NCERT Solutions Class 9 Science Chapter 9 Force and Laws of Motion
NCERT Solutions Class 9 Science Chapter 9 Force and Laws of Motion: National Council of Educational Research and Training Class 9 Science Chapter 9 Solutions – Force and Laws of Motion. NCERT Solutions Class 9 Science Chapter 9 PDF Download.
NCERT Solutions Class 9 Science Chapter 9: Overview
Board |
NCERT |
Class |
9 |
Subject |
Science |
Chapter |
9 |
Chapter Name |
Force and Laws of Motion |
Topic |
Exercise Solutions |
NCERT Solutions Class 9 Science Chapter 9 – Force and Laws of Motion
Part 1
1.) Which of the following has more inertia:
(a) a rubber ball and astone of the same size?
(b) a bicycle and a train?
(c) a five-rupees coin and a one rupee coin?
Answer: inertia is a property which resist to change in its state so it depend on mass.
a) The density of stone is greater than rubber ball but volume is same therefore mass of stone is greater than rubber ball thus stone has greater inertia than rubber ball
b) Train has more mass thenbicycle so inertia of train is greater than bicycle.
c) Five rupees coin has more mass then one rupee coin so inertia of five rupees is greater than 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 football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player ofhis own team”. Also identify the agent supplying the force in each case.
Answer: Given example is of football match. As we know that, forces changes the velocity (mass of football is constant). In this match players are the agents for supplying the force in each case.
Initially, first player kick the ball and change the velocity. After that another player also changes the velocity of football second times. Third time goalkeeper of opposite team changes velocity. At fourth times, player of opposite team changes the velocity. In this way velocity changed 4 time in this example.
3.) Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
As we know the concept of inertia. Leaves have inertia because of mass. When we shake tree then leaves do not accept this change and oppose it. But tree accelerate and some leaves which loosely connect separate 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?
Answer: We are also move with same speed of bus. But when driver brakes, then our body does not accept this change because of inertia and we fall forward.
When bus in state of rest then we also in state of rest. But when bus suddenly accelerate forward then body does not accept this change and do not easily move with bus because of inertia. So the body trying to maintain in it’s state of rest but bus moves forward so we fall backwards.
Part 2
1.) If action is always equal to the reaction, explain how a horse can pull a cart.
Answer: Horse push the earth backward is an action and earth push the horse forward is a reaction. In this way equal and opposite reaction happened and horse pull the cart.
2.) Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water ata high velocity.
Answer: When large amount of water moves through a hose suddenly then it’s momentum increases and exert a reaction force in opposite direction. The firemen must hold hose hardly to neutral this reacting force. this fireman must hold a hose hardly.
3.) From a rifle of mass 4 kg, abulletof mass 50 g is fired with aninitial velocity of 35 m/s. Calculate the initial recoil velocityof the rifle.
Answer: Given, mass of rifle = M = 4 kg,
mass of bullet = m = 50 gm = 0.05 kg,
Velocity of bullet = v = 35 m/s.
We have to calculate velocity of rifle = V.
But rifle moves backward so velocity of rifle = -V.
As we know that conservation of linear momentum.
Initial momentum = final momentum. ( 1)
But initially rifle and bullet are in rest so total Initial momentum is zero.
Final momentum= momentum of rifle + momentum of bullet.
Put the value of initial momentum and final momentum in equation 1.
0 = momentum of rifle + momentum of bullet
0 = M× (-V) + mv
MV = mv
4 ×V = 0.05 × 35
V = 1.75/4
V = 0.44 m/s.
The recoil velocity of rifle is 0.44 m/s.
4.) Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1 m/s, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. Determinethe velocity of the second object.
Answer: Given,
Mass of first object = m = 100 g= 0.1 kg
Mass of second object = M = 100 g= 0.2 kg
Initial velocity of first object = u = 2 m/s.
Final velocity of first object = v = 1.67 m/s.
Initial velocity of second object = U = 1 m/s.
Final velocity of second object = V m/s.
As we know the law of conservation of linear momentum.
Total initial momentum = Total final momentum.
Initial momentum of first object + initial momentum of second object
= final momentum of first object + final momentum of second object
m × u + M × U = m × v + M × V
0.1 × 2 + 0.2 × 1 = 0.1 × 1.67 + 0.2 × V
0.2 + 0.2 = 0.167 + 0.2V
0.4 – 0.167= 0.2V
V = 0.233/0.2
V = 1.165 m/s.
The final velocity of object second is 1.165 m/s.
Exercise Solution
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.
Answer: Yes it is possible that the object travels with a non-zero velocity if it experiences a net zero external unbalanced force.
Newton’s first law state that, the object remains in the state of rest or state of uniform motion until we act an external unbalanced force on it.
If a stone is released from certain height then it fall under force of gravity and air resistance opposes the motion. Therefore magnitude of net force is zero still the stone fall with constant velocity. The direction of velocity is same as before and magnitude of velocity remains constant.
2.) When a carpet is beaten with a stick, dust comes out of it.Explain.
Answer: When a carpet is beaten with a stick then the carpet accelerate but the dust particles are trying to remain in state of rest because of inertia. Therefore they separate from carpet.
3.) Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer: The journey of bus is not uniform. It accelerate continuously. Therefore luggage also accelerate with the bus but it is trying to remains in its state of rest because of inertia. In this process luggage may fall on passenger. Thus it is advised to tie it.
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 wouldwant to come to rest.
Answer: 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
(c) there is a force on the ball opposing the motion.
5.) A truck starts from rest and rolls down a hill with a constantacceleration. It travels a distance of 400 m in 20 s. Find itsacceleration. Find the force acting on it if its mass is7 tonnes (Hint: 1 tonne = 1000 kg.)
Answer: Given, bus start from rest so initial velocity = 0 m/s, distance = s = 400 m,
Time = t = 20 s, Mass of bus = m = 7 tonnes.
But 1 tonne = 1000 kg,
m = 7 tonnes = 7000 kg.
We calculate acceleration,
As we know, second equation of motion,
S = ut + ½ at²
400 = 0 ×20 + ½ × a × 20²
400 = 200a
a = 400/200
a = 2 m/s².
The acceleration of truck is 2 m/s².
Now we calculate force exerted on truck.
According to second law of motion,
Force = mass × acceleration
Force = 7000 × 2
Force = 14000 N.
Force acting on truck is 14000 N.
6.) A stone of 1 kg is thrown with a velocity of 20 m/s acrossthe frozen surface of a lake and comes to rest after travellinga distance of 50 m. What is the force of friction between thestone and the ice?
Answer: Given, Mass = m = 1 kg ,
Initial velocity = u = 20 m/s,
Finally it stop so final velocity = v = 0 m/s,
Distance = s = 50 m.
Firstly we calculate acceleration,
As we know,
Third equation of motion,
v² = u² + 2as
0² = 20² + 2a × 50
a = -400/100
a = -4 m/s².
Where minus sign indicates deceleration. The Frictional force stop stone.
According to second law of motion,
Force = mass × acceleration
Force = 1 × (-4 )
Force = -4 N.
The Frictional force is -4 N.
7.) A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg,along a horizontal track. If the engine exerts a force of 40000 Nand the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force and
(b) the acceleration of the train.
Answer: Given, Mass of engine = 8000 kg,
Mass of a wagons = 2000 kg,
Mass of 5 wagons = 5 × 2000 = 10000 kg.
Total mass = 8000 + 10000 = 18000 kg,
Force in forward direction = 40000 N,
Force in backward direction = 5000 N,
The diagram illustrated force acting on train.
Net force acting on train = forward force – backward force
Net force acting on train = 40000- 5000
Net force acting on train = 35000 N.
From second law of motion,
Net force = mass × acceleration
Acceleration = net force/mass
Acceleration = 35000/18000
Acceleration = 1.944 m/s².
The acceleration of train is 1.944 m/s².
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 m/s²?
9.) What is the momentum of an object of mass m, moving withavelocity v?
(a) (mv)2
(b) mv2
(c) ½ mv2
(d) mv
Answer: The momentum of an object of mass m, moving withavelocity v is (d) mv.
In case you are missed :- Previous Chapter Solution
10.) Using a horizontal force of 200 N, we intend to move a woodencabinet across a floor at a constant velocity. What is the frictionforce that will be exerted on the cabinet?
Answer: Given, forward force = 200 N.
According to first law of motion, if a body moves with constant velocity then net force acting on that body is zero.
Forward force + backward force = 0
But backward force is Frictional force.
Forward force + Frictional force = 0
2000 N + Frictional force = 0
Frictional force = – 2000 N
Minus sign indicated that Frictional force opposes the motion.
The value of Frictional force is 2000 N.
11.) Two objects, each of mass 1.5 kg, are moving in the samestraight line but in opposite directions. The velocity of eachobject is 2.5 m/s before the collision during which theystick together. What will be the velocity of the combinedobject after collision?
Answer: Mass of first object = mass of second object = M= 1.5 kg,
Magnitude of velocity of both objects is same but direction is opposite,
Velocity of 1st object = u = 2.5 m/s
Velocity of second object is same but direction is opposite,
Velocity of 2nd object = U = -2.5 m/s
After collision they stick together so the velocity of system after collision = v m/s.
According to law of conservation of momentum,
Initialmomentum = final momentum
Mu + MU = v (M+M)
1.5 × 2.5 + 1.5 ×(-2.5) = v (1.5 + 1.5)
3.75 – 3.75 = v (3)
0 = v×3
v = 0 m/s.
The velocity of combination becomes zero after collision.
12.) According to the third law of motion when we push on an object,the object pushes back on us with an equal and opposite force.If the object is a massive truck parked along the roadside, itwill probably not move. A student justifies this by answeringthat the two opposite and equal forces cancel each other.Comment on this logic and explain why the truck does notmove.
Answer: If the object is massive and not moved by exerting force which applied by us then we can explain this on the basis of first law and frictional force.
1) According to 1st law of motion, every object has a property to opposes the change in it’s state. This property is called inertia and inertia is depend mass of the object. The force applied by us to the heavy truck is not sufficient to change in it’s state.
2) The applied force is cancel out by Frictional force. So the net force in truck becomes zero and it does not move.
13.) A hockey ball of mass 200 g travelling at 10 m/s is struck bya hockey stick so as to return it along its original path with avelocity at 5 m/s. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by theforce applied by the hockey stick.
Answer: Given, Mass of hockey ball= m = 200 g = 0.2 kg,
Initial velocity = u = 10 m/s,
After struck it return to its initial path so, Final velocity becomes negative,
Final velocity = v = – 5 m/s.
We know that,
Change in momentum = final momentum – initial momentum,
Change in momentum = mu – mv
Change in momentum = 0.2 × 10 – 0.2 ×(-5)
Change in momentum = 2 +1
Change in momentum = 3 kg m/s
The change in momentum by applied force is 3 kg m/s.
14.) A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to restin 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exertedby the wooden block on the bullet.
Answer: Given, Mass of bullet = 10 g = 0.01 kg,
Initial velocity = u = 150 m/s, Finally it stop after penetration so final velocity becomes zero.
Final velocity = 0 m/s.
Time = 0.03 s.
We calculate acceleration using definition of acceleration,
Acceleration = (v-u)/t
Acceleration =(0- 150)/0.03
Acceleration = -150/0.03
Acceleration = – 5000 m/s².
Minus sign indicate retardation.
Now we calculate distance using second equation of motion,
s = ut + ½ at²
s = 150 × 0.03 + ½ ×(-5000 )× (0.03)²
s = 4.5 – 2500× 0.0.0009
s = 4.5 –2.25
s = 2.25m
Thus the bullet cover 2.25 m in the stationary wooden block.
Now we calculate force exerted by wooden block.
According to second law of motion,
Force = mass × acceleration
Force = 0.01 × (-5000)
Force = -50 N
Where minus sign indicates that the force opposes the motion.
The wooden block exerts 50 N on bullet.
15.) An object of mass 1 kg travelling in a straight line with a velocityof 10 m/s collides with, and sticks to, a stationarywoodenblock of mass 5 kg. Then they both move off together in thesame straight line. Calculate the total momentum just beforethe impact and just after the impact. Also, calculate the velocityof the combined object.
Answer: Given, Mass of object = m = 1 kg,
Mass of wooden block = M = 5 kg,
Initial velocity of object = u = 10 m/s,
Initial velocity of wooden block = U = 0 m/s.
Let combined velocity of object = v.
Initial momentum = momentum of object+ momentum of wooden block.
Initial momentum = mu + MU
Initial momentum = 1 × 10 + 5 × 0
Initial momentum = 10 kg m/s.
According to conservation of momentum, momentum is conserved,
Therefore Final momentum also 10 kg m/s.
Now we calculate velocity of combined object,
Initial momentum = final combined momentum
10 = v (m + M)
10 = v ( 1 + 5)
10 = 6 × v
v = 10/6
v = 5/3 m/s.
The combined velocity of the object is 5/3 m/s.
16.) An object of mass 100 kg is accelerated uniformly from a velocityof 5 m/s to 8 m/s in 6 s. Calculate the initial and finalmomentum of the object. Also, find the magnitude of the forceexerted on the object.
Answer: Given, Mass of object = m = 100 kg,
Initial velocity = u = 5 m/s,
Final velocity = v = 8 m/s,
Time = 6 s,
As,
Initial momentum = mass × initial velocity
Initial momentum = 100 × 5
Initial momentum = 50p kg m/s
Similarly we calculate Final momentum,
Final momentum = mass × Final momentum
Final momentum = 100 × 8
Final momentum = 800 kg m/s.
Now, we calculate force,
As we know that,
Force = (final momentum-initial momentum)/time,
Force = (800 – 500)/6
Force = 300/6
Force = 50 N.
The value of
Final momentum = 500 kg m/s
Initial momentum = 800 kg m/s
Force = 50 N.
17.) Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insecthit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggestedthat the insect suffered a greater change in momentum ascompared to the change in momentum of the motorcar (becausethe change in the velocity of the insect was much more thanthat of the motorcar). Akhtar said that since the motorcar wasmoving with a larger velocity, it exerted a larger force on theinsect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and theinsectexperienced the same force and a change in theirmomentum. Comment on these suggestions.
Answer: We overview the explanation made by Akhtar, Kiran and Rahul.
Kiran’s explanation: He 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 thanthat of the motorcar).
The change in velocity of insect is greater but mass is negligible as compared to car. Thus Kiran’s explanation is wrong.
Akhtar’s explanation: He 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.
Force is not depend on magnitude of velocity. Thus Akhtar’s observation is also wrong.
Rahul’s observation: Rahul explained that both the motorcar and the insect experienced the same force and a change in theirmomentum.
According to third law of motion action and reaction has same values. Thus is explanation is right.
18.) How much momentum will a dumb-bell of mass 10 kgtransfer to the floor if it falls from a height of 80 cm? Take its downwardacceleration to be 10 m/s².
Answer: Given, Mass = m = 10 kg, vertical displacement = 80 cm
Vertical displacement = 0.8 m,
Initial velocity = u= 0 m/s.
Acceleration = 10 m/s².
We calculate final velocity using 3rd law,
v² = u² + 2as
v² = 0² + 2 × 10 ×0.8
v² = 16
Taking square root on both side,
v = 4 m/s.
The final velocity of dumb-bell is 4 m/s.
Now we calculate momentum,
As,
Momentum = mv – mu
Momentum = 10× 4 – 10 ×0
Momentum = 40 kg m/s.
The momentum of dumb-bell is 40 kg m/s.
In case you are missed :- Next Chapter Solution