{CH} = – F_{GH} [equal and opposite]

Force exerted by the horse on cart = Force exerted by cart on the horse

F_{HC} = – F_{CH} [equal and opposite]

When, horse starts moving. Cart also starts moving forward.

The condition for moving forward for horse is

Net forward force on horse > net backward force on the horse

F_{GH} > F_{CH}

The condition for moving forward for cart is

Net forward force on cart > net backward force on the cart

F_{HC} > F_{GC}

The condition for moving forward for whole system [horse cart] is

Net forward force on horse cart > net backward force on the horse cart

F_{HC} + F_{GH} > F_{GC} + F_{CH}

But F_{HC} = – F_{GC} [equal and opposite forces by Newton’s third law]

Therefore,

F_{GH} > F_{GC }[ without considering negative sign]

Net forward force on horse > net backward force on the cart

or

Friction force of ground on horse > Friction force of ground on cart

Friction force of ground on horse should be greater than the friction force of ground on cart to move horse cart forward. Friction force of ground on wheel is very small as wheel is rolling and friction of ground on horse is very large as it is static friction.

The other forces like upward force by ground on cart and horse, downward force of gravity do not contribute in horizontal movement.

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

**Ans-** When water from hose ejected, it exerts a backward force on the fireman. Water is coming out with high velocity which causes high momentum. This high momentum of the water exerts high force on hose. This force also exerts equal force on fireman in opposite direction. Therefore, fireman feels difficulty in holding hose.

**(3) From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35m/s. Calculate the initial recoil velocity of the rifle.**

**Ans-**

**(4) Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 2m/s and 1m/s, respectively. They collide and after the collision, the first object moves at a velocity of 1.67m/s. Determine the velocity of the second object.**

**Ans-**

Mass of the first object m_{1}= 100 g

= 0.1kg [1 kg = 1000 g]

Mass of the second object m_{2} = 200 g

= .5kg [1 kg = 1000 g]

Initial velocity of first object u_{1} = 2 m/s

Initial velocity of second object u_{2}= 1 m/s

After collision, velocity of first object v_{1} = 1.67 m/s

Let after collision velocity of second object = v_{2} m/s

**Exercise**

**(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 the velocity. If no, provide a reason?**

**Ans-**Yes, an object can travel with a non zero velocity even if it experiences a net zero external unbalanced force. The object can move with constant velocity in a particular direction. The object will stop only if it experience external net non zero unbalanced force.

**(2) When a carpet is beaten with a stick, dust comes out of it. Explain?**

**Ans-** Due to inertia dust particle wants to remain in their position of rest but carpet has some movement when beaten. Therefore, dust particle come out of the carpet.

**(3) Why is it advised to tie any luggage kept on the roof of a bus with a rope?**

**Ans-** When bus starts movement, the luggage at the top of the roof wants to remain position of rest. When bus move forward, due to inertia luggage do not move and fall off from roof of the bus. Therefore, it is advised to tie any luggage kept on the roof of a bus with a rope.

**(4)A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because**

**(a)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 the ball would want to come to rest.**

**Ans-**The ball stops due to the friction force between the field and ball. Friction force always opposes the motion.

**(5)A truck starts from the rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tons (Hint:1 metric ton = 1000 kg)**

**Ans-**

**(6)** **A stone of 1kg is thrown with a velocity of 20m/s across the frozen surface of a lake and comes to rest after traveling a distance of 50m.What is the force of friction between the stone and the ice?**

**Ans-**

**(7) A 8000 kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40000N and the track offers a friction force of 5000N, then calculate:**

**(a) The net accelerating force**

**(b) The acceleration of the train**

**(c) The force of wagon 1 on wagon 2**

**Ans-**

Mass of the engine = 8000kg

Mass of one wagons = 2000kg

Number of wagons = 5

Force exerted by engine = 40000N

Friction force offered by track = 5000N

**(c)**

Net acceleration of the train = 1.944m/s^{2}

Force of wagon 1= Mass of the wagon 1 × Net acceleration of the train

= 2000 × 1.944

= 3888 N

Force of engine = Mass of the engine × Net acceleration of the train

=8000 × 1.944

= 15552N

Net accelerating force of train is 35000N. Engine is pulling all wagons. So it is exerting force on all wagons through first wagon to second wagon and second wagon is passing force to third wagon and so on. When engine pulls first wagon with force (=ma), first wagon accelerates. Wagon first needs force equal to multiplication of its mass and acceleration to move forward. Rest of force it passes to second wagon. Again second wagon passes force to third wagon and this process continues till last wagon. As the result, all wagons start accelerating.

Every wagon and engine experienced equal and opposite force. Therefore, all forces will cancel out because of equal and opposite in direction [Newton’s third law of motion]except force of engine, force of all wagons. Addition of force of engine(8000 1.94 = 15552N) and force of all wagons ( 5 × 2000 × 1.944 = 19440) will give us net force of train( approx. 35000N).Friction of wheels of wagons will not included because they are rolling and not driving force. Other forces on wagons like weight force and normal force will cancel each other as they are equal and opposite in direction.

**(8) An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7m/s ^{2}?**

**Ans-**

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

**(a)(mv)2**

**(b) mv _{2 }**

**(c)1/2mv _{2}**

**(d)mv**

**Ans-**

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

**Ans-** When we move a wooden cabinet across floor with a horizontal force 200 N, according to third law of motion equal and opposite force acted on different objects. Therefore, a friction force equal to horizontal force 200N will be exerted on cabinet when we intended it to move.

**(11)**Two objects , each of mass 1.5 kg, are moving in the same straight line but in the opposite direction. The velocity of each object is 2.5 m/s before collision during which they stick together. What will be the velocity of the combined object after collision?

**Ans-**

**(12) According to the third law of motion when we push on an object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.**

**Ans-**A student justification that the two opposite and equal forces cancel each other is correct. Truck will not move because a truck is pushing ground downward and ground is also pushing it upward with equal force [Newton’s third law].Truck will move when push force on truck will be greater than the friction force.

**(13)A hockey ball of mass 200g travelling at 10m/s is struck by a hockey stick so as return it along its original path with a velocity at 5 m/s. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.**

**Ans-**

**(14) A bullet of mass 10g travelling horizontally with a velocity of 150m/s strikes a stationary wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by wooden block on the bullet.**

**Ans-**

Mass of the bullet = 10g = .01kg

Initial velocity of the bullet u= 150m/s

Final velocity of the bullet v = 0m/s

Time taken by bullet to come to rest t = .03s

Let distance of penetration = s m

We know

v = u + at

Where

v = final velocity of object

u = initial velocity of the object

a = acceleration of the object

t = time taken by the object

Putting values

0 = 150 + a × .03

-.03a = 150

a =- 5000m/s^{2}[-ve sign shows that velocity of the object is decreasing]

we know

s = ut + ½ at^{2}

Putting values

**(15)An object of mass 1kg travelling in a straight line with a velocity of 10 m/s 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.**

**Ans-**

**(16)An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 seconds. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.**

**Ans-**

Mass of the object m = 100 kg

Initial velocity of the object u = 5 m/s

Final velocity of the object v = 8 m/s

Time taken by the object t = 6s

Initial momentum of the object = mu

= 100 × 5

= 500 kg m/s

Final momentum of the object = mv

= 100 × 8

= 800 kg m/s

We know

V = u + at

8 = 5 + a × 6

6a = 3

a = .5 m / s^{2}

Force = ma

= 100 .5

=50 N

Hence, Initial momentum of the object is 500 kg m/s and final momentum of the object is 800 kg m/s. Force exerted on the object is 50 N.

**(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 windshield. Akhtar and Kiram started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change of 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. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and change in their momentum. Comment on these suggestions.**

**Ans-**

Let simplify this question mathematically.

Kiran suggestion that the insect suffered a greater change in momentum as compared to the change of momentum of the motorcar is not true. But, her statement regarding the change in the velocity of the insect was much more than that of the motorcar is correct.

Akhtar comment that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect is incorrect. Because both car and insect experience same amount of force in opposite directions [Newton’s third law of motion]. The force is very high for insect, therefore he dies. But force is not very high for car, there it keeps moving forward.

Rahul statement is right. Due to law of conservation of momentum change in momentum after collision is same for both insect and momentum. According Newton’s third law of motion, both will experience same force.

**(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 download acceleration to be 10 m/s ^{2}.**

**Ans-**

Mass of the dumb-bell m = 10 kg

Height (distance) s = 80 cm = .8 m

Download acceleration a = 10 m/s^{2}

Let final velocity of the dum-bell = v m/s

v^{2} = u^{2 }+2as

Where

v = Final velocity of the object

u = Initial velocity of the object

a = acceleration of the object

s =distance travelled by object

Putting values

v^{2} = 0 + 2 × 10 × .8

=16

v = 4 m/s

Momentum transfer by dumb-bell = mv

= 10 × 4

=40 kg m/s

Hence,40 kg m/s momentum will dumb – bell will transfer to the floor.

**Helping Topics**