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0 (same as initial momentum)
Therefore, m1v1 + m2v2 = 0

The negative sign indicates that recoil velocity is opposite to the bullet’s motion
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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
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67 ms–1
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Solution

NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
Assuming that the first object is object A and the second one is object B, it is given that:
Mass of A (m1) = 100g
Mass of B (m2) = 200g
Initial velocity of A (u1) = 2 m/s
Initial velocity of B (u2) = 1 m/s
Final velocity of A (v1) = 1
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m
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v2 = 1
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165 meters per second
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An object experiences a net zero external unbalanced force
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If no, provide a reason
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An object moving in some direction with constant velocity will continue in its state of
motion as long as there are no external unbalanced forces acting on it
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2
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Explain
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The inertia of the dust particles residing on the carpet resists the change in the motion of the carpet
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This is why the dust comes out of the carpet when beaten
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Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Solution
When some luggage is placed on the roof of a bus which is initially at rest, the acceleration of the bus in
the forward direction will exert a force (in the backward direction) on the luggage
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Depending on the mass of the luggage and the magnitude of the force, the luggage may fall off the bus
due to inertia
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4
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After covering a short
distance, the ball comes to rest
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(b) velocity is proportional to the force exerted on the ball
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(d) there is no unbalanced force on the ball, so the ball would
want to come to rest
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This frictional force eventually stops the ball
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If the surface of the level ground is lubricated (with oil or some other lubricant), the friction that arises
between the ball and the ground will reduce, which will enable the ball to roll for a longer distance
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A truck starts from rest and rolls down a hill with a constant acceleration
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Find its acceleration
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)
Solution
Given, distance covered by the truck (s) = 400 meters

NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
Time taken to cover the distance (t) = 20 seconds
The initial velocity of the truck (u) = 0 (since it starts from a state of rest)

6
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What is the force of friction between the stone
and the ice?
Solution
Given, Mass of the stone (m) = 1kg
Initial velocity (u) = 20m/s
Terminal velocity (v) = 0 m/s (the stone reaches a position of rest)
Distance travelled by the stone (s) = 50 m
As per the third equation of motion
v² = u² + 2as
Substituting the values in the above equation we get,
0² = (20)² + 2(a)(50)
-400 = 100a
a = -400/100 = -4m/s² (retardation)
We know that
F = m×a
Substituting above obtained value of a = -4 in F = m x a
We get,
F = 1 × (-4) = -4N
Here the negative sign indicates the opposing force which is Friction
7
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If the
engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force and (b) the acceleration of the train

NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
Solution
(a) Given, the force exerted by the train (F) = 40,000 N
Force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction)
Therefore, the net accelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N
(b) Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg
The total mass of the train is 18000 kg
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94 ms-2
The acceleration of the train is 1
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s-2
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An automobile vehicle has a mass of 1500 kg
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7 ms-2?
Solution
Given, mass of the vehicle (m) = 1500 kg
Acceleration (a) = -1
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7 ms-2) = -2550 N
Hence, the force between the automobile and the road is -2550 N, in the opposite direction of the
automobile’s motion
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What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv
Solution
The momentum of an object is defined as the product of its mass m and velocity v
Momentum = mass x velocity
Hence, the correct answer is mv i
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Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a
constant velocity
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Therefore, the effective force
acting on it is also zero
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Therefore, the total friction force is -200 N
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Two objects, each of mass 1
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The velocity of each object is 2
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What will be the velocity of the combined object after collision?

NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
Solution
Given
Mass of first object, m1 = 1
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5 kg
Velocity of first object before collision, v1 = 2
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5 m/s
We know that,
Total momentum before collision = Total momentum after collision
m1v1 + m2v2 = (m1 + m2)v
1
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5) + 1
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5) = (1
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5)v
3
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75 = 3v
v=0
Therefore, the velocity of the combined object after the collision is 0 m/s
12
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If the object is a massive truck parked along the roadside,
it will probably not move
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Comment on this logic and explain why the truck does not move
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When
pushing the truck with a small force, the frictional force cancels out the applied force and the truck does
not move
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Therefore, the student’s logic
is correct
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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
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Solution
Given, mass of the ball (m) = 200g
Initial velocity of the ball (u) = 10 m/s
Final velocity of the ball (v) = – 5m/s
Initial momentum of the ball = mu = 200g × 10 ms-1 = 2000 g
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s-1
Final momentum of the ball = mv = 200g × –5 ms-1 = –1000 g
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s-1
Therefore, the change in momentum (mv – mu) = –1000 g
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s-1 – 2000 g
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s-1 = –3000 g
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s-1
This implies that the momentum of the ball reduces by 1000 g
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s-1 after being struck by the hockey
stick
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A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikes a stationary
wooden block and comes to rest in 0
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Calculate the distance of penetration of the bullet
into the block
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Solution
Given, mass of the bullet (m) = 10g (or 0
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03 s
To find the distance of penetration, the acceleration of the bullet must be calculated
Let the distance of penetration be s
As per the first law of motion
v = u + at
0 = 150 + a (0
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25 m
As per the second law of motion, F = ma
F = 0
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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
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Calculate the total momentum just before the impact and just after the
impact
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Solution
Given, mass of the object (m1) = 1kg
Mass of the block (m2) = 5kg
Initial velocity of the object (u1) = 10 m/s
Initial velocity of the block (u2) = 0
Mass of the resulting object = m1 + m2 = 6kg
Velocity of the resulting object (v) =?
Total momentum before the collision = m1u1 + m2u2 = (1kg) × (10m/s) + 0 = 10 kg
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s-1

NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
As per the law of conservation of momentum, the total momentum before the collision is equal to the
total momentum post the collision
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m
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m
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66 meters per second
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An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1 to 8 ms–1 in 6 s
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Also, find the magnitude of the force
exerted on the object
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m
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m
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5 ms-2
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5 ms-2) = 50 N
Therefore, a force of 50 N is applied on the 100kg object, which accelerates it by 0
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17
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Akhtar and
Kiran started pondering over the situation
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Akhtar said that
since the motorcar was moving with a larger velocity, it exerted a larger force on the insect
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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
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Solution
As per the law of conservation of momentum, the total momentum before the collision between the
insect and the car is equal to the total momentum after the collision
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Akhtar’s assumption is partially right
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NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
Kiran’s statement is false
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The velocity of insect changes accordingly due to its mass as it is very
small compared to the motorcar
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Rahul’s statement is completely right
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However, Rahul’s suggestion
that the change in the momentum is the same contradicts the law of conservation of momentum
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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
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8m
Initial velocity (u) = 0 (it is dropped from a position of rest)
Acceleration (a) = 10ms-2
Terminal velocity (v) =?
Momentum of the dumb-bell when it hits the ground = mv
As per the third law of motion
v2 – u2 = 2as
Therefore, v2 – 0 = 2 (10 ms-2) (0
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m
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The following is the distance-time table of an object in motion:

Time (seconds)

Distance (meters)

0

0

1

1

2

8

3

27

4

64

5

125

6

216

7

343

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing,
or zero? (b) What do you infer about the forces acting on the object?
Solution
(a) The distance covered by the object at any time interval is greater than any of the distances covered
in previous time intervals
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(b) As per the second law of motion, force = mass × acceleration
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Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level
road
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2 ms 2

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2 ms-2
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2 ms-2 = 240N
The force applied by the third person on the car is 240 N
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NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
3
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The nail stops the hammer in a
very short time of 0
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What is the force of the nail on the hammer?
Solution
Given, mass of the hammer (m) = 500g = 0
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Time period (t) = 0
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5kg) * (-5000 ms-2) = -2500 N
As per the third law of motion, the nail exerts an equal and opposite force on the hammer
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4
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Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force
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Also calculate the magnitude of the force required
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Initial momentum of the car = m × u = (1200kg) × (25m/s) = 30,000 kg
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s-1
Final momentum of the car = m × v = (1200kg) × (5m/s) = 6,000 kg
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s-1
Therefore, change in momentum (final momentum – initial momentum) = (6,000 – 30,000) kg
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s-1
= -24,000 kg
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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



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