We suggest reading Force - Class 09 Notes,
before having a look at its exercises.
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Group - A
1. What is force? Write down its SI unit.
Answer: Force is the push or pull acting on an object that changes or
ties to change the position of an object.
Its SI unit is Newton (N).
2. Define 1N force. Write down CGS unit of Force.
Answer: 1N force can be defined as the 1 m/s² acceleration produced on
a body of 1kg mass.
The CGS unit of Force is Dyne.
3. What is inertia?
Answer: The inability of a body at a state of rest or motion, to change
its state by itself is called inertia.
4. What is inertia of rest? Write with an example.
Answer: Inertia of rest is the inability of a body at rest to change
its state of rest by itself.
For example: the dust on a carpet fall when the carpet is beaten with a
stick.
5. Define inertia of motion with any one example.
Answer: Inertia of motion is the inability of a body at motion to
change its state of motion by itself.
For example: passengers fall forwards when a moving bus is stopped.
6. What factors affects inertia of a body?
Answer: The mass 'm' of the body is the factor that affects inertia of
the body.
7. What is the relation between mass and inertia of a body?
Answer: Mass and Inertia of a body are direct proportional to each
other.
i.e. mass of a body ∝ inertia of the body
8. What is speed? Write down its formula.
Answer: Speed is the distance travelled by a body per unit time.
Formula of speed = distance travelled(s) / time taken(t)
9. What is a scalar quantity?
Answer: Physical quantity that has magnitude but no fixed direction is
called a scalar quantity.
10. Define vector quantity with one example.
Answer: Physical quantity that has both magnitude and fixed direction
is called vector quantity.
One example of vector quantity is Velocity.
11. What is displacement? Write its SI unit.
Answer: The distance travelled in a fixed direction is called
displacement.
Its SI unit is metre (m).
12. What is average velocity? How is it calculated?
Answer: Average velocity is the mathematical mean of initial velocity
and final velocity of a body.
It is calculated using the formula:
Average velocity = (u+v)/2
Adver tisement
13. What is acceleration? Write down its SI unit.
Answer: Acceleration is the rate of change win velocity of a moving body.
Its SI unit is m/s².
14. What is retardation? In which condition is it possible?
Answer: The decrease in the of velocity of a moving body per unit time
is called retardation.
It is possible when the velocity of a moving body is decreasing.
15. Define equations of motion.
Answer: Three mathematical equations that are true for every cases of a
body moving with uniform acceleration is defined as equations of motion.
16. Define retardation. In which unit is it measured?
Answer: Retardation can be defined as the decrease in the rate of
velocity of a moving body.
It is measured in m/s².
17. State Newton's second law of motion.
Answer: Newton's second law of motion states that, "Acceleration
produced on a body is directly proportional to the force applied on it and
inversely proportional to its mass."
18. State Newton's third law of motion.
Answer: Newton's third law of motion states that, "To every action
there is an equal but opposite reaction."
19. "Every action has equal but opposite reaction." Which law of Newton
is stated by this statement?
Answer: "Every action has equal but opposite reaction." This statement
states Newton's third law of motion.
20. What is velocity-time graph?
Answer: The geometrical relationship between the velocity of a body and
the time taken is called velocity-time graph.
21. Which factor does the inertia of rest depend on?
Answer: The Inertia of rest depends on the mass of the body.
22. What is the relationship between Newton's first law of motion and
inertia?
Answer: The relationship between Newton's first law of motion and
inertia is that both of them state the same law.
23. What is the relationship among initial velocity, distance covered,
acceleration produced and final velocity of a moving body?
Answer: The relationship between initial velocity, distance covered,
acceleration produced and final velocity of a moving body is:
v² = u² + 2as
Where, 'v' is the final velocity, 'u' is the initial velocity, 'a' is the
acceleration and 's' is the distance covered.
24. Give one example which describes Newton's first law of motion.
Answer: One example of Newton's first law of motion is, a moving
football doesn't stop until an external force (friction, human being, etc)
stops it.
25. Give one example which describes Newton's third law of motion.
Answer: One example that describes Newton's third law of motion is:
When we fire a bullet from a gun, the bullet goes forward with high velocity.
And, the gun gives a backward jerk (recoil) to one's shoulder, which is the
opposite reaction to the bullet fired.
26. What is balanced force? Give one example.
Answer: If a force acting on a body doesn't change the state of the
body, it is called balanced force.
One example of balanced force is tug of war.
27. What is unbalanced force? Give one example.
Answer: If a force acting on a body changes the state of the body, it
is called unbalanced force.
One example of unbalanced force is kicking a football.
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Group - B
1. Write any two differences between vectors and scalars.
Answer: Two differences between vectors and scalar are:
| Vectors | Scalars |
|---|---|
| Vectors have both fixed direction and magnitude. | Scalars have magnitude only. |
| Example are: displacement, Velocity, etc. | Examples are: speed, distance, etc. |
2. Speed is called a scalar quantity but velocity is called a vector
quantity, why?
Answer: Speed is called a scalar quantity because it has magnitude only
but no fixed direction. Velocity is called a vector quantity because it has
both magnitude and fixed direction.
3. Write any two differences between distance and displacement.
Answer: Two differences between distance and displacement are:
| Distance | Displacement |
|---|---|
| Distance is the total length between any two points. | Displacements is the shortest distance travelled by a body in a straight line. |
| Distance is a scalar quantity. | Displacement is a vector quantity. |
4. Write any two differences between acceleration and retardation.
Answer: Two differences between acceleration and retardation are:
| Acceleration | Retardation |
|---|---|
| Acceleration is the rate of change in velocity of a moving body. | Retardation is the negative rate of change of velocity of a moving body. |
| Due to acceleration, the velocity of a body increases. | Due to retardation, the velocity of a body decreases. |
5. Blades of a running fan continue to spin for some time even after the
electricity is switched off. Why?
Answer: The blades of a running fan are in a state of motion. When the
electricity is switched off the blades continue to spin for sometime due to
inertia of motion.
6. When a carpet is beaten with a stick, the dust particles fall off.
Give reason.
Answer: The dust particles stuck in the carpet and the carpet, both are
in a state of rest. When the carpet is beaten, the carpet comes in motion but
the stick remains in state of rest, because of inertia of rest. Due to this,
the dust particles fall of when the carpet is beaten with a stick.
7. It is dangerous to jump out of a moving vehicle, why?
Answer: When we are travelling in a moving vehicle, we are in a state
of motion. When we jump out of the vehicle and touch the ground, our lower
body part comes in rest but the upper part continues to stay in motion due to
inertia of motion. This can lead us to heavy injuries. So, it is dangerous to
jump out off a moving vehicle.
8. The passengers standing on a stationary bus falls backward when the
bus suddenly moves forward. Give reason.
Answer: The passengers standing on a stationary bus are in a state of
rest. When the bus suddenly moves, their lower body part comes in motion but
the upper body part remains in a state of rest due to inertia of rest.
Therefore, they fall backward in that condition.
9. An athlete runs some distance before taking a long jump, why?
Answer: In a long jump, athletes want to land as far as they can. When
athletes run some distance before taking along jump, their body comes in a
state of motion that continues until their feet touches the ground once they
jump. As a result of inertia of motion, they end up covering more distance.
10. A gun recoils while firing a bullet. Give reason.
Answer: When a gun is fired, the bullet comes out of the gun with high
velocity pushing the gun backward. The backward push, also known as recoil is
exerted on the shoulder of the gunman. This whole condition follows Newton's
third law of motion so, a gun recoils when firing a bullet.
11. A batsman often ducks to a bouncer, why?
Answer: The bouncer balls often have higher velocity. Because of higher
velocity, the momentum of the ball increases. So, to avoid any injuries by
hitting the ball, a batsman often ducks to a bouncer.
12. When we jump out of a boat, the boat moves slightly backward. Give
reason.
Answer: Newton's third law of motion states that, "To every action
there is an equal but opposite reaction." So, when we move out of the boat,
our feet exert pressure on the boat (action) and the boat exert pressure on
our feet (reaction). Because we move forward, the boat moves slightly
backward.
13. We can catch a rolling ball but not a flying bullet. Why?
Answer: The mass of a ball might be greater than that of a flying
bullet. But, the velocity of a rolling ball is very less than that of a flying
bullet. Due to momentum (mass*velocity), we can catch a rolling ball but not a
flying bullet. As the rolling ball has less momentum whereas the flying bullet
has larger momentum.
14. A cricket player lowers his hand while catching the ball, why?
Answer: The ball coming down from a height has higher velocity. Due to
this, the momentum of the ball is huge. By lowering the hand, players allow
the ball to take time before landing and decrease it's velocity that directly
reduces its momentum. Therefore, the cricket player lowers his hand while
catching the ball.
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Group - C
1. What is distance-time graph? Draw a distance-time graph to represent a
body moving with uniform speed.
Answer: The graphical representation of the distance travelled by a
body and the time taken is called distance-time graph.
2. What is velocity-time graph? Draw a velocity-time graph to represent a
body with uniform acceleration.
Answer: The graphical representation of the velocity of a body and the time taken is called velocity-time graph.
3. Draw the graphical plot to describe the following motion.
i) A body at rest
ii) A body moving with non-uniform acceleration
Answer:
Check the graph yourself!
4. Prove that: v= u+at
Solution:
Let us consider a body having initial velocity 'u' is moving with the
acceleration 'a' and after certain time 't' reaches the final velocity
'v'. Then,
Acceleration (a) = (v-u)/t
or, at = v-u
or, v = at + u
or, v = u + at #proved
5. Prove that: F = ma
Solution:
From Newton's Second Law of Motion:
a ∝ F ..... (1)and,
a ∝ 1/m .... (2)
Combining (i) and (ii), we get,
a ∝ F/m
or, F∝ m/a ..... (3)
or, F = kma [here 'k' is a constant of proportionality]
Let, F= 1N, m =1 kg and a= 1m/s^2 then
,
k = 1
Now, put the value of k in equation (3),
F= 1.ma
So, F = ma #proved
6. State Newton's third law of motion with an example. It is also
called the law of action and reaction. Why?
Answer: Newton's third law of motion states, "To every action
there is an equal but opposite reaction." Example of Newton's third law of
motion is when a gun is fired, it recoils.
It is also called the law of action and reaction because in this law, two
forces (exerted by two bodies) act on each other. that are action and reaction
forces.
7. Prove that: v2 = u2as
Solution:
Let us consider a body having initial velocity 'u' is moving with the
acceleration 'a' and after certain time 't' reaches the final velocity
'v'. Then,
Average velocity = (v + u) /2
We know, Distance traveled (s)= Average velocity (v av) x Time taken (t)
or, s = v av x t
or, s = (v + u) /2 x t
or, s = (v + u)/ 2 x (v-u) / a [ i.e. t = (v-u) / a]
or, s = v^2 - u^2 / 2a
or, 2as = v^2 - u^2
so, v^2 = u^2 + 2as #proved
8. If a body starts from rest and attains a velocity of 20 m/s in 8
seconds, find the acceleration produced on the body.
Solution:
Given,
initial velocity of the body (u) = 0 m/s
final velocity of the body (v) = 20 m/s
time taken (t) = 8 s
Now,
acceleration produced on the body (a) = (v-u) /t
= (20-0)/8
= 2.5 m/s2
9. A vehicle is running at a speed of 45 km/h. If it is stopped in 3
seconds by applying brakes, calculate the retardation of the vehicle and
the distance travelled before it stops.
Solution:
initial velocity of the body (u) = 45km/h
= 45*1000m/3600s
= 12.5 m/s
final velocity (v) = 0m/s
time taken (t) = 3 s
Now,
retardation of the vehicle (-a) = (u-v) /t
= (12.5-0)/3
= 4.16 m/s2
And,
distance travelled by the vehilce (s) = (u+v) /2 * t
= (12.5 + 0)/2 * 3
= 18.75 m
Hence, the retardation of the vehicle before it stops is
4.16m/s2 and distance travelled is 18.75m.
10. The velocity of a moving body increases from 10m/s to 15 m/s in 5
seconds. Calculate its acceleration.
Solution:
Given,
initial velocity of a body (u) = 10m/s
final velocity of a body (v) = 15 m/s
time taken (t) = 5 s
Now,
acceleration of the body (a) = (v-u) /t
= (15-10)/5
= 1m/s2
Hence, the acceleration of the body is 1m/s2.
11. A body moving along a straight path at a velocity of 20m/s
attains an acceleration of 4m/s2. Calculate its velocity after
2 seconds.
Solution:
initial velocity of a body (u) = 20m/s
acceleration of a body (a) = 4m/s2
time taken (t) = 2s
Now,
a = (v-u) /t
or, 4 = (v-20) /2
or, 8 = v-20
So, v = 28 m/s
Hence, the velocity of the body after 2 seconds is 28 m/s.
12. A body of mass 10kg is pulled by a force of 8N. Calculate the
acceleration and the final velocity of the body after 5
seconds.
Solution:
Given,
mass of the body (m) = 10 kg
Force (F) = 8 N
initial velocity of the body (u) = 0 m/s
time taken (t) = 5s
Now,
acceleration (a) = F/m
= 8/10
= 0.8 m/s2
And,
final velocity of the body (v) = a*t + u
= 0.8 * 5 + 0
= 4 m/s
Hence, the acceleration of the body is 0.8m/s2 the final
velocity of the body after 5 seconds is 4 m/s.
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Group - D
1. A car was moving at a speed of 90 km/h. The driver jammed on the
brakes when he saw a baby on the road 20m ahead and the car came to rest
at a distance of 15 m. What is the retardation and how long does it take
to come at rest?
Solution:
Given,
Initial velocity of a car (u) = 90km/h
= 90*1000m/3600s
= 25 m/s
Final velocity (u) = 0 m/s
Distance (s) = 20m
Distance covered (s1) = 15m
Now,
Let retardation be '-a'
v² = u² + 2as
or, 0² = 25² + 2a*15
or, -625 = 30a
or, a = -20.83
So, -a = 20.83 m/s²
And,
Let time taken be 't'
t = (v-u)/a
= (0-25) / (-20.83)
= 25/20.83
= 1.2 s
Hence, the retardation of the car is 20.83 m/s² and the time taken to come at
rest is 1.2 seconds.
2. A body starts moving from rest and attains the acceleration of 0.5
m/s². Calculate the velocity at the end of 3 minutes and also find the
distance travelled by it during that time.
Solution:
Given,
initial velocity of the body (u) = 0m/s
acceleration of the body (a) = 0.5 m/s²
time taken (t) = 3 minutes = 3*60 s = 180s
Now,
final velocity (v) = at + u
= 0.5*180 + 0
= 90 m/s
And,
Distance travelled (s) = ut + 1/2 at2
= 0*180 +1/2 *0.5*180²
= 8100 m
Hence, the velocity of the body at the end of 3 minutes is 90m/a and the
distance travelled by it during that time is 8100m.
3. A car weighing 1600kg is accelerated to 30m/s from start in 20s.
Calculate the acceleration of the car and the magnitude of force
applied.
Solution:
Given,
mass of the car (m) = 1600kg
initial velocity (u) = 0m/s
final velocity (v) = 30m/s
time taken (t) = 20s
Now,
Acceleration of the car (a) = (v-u)/t
= (30-0)/20
= 1.5 m/s²
Force (F) = m * a
= 1600 * 1.5
= 2400 N
Hence, the acceleration of the car is 1.5 m/s² and the magnitude of force
applied is 2400 N.
5. Calculate the momentum of a truck of mass 25000 kg when
i) it is at rest.
ii) it is moving with a uniform velocity of 90km/h.
Solution:
Given,
mass of a truck (m) = 25000kg
(i) velocity (v) = 0m/s
Now,
Momentum of the truck under condition (i) (p) = m * v
= 25000 * 0
= 0 kgm/s
(ii) velocity (v) = 90km/h
= 90*1000m/3600s
= 25 m/s
And,
Momentum of the truck under condition (ii) (p) = m*v
= 25000* 25
= 625000 kgm/s
Hence, the required momentum of the truck are (i) 0kgm/s and (ii) 625000
kgm/s.
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