Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion

Intext Questions – 1

Page: 118

1
...
The following objects hold greater inertia because of their mass
...
Stone
2
...
Five-Rupee coin
2
...
The goalkeeper of the opposite team collects the football and kicks it towards a player
of his own team”
...

Solution
The velocity of football changes four times
...
Third when the goalkeeper stops the football
...

Agent supplying the force:
a) The First case is the First player
b) The Second case is the Second player
c) The Third case is Goalkeeper
d) The Fourth case is Goalkeeper
3
...

Solution
When the branch of the tree is shaken, the branch moves in a to-and-fro motion
...
Therefore, the leaves that are
weakly attached to the branch fall off due to inertia whereas the leaves that are firmly attached to the
branch remain attached
...
Why do you fall in the forward direction when a moving bus brakes to a stop and fall
backwards when it accelerates from rest?
Solution
Initially, when the bus accelerates in a forward direction from a state of rest, the passengers experience
a force exerted on them in the backward direction due to their inertia opposing the forward motion
...
When the
brakes are applied, the bus moves towards a position of rest
...
This
causes the passengers to fall forwards when the brakes are applied
...
If action is always equal to the reaction, explain how a horse can pull a cart
...
An equal force in the opposite direction (forward direction) is applied on the horse by the
Earth
...
As a result, the cart moves forward
...
Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at
a high velocity
...
This is because of
Newton’s third law of motion
...

Hence, it is difficult for him to remain stable while holding the hose
...
From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s–1
...

Solution
Given, the Bullet’s mass (m1) = 50 g
The rifle’s mass (m2) = 4kg = 4000g
Initial velocity of the fired bullet (v1) = 35 m/s
Let the recoil velocity be v2
...

4
...
They collide and after the collision, the first object
moves at a velocity of 1
...
Determine the velocity of the second object
...
67 m/s
Final velocity of B (v2) =?
Total initial momentum = Initial momentum of A + initial momentum of B
= m1u1 + m2u2
= (100g) × (2m/s) + (200g) × (1m/s) = 400 g
...
sec-1
As per the law of conservation of momentum, the total momentum before collision must be equal to the
total momentum post collision
...
165 m/s
Therefore, the velocity of object B after the collision is 1
...


NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion

Exercises

Page: 128,129

1
...
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
...

Solution
Yes, it is possible
...
In order to change the motion of
the object, some external unbalanced force must act upon it
...
When a carpet is beaten with a stick, dust comes out of it
...

Solution
When the carpet is beaten with a stick, the stick exerts a force on the carpet which sets it in motion
...

Therefore, the forward motion of the carpet exerts a backward force on the dust particles, setting them
in motion in the opposite direction
...

3
...
In a similar manner,
when a bus which is initially in a state of motion suddenly comes to rest due to the application of
brakes, a force (in the forward direction) is exerted on the luggage
...
Tying up the luggage will secure its position and prevent it from falling off the bus
...
A batsman hits a cricket ball which then rolls on a level ground
...
The ball slows to a stop because (a) the batsman did not hit the
ball hard enough
...
(c) there is a force
on the ball opposing the motion
...

Solution
When the ball rolls on the flat surface of the ground, its motion is opposed by the force of friction (the
friction arises between the ground and the ball)
...

Therefore, the correct answer is (c)
...

5
...
It travels a distance
of 400 m in 20 s
...
Find the force acting on it if it’s mass is 7 tonnes (Hint: 1
tonne = 1000 kg
...
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
...
An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track
...

As per the second law of motion, F = ma (or: a = F/m)
Therefore, acceleration of the train = (net accelerating force) / (total mass of the train)
= 35,000/18,000 = 1
...
94 m
...

8
...
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
As per the second law of motion, F = ma
F = 1500kg × (-1
...

9
...
e option (d)
10
...
What is the friction force that will be exerted on the cabinet?
Solution
Since the velocity of the cabinet is constant, its acceleration must be zero
...
This implies that the magnitude of opposing frictional force is equal to the force
exerted on the cabinet, which is 200 N
...

11
...
5 kg, are moving in the same straight line but in opposite
directions
...
5 ms-1 before the collision during which they stick
together
...
5 kg
Mass of second object, m2 = 1
...
5 m/s
The velocity of the second object which is moving in the opposite direction, v 2 = -2
...
5(2
...
5 (-2
...
5 + 1
...
75 – 3
...
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
...
A student justifies this by answering that the two opposite and equal
forces cancel each other
...

Solution
Since the truck has a very high mass, the static friction between the road and the truck is high
...
This implies that the two forces are equal in magnitude but opposite in direction (since the
person pushing the truck is not displaced when the truck doesn’t move)
...

13
...
Calculate the magnitude of change of
momentum occurred in the motion of the hockey ball by the force applied by the hockey stick
...
m
...
m
...
m
...
m
...
m
...
m
...


NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
14
...
03 s
...
Also calculate the magnitude of the force exerted by the wooden block on the
bullet
...
01 kg)
Initial velocity of the bullet (u) = 150 m/s
Terminal velocity of the bullet (v) = 0 m/s
Time period (t) = 0
...
03)
a = -5000 ms-2
v2 = u2 + 2as
0 = 1502 + 2 x (-5000)s
s = 2
...
01kg × (-5000 ms-2)
F = -50 N
15
...
Then they both move off together in the
same straight line
...
Also, calculate the velocity of the combined object
...
m
...
Therefore, the total momentum post the collision is also 10 kg
...
s -1
Now, (m1 + m2) × v = 10kg
...
s-1

`
The resulting object moves with a velocity of 1
...

16
...

Calculate the initial and final momentum of the object
...

Solution
Given, mass of the object (m) = 100kg
Initial velocity (u) = 5 m/s
Terminal velocity (v) = 8 m/s
Time period (t) = 6s
Now, initial momentum (m × u) = 100kg × 5m/s = 500 kg
...
s-1
Final momentum (m × v) = 100kg × 8m/s = 800 kg
...
s-1

Therefore, the object accelerates at 0
...
This implies that the force acting on the object (F = ma) is
equal to:
F = (100kg) × (0
...


NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
3
...
The nail stops the hammer in a
very short time of 0
...
What is the force of the nail on the hammer?
Solution
Given, mass of the hammer (m) = 500g = 0
...

Time period (t) = 0
...
5kg) * (-5000 ms-2) = -2500 N
As per the third law of motion, the nail exerts an equal and opposite force on the hammer
...

4
...

Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force
...
Also calculate the magnitude of the force required
...

Initial momentum of the car = m × u = (1200kg) × (25m/s) = 30,000 kg
...
s-1
Final momentum of the car = m × v = (1200kg) × (5m/s) = 6,000 kg
...
s-1
Therefore, change in momentum (final momentum – initial momentum) = (6,000 – 30,000) kg
...
s-1
= -24,000 kg
...
s-1
External force applied = mass of car × acceleration = (1200kg) × (-5 ms-2) = -6000N
Therefore, the magnitude of force required to slow down the vehicle to 18 km/hour is 6000 N



---