Class 9 Science Chapter 9 – Force and Laws of Motion
Best NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion .”Force and Laws of Motion,” Chapter 8 of Class 9 Science, is an introductory chapter that covers important physics ideas pertaining to how and why objects move.
EXERCISE 1
1. Which of the following has more inertia:
(a) a rubber ball and a stone of the same size?
Answer: An object’s mass determines its inertia. Greater inertia is provided by heavier or more massive objects.
Compared to a rubber ball of the same size, stone is heavier. e. As a result, the stone’s inertia is higher than a rubber ball’s.
(b) a bicycle and a train?
Answer: A train is more substantial than a bicycle. Therefore, the train’s inertia is larger than the bicycle.
(c) a five-rupees coin and a one-rupee coin?
Answer: A coin worth five rupees weighs more than one rupee. As a result, the five rupee coin has more inertia than the 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 of his own team”. Also identify the agent supplying the force in each case.
Answer: There are four velocity alterations for the ball.
First change: As the football player kicks the ball, its speed increases from 0 to a predetermined figure. The ball’s velocity changes as a result.
Second change: In the second change, a different player is kicking the ball to the goal post. The ball’s direction is altered as a consequence. Consequently, its speed fluctuates. In this instance, the player changed the ball’s velocity by using force.
Third change: During the third change, the goaltender is gathering the ball. The ball eventually stops moving. Its speed is consequently reduced from a given value to zero. The ball is moving at a different speed. In this instance, To change the ball’s pace or slow it down, the goalie applied a counterforce.
3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer: When a branch is swiftly moved, the leaves attached to it have a tendency to remain in their resting position due to the inertia of rest. This causes a great deal of stress to be placed on the leaves and branch junctions. A portion of the leaves on this strain fall off the branch.
Fourth substitution: The goalie kicks the ball to his colleagues. The ball’s velocity rises from zero to a specific value as a result. Its velocity changes again as a result. The goalie in this instance changed the ball’s velocity by using force.
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: Since both the bus and our upper body are moving when the bus is travelling and our body is trying to stay in motion due to inertia of motion when the bus is breaking, we experience a forward push when the bus is braking. In a similar vein, passengers tend to fall backwards as the bus accelerates away from them.
This is due to the fact that when the bus accelerates, the inertia of the passengers tends to oppose the bus’s forward motion. As a result, the passenger often falls backwards when the vehicle accelerates.
EXERCISE 2
1. If action is always equal to the reaction, explain how a horse can pull a cart.
Answer: A horse pushes the ground backward with his foot. The Earth applies a reaction force to the horse that propels it forward in accordance with Newton’s third law of motion. The waggon moves forward as a result.
2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer: Newton’s Third Law of Motion indicates that water pulls the hose backwards with the same force when it is discharged from a hose at a high speed. Because of this, firefighters find it challenging to hold onto a hose that rapidly ejects a large amount of water.
3. From a rifle of mass 44 kg, a bullet of mass 5050 g is fired with an initial velocity of 3535 ms−1ms−1. Calculate the initial recoil velocity of the rifle.
Given:
Mass of the rifle, m1=4m1=4 kg
Mass of the bullet, m2=50m2=50 g =0.05=0.05 kg
Recoil velocity of the rifle =v1=v1
Initial velocity of bullet, v2=35v2=35 m/s
Answer: As, the riffle is at rest, its initial velocity, v=0v=0
Total initial momentum of the rifle and bullet system =(m1+m2)v=0=(m1+m2)v=0
Total momentum of the rifle and bullet system after firing:
=m1v1+m2v2=m1v1+m2v2
=4(v1)+0.05×35=4(v1)+0.05×35
=4v1+1.75=4v1+1.75
According to the law of conservation of momentum,
Total momentum after the firing = Total momentum before the firing
4v1+1.75=04v1+1.75=0
4v1=−1.754v1=−1.75
v1=−1.754v1=−1.754
v1=−0.4375v1=−0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity v1=−0.4375v1=−0.4375 m/s
4. Two objects of masses 100100 g and 200200 g are moving along the same line and direction with velocities of 22 ms−1ms−1 and 11 ms−1ms−1, respectively. They collide and after the collision, the first object moves at a velocity of 1.671.67 ms−1ms−1. Determine the velocity of the second object. Best NCERT Solutions for Class 9 Science Chapter 8 – Force and Laws of Motion
Given:
Mass of one of the objects, m1=100m1=100 g =0.1=0.1 kg
Mass of the other object, m2=200m2=200g =0.2=0.2 kg
Velocity of m1 before collision, v1=2v1=2 m/s
Velocity of m2 before collision, v2=1v2=1 m/s
Velocity of m1 after collision, v3=1.67v3=1.67 m/s
Answer: Velocity of m2 after collision =v4=v4
According to the law of conservation of momentum:
Total momentum before collision == Total momentum after collision
m1v1+m2v2=m3v3+m4v4m1v1+m2v2=m3v3+m4v4
(0.1)2+(0.2)1=(0.1)1.67+(0.2)v4(0.1)2+(0.2)1=(0.1)1.67+(0.2)v4
0.2+0.2=0.167+0.2v40.2+0.2=0.167+0.2v4
0.4=0.167+0.2v40.4=0.167+0.2v4
0.4−0.167=0.2v40.4−0.167=0.2v4
0.233=0.2v40.233=0.2v4
v4=0.2330.2v4=0.2330.2
v4=1.165v4=1.165 m/s
Hence, the velocity of the second object becomes 1.1651.165 m/s after the collision.
NCERT 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.
Answer: yes. Even in the presence of a net zero external unbalanced force, an item can nonetheless move at a non-zero velocity. This is only feasible if the object travels in a predetermined direction at a constant speed. The body is therefore not exposed to any net unbalanced forces. The object will keep moving at a speed higher than zero. To alter the item’s state of motion, an external unbalanced force that is net non-zero must be applied. Indeed.
Even in the presence of a net zero external unbalanced force, an item can nonetheless move at a non-zero velocity. This is only feasible if the object travels in a predetermined direction at a constant speed. The body is therefore not exposed to any net unbalanced forces. To alter the item’s state of motion, an external unbalanced force that is net non-zero must be applied.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer: When a carpet is beaten with a stick, the dust particles that are caught in its pores want to remain still because inertia prevents an object from changing from its resting or moving state. This causes the carpet to move more swiftly. Newton’s first law of motion states that the dust particles stay at rest as the carpet moves. Dust particles consequently come out of the carpet.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope? Best NCERT Solutions for Class 9 Science Chapter 8 – Force and Laws of Motion
Answer: First Law of Motion states that when a bus is moving, luggage on its roof will usually continue to move in the direction of motion and remain at rest. Luggage on the roof may fall to preserve the resting place when the bus resumes its journey after a stop. Similar to this, when a driving bus comes to a stop, its luggage on the roof will fall forward due to inertia of motion. Any luggage stored on a bus’s roof ought to be secured with a rope to prevent this.
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.
Answer: Option(c)-
Friction between the ball and the ground causes the ball to move against its own momentum until it eventually comes to a stop.
5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400400 m in 2020 s. Find its acceleration. Find the force acting on it if its mass is 77 metric tonnes (Hint: 11 metric tonne =1000=1000 kg).
Given:
Initial velocity of the truck , u=0u=0 (since the truck is initially at rest)
Distance travelled, s =400=400 m
Time taken, t =20=20 s
Answer: According to the second equation of motion:
s=ut+12at2s=ut+12at2
400=0+12a(20)2400=0+12a(20)2
400=12a(400)400=12a(400)
400=a(200)400=a(200)
m/s2
metric tonne
metric tonnes
Mass of truck,
From Newton’s second law of motion:
Force, F = Mass × Acceleration
F = ma
F=
Hence, the acceleration of the truck is22
m/s2 and the force acting on the truck F
6. A stone of11
kg is thrown with a velocity of
across the frozen surface of a lake and comes to rest after travelling a distance of
m. What is the force of friction between the stone and the ice?
Given:
Initial velocity of the stone, u
Final velocity of the stone, v
(finally the stone comes to rest)
Distance covered by the stone, s
Answer: According to the third equation of motion:
v2=u2+2asv2=u2+2as
02=(20)2+2×a×5002=(20)2+2×a×50
0=400+100a0=400+100a
100a=−400100a=−400
a=−400100a=−400100
a=−4a=−4
a = −4 ms2ms2
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, m =1=1 kg
From Newton’s second law of motion:
Force, F = Mass × Acceleration
F = ma
F =1×−4=1×−4
F =−4=−4 N
Hence, the force of friction between the stone and the ice F =−4=−4 N .
7. A 80008000 kg engine pulls a train of 55 wagons, each of 20002000 kg, along a horizontal track. If the engine exerts a force of 4000040000 N and the track offers a friction force of 50005000 N, then calculate:
(a) the net accelerating force;
Given:
Force exerted by the engine, F =40000=40000 N
Frictional force offered by the track, Ffraction=5000Ffraction=5000 N
Answer : Net accelerating force,
Fnet=F−FfrictionFnet=F−Ffriction
Hence, the net accelerating force
(b) the acceleration of the train; and
Given:
The engine exerts a force of
N on all the five wagons.
Net accelerating force on the wagons,
Mass of a wagon
Number of wagons
Formula:
Total Mass of the wagons,
m = Mass of a wagon × Number of wagons
Answer:Total Mass of the wagons,m
Mass of the engine, m′
Total mass, M = m + m′
From Newton’s second law of motion:
Hence, the acceleration of the wagons and the train
(c) The force of wagon 1 on wagon 2.
Answer: The force of wagon 1 on wagon 2 = mass of four wagons x acceleration
Mass of 4 wagons
F
8. An automobile vehicle has a mass of
kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of
Given:
Mass of the automobile vehicle, m
Final velocity,
Acceleration of the automobile, a
Answer: From Newton’s second law of motion,
Force = Mass × Acceleration
N
Hence, the force between the automobile and the road
N.
Negative sign shows that the force is acting in the opposite direction of the vehicle.
9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2
(b) mv2
(c)1/2 mv2
(d) mv
Answer:
(d) mv Mass of the object
Velocity
Momentum = Mass × Velocity
Momentum
10. Using a horizontal force of 200200
N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer: The third law of motion by Newton states that an equivalent amount of force will act in the opposite direction.
This force is known as friction. Consequently, the cabinet faces a 200N force of friction.
11. 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, 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.
Answer : The truck’s enormous size creates a significant static friction force. It is impossible for the student to move the truck since their effort is not strong enough to overcome the static friction. Because there is no net uneven force acting in either direction in this situation, there is no movement. The force generated by static friction and the force applied by the learner cancel each other out.
Therefore, the student’s assertion that the two equal but opposing forces cancel each other out is accurate.
12. A hockey ball of mass
g travelling at
is struck by a hockey stick so as to return it along its original path with a velocity at 5
. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer: Mass of the hockey ball, m
velocity of the ball ,
Initial momentum
velocity of the ball after struck by the stick,
Final momentum
Change in momentum
Hence, the change in momentum of the hockey ball
13. A bullet of mass
g travelling horizontally with a velocity of
strikes a stationary wooden block and comes to rest in
s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Given:
Initial velocity of the bullet , u
Final velocity,
Time,
Answer: According to the first equation of motion,
Acceleration of the bullet, a
m/s2
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
Hence, the distance of penetration of the bullet into the block
From Newton’s second law of motion:
Force, F = Mass × Acceleration
Mass of the bullet, m
Acceleration of the bullet, a
F = ma
Hence, the magnitude of force exerted by the wooden block on the bullet
but it acts in opposite direction.
14. An object of mass
kg travelling in a straight line with a velocity of
collides with, and sticks to, a stationary wooden block of mass
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.
Given:
Mass of the object,
kg
Velocity of the object before collision,
m/s
Mass of the wooden block,
kg
Velocity of the wooden block before collision,
m/s
Answer: Best NCERT Solutions for Class 9 Science Chapter 8 – Force and Laws of Motion
∴ Total momentum before collision
=m1v1+m2v2=m1v1+m2v2
It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system,
Velocity of the combined object
According to the law of conservation of momentum:
Total momentum before collision
Total momentum after collision
⇒m1v1+m2v2⇒m1v1+m2v2 =(m1+m2)v=(m1+m2)v
⇒1(10)+5(0)=(1+5)v⇒1(10)+5(0)=(1+5)v
⇒10=6v⇒10=6v
⇒v=106⇒v=106
⇒v=53⇒v=53 m/s
v=1.66v=1.66 m/s
Total momentum after collision
The total momentum after collision is also
Total momentum just before the impact
Total momentum just after the impact
Hence, velocity of the combined object after collision
15. An object of mass
kg is accelerated uniformly from a velocity of
to
in
s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Given:
Initial velocity of the object, u
Final velocity of the object, v
Mass of the object, m
Time taken by the object to accelerate, t
Answer:
Initial momentum
Final momentum
Force exerted on the object, F
mv-mu/t F
Initial momentum of the object is
.
Final momentum of the object is
.
Force exerted on the object is
16. Akhtar, Kiran and Rahul were riding in a motor car 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. 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 a change in their momentum. Comment on these suggestions.
Answer: As a result, the vehicle and insect systems have no change in momentum.
The insect’s velocity changes more in this instance, leading to a larger shift in momentum. In this sense, Kiran’s appraisal is accurate.
The motorcar is larger and moves more quickly than the bug.
In addition, the fact that the car keeps going in the same direction after the impact shows that the bug has changed its momentum more than the motorcar has. Thus, Akhtar is also correct in what he says.
Rahul’s observation is also accurate because the momentum gained by the bug is equivalent to the momentum lost by the motorcar. The conservation of momentum law concurs with this as well. But since there’s a bug in the system, he made a mistake. There is no change in momentum after the accident since the momentum prior to the collision is the same as the momentum following the hit.
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17. How much momentum will a dumbbell of mass
kg transfer to the floor if it falls from a height of
cm? Take its downward acceleration to be
.
Given:
Mass of the dumbbell, m
Distance covered by the dumbbell, s
Acceleration in the downward direction, a
Initial velocity of the dumbbell, u=0=0
Best NCERT Solutions for Class 9 Science Chapter 9 – Force and Laws of Motion
Answer:
Final velocity of the dumbbell v = ?
According to the third equation of motion:
Hence, the momentum with which the dumbbell hits the floor is
Best NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion