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Question (1):
Define force.
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Answer:
Force is an external agent or influence, which changes or tends to change the state of rest or state of uniform motion of a body in a straight line.
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Question (2):
What are balanced and unbalanced forces?
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Answer:
Balanced forces are those forces whose resultant is zero. Unbalanced forces are those forces whose resultant is not equal to zero.
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Question (3):
What is the effect of force in the following cases when -
a) A fielder catches a cricket ball.
b) Brakes are applied to a moving car.
c) A player hits an in-coming ball with a hockey stick.
d) A soft rubber ball is squeezed between your hands.
e) A spring is stretched.
f) A football lying on ground is kicked.
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Answer:
a) The speed of the cricket ball becomes zero.
b) The speed of the car decreases.
c) The speed changes and direction may also change.
d) The shape of the softball changes.
e) The shape and size of the spring change.
f) The speed of the football increases.
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Question (4):
Define Inertia.
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Answer:
Inertia is the property of matter by virtue of which it opposes any change in its state of rest or state of uniform motion along a straight line.
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Question (5):
Define the following types of inertia by giving suitable examples.
a) Inertia of rest b) Inertia of motion c) Inertia of direction
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Answer:
 Inertia of rest is the property of matter by virtue of which it opposes any change in its state of rest
Example:
The dust particles on a carpet fall when we beat the carpet with a stick. This is because the dust particles due to inertia tend to remain in the same position where as the carpet moves.
 Inertia of motion is the property of matter by virtue of which it opposes any change in its state of motion
Example:
A man carelessly alighting from a moving train falls forward.
 Inertia of direction is the property of matter by virtue of which it opposes any change in its direction
Example:
The force experienced by the passengers whenever a driver is negotiating a curve, is an example of inertia of direction.
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Question (6):
Identify the type of inertia in the following cases.
(a) Falling of fruits when a branch of a tree is shaken
(b) An athlete runs a distance before taking a leap in a high jump
(c) A passenger moving forward when a train stops
(d) A passenger jumping out of a moving bus falls forward if he does not run forward
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Answer:
(a) Inertia of rest
(b) Inertia of motion
(c) Inertia of motion
(d) Inertia of motion
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Question (7):
Who first stated the law of inertia? Describe a simple experiment to demonstrate the inertia of rest.
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Answer:
Galileo first gave the law of inertia.
Place a card on an empty tumbler and a coin on the card as show in the figure. Flick the card with a finger. The coin falls into the tumbler. When we flick the card, the card moves fast but the coin continues in its state of rest and hence falls into the tumbler.
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Question (8):
Why does a passenger standing in a bus move forward when the brakes are applied?
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Answer:
A passenger standing in a bus move forward when the brakes are applied due to inertia of motion.
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Question (9):
State the law of inertia (Newton's first law of motion).
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Answer:
Every body continues to be in a state of rest or of uniform motion along a straight line unless an external force acts upon it.
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Question (10):
It is possible to remove a tablecloth from a table without disturbing the dishes on it. Why?
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Answer:
Due to inertia of rest the dishes continue to remain on the table while the tablecloth is being removed. This is possible only if the tablecloth is removed very quickly.
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Question (11):
Seat belt normally reduces the possibility of injury during an automobile accident. Why?
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Answer:
When an accident takes place the driver applies brakes all of a sudden and due to inertia of motion passengers are jerked or thrown forward and may get hurt due to banging into a hard surface in front of them. This can be avoided if the passengers fasten their seat belts.
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Question (12):
State the law, which gives a quantitative definition of force.
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Answer:
Newton's second law gives a quantitative definition of force. According to this law the rate of change of momentum of a body is directly proportional to the force and takes place in the same direction as the force.
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Question (13):
Define one Newton.
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Answer:
One Newton is that force which when acting on an object of mass 1 kg produces an acceleration of 1m/s2 in it.
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Question (14):
Define the momentum of a body. What is its unit?
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Answer:
The momentum of a body is defined as the product of its mass and velocity. The SI unit of momentum is kg m/s or N s.
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Question (15):
Name the physical quantity, which is equal to the product of force and time.
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Answer:
Impulse is the physical quantity, which is equal to the product of force and time.
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Question (16):
What force will produce an acceleration of 7 m/s2, in a body of mass 21 kg?
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Answer:
F = ma
Mass of the body (m) = 21 kg
Acceleration (a) = 7 m/s2
F = ma
= 21 x 7
=147 N
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Question (17):
Calculate the mass of an object when a force of 1000 N produces an acceleration of 10 m/s2.
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Answer:
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Question (18):
Find the force acting on a 10 kg mass accelerating at 2 m/s2.
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Answer:
Mass of the body = 10 kg
Acceleration = 2 m/s2
Force (F) = ma
= 10 x 2 = 20 N
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Question (19):
Give the mathematical representation of Newton's second law of motion.
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Answer:
Force (F) applied on an object is directly proportional to the rate of change of momentum.
Where
mv is the final momentum
mu is the initial momentum and
t is the time for which the force is applied.
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Question (20):
A force acting on a stationary body of mass 100 kg for 5 seconds moves it through a distance of 100 m in the next 10 seconds. Calculate the force applied.
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Answer:
Mass of the body (m) = 100 kg
Time for which the force is acting (t) = 5 seconds
[The force is applied only for 5 seconds and after that no force is acting on the body]
Distance covered (S) = 100 m
We have to calculate the acceleration using the II equation of motion.
= 100 x 8
= 800 Newtons
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Question (21):
A constant retarding force of 100 N is acting on a body of mass 20 kg moving initially with a speed of 15 m/s. How long does the body take to stop?
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Answer:
Retarding Force (F) = 100 N or Force = - 100 N
Mass (m) = 20 kg
Initial velocity (u) = 15 m/s, Final velocity (v) = 0
F = ma
Time required to stop the body = 3 seconds.
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Question (22):
What is the acceleration produced by a force of 10 N acting on a body of mass 20 kg?
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Answer:
Force acting on the body (F) = 10 N
Mass of the body (m) = 20 kg
Acceleration produced = a
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Question (23):
A car of mass 200 kg moving at 36 km/h is brought to rest after it covered a distance of 10 m. Find the retarding force acting on the car.
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Answer:
Mass of the car (m) = 200 kg
Final velocity (v) = 0
Distance covered (S) = 10 m
F = ma
First calculate a, using III equation of motion
v2 - u2 = 2aS
- 100 = 20 a
= -1000 N
\ Retarding force = 1000 N
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Question (24):
A force of 4 x 10-3 N acts on a mass of 0.04 kg over a distance of 20 m. Assuming the mass is initially at rest, find the final velocity.
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Answer:
Mass (m) = 0.04 kg
Distance (S) = 20 m
Initial velocity (u) = 0
Use III equation of motion to find final velocity.
v2 = 4
v = 2 m/s
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Question (25):
What will be the change in acceleration of a sliding block, if its mass is doubled while a constant force is acting on it?
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Answer:
Force exerted on the block (F) = ma
Let force acting on the object when the mass is doubled be equal to F1
i.e., Mass (m1) = 2 m
Acceleration produced = a1
F1 = 2m x a1
Given F = F1
a = 2 a1
a1 = a/2
i.e., acceleration is reduced to half.
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Question (26):
What is the magnitude of the force which when it acts on a mass of 10 kg gives it a velocity of 5 m/s in one minute?
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Answer:
Initial velocity (u) = 0
Final velocity (v) = 5 m/s
Time (t) = 1 min = 60 seconds.
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Question (27):
A driver accelerates the car at the rate of 2 m/s2 and after some time the car starts accelerating at the rate of 4 m/s2. Calculate the ratio of forces.
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Answer:
Mass of the car remains the same. Let m be the mass of the car. Since the acceleration produced is different we conclude that the forces acting on the car are of different magnitude. Let F1 and F2 be the forces applied on the car.
\ Ratio of the forces is calculated by dividing equation (1) by equation (2)
F1 : F2 = 1 : 2
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Question (28):
A force of 600 N applied to a mass produces an acceleration of 10 m/s2. What is the distance covered by the object at the end of 10 seconds?
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Answer:
Foce (F) = 600 N
Acceleration (a) = 10 m/s2
Distance covered (S) = ?
Time (t) = 10 s
Initial velocity (u) = 0
The distance covered is determined using the II equation of motion.
= 0 + 5 x 100
= 500 m
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Question (29):
The figure below show a velocity time graph for a scooterist having a total mass of 150 kg. From the graph calculate -
a) The acceleration in first 4 seconds
b) The distance covered in the first 4 seconds.
c) The force acting in the first 4 seconds.
d) The retarding force.
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Answer:
a) The acceleration in the first four seconds is given by the slope of the graph AB.
b) The distance covered in the first four seconds = area of D ABF
c) Force acting on the body in the first four seconds
= ma
= 150 x 5
= 750 N
d) The retarding force = ma
F = ma
The object starts retarding from the 10th second.
 =- 5 m/s2
Force acting on the body F = ma
= 150 x -5
= - 750 N
The retarding force acting on the body = 750 N.
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Question (30):
The figure below shows the v-t graph of a car of mass 1500 kg. Calculate the retarding force.
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Answer:
Retarding force F = ma
Mass of the car m = 1500 kg
We have to calculate a. The acceleration is zero till 20 seconds.
=1500 x -4
= - 6000 N
Retarding force = + 6000 N
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Question (31):
A certain force exerted for 1.2 seconds raises the speed of an object from 1.8 m/s to 4.2 m/s. Later the same force is applied for 2 seconds. How much does the velocity change in 2 seconds?
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Answer:
Initial velocity (u) = 1.8 m/s.
Final velocity (v) = 4.2 m/s
Time (t) = 1.2 seconds
First calculate acceleration
As the same force acts for the next two seconds the acceleration produced will be the same. The final velocity in the first case will now become the initial velocity. We have to calculate the final velocity at end of 2 seconds.
Acceleration (a) = 2 m/s2
Intial velocity (u) = 4.2 m/s, t = 2 s
Final velocity (v) = ?
 [First equation of motion]
Change in velocity in two seconds = 8.2 - 4.2 = 4 m/s.
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Question (32):
Two blocks made of different metals, identical in shape and size are acted upon by equal forces, which cause them to slide on a horizontal surface. The acceleration of the second block is found to be 5 times that of the first.What is the ratio of the mass of second block to that of the first.
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Answer:
Let m1 and m2 be the mass of the first and second block respectively and a be the acceleration.
Given :
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Question (33):
Brakes are applied on a car of mass 1000 kg moving with a velocity of 54 km/h. The car covers a distance of 50 m before coming to rest. Calculate the force applied on the car.
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Answer:
Initial velocity u = 54 km/h
= 15 m/s
 Final velocity v = 0 (because the car is brought to rest)
Distance covered S = 50 m
We have to calculate force applied, i.e., F = ma. But we do not know the value of a. Making use of III equation of motion we get,
F = ma
Mass = 1000 kg
Force = 1000 x (- 2.25)
= - 2250 N
Retarding force = 2250 N
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Question (34):
A car of mass 800 kg, moving at 108 km/h is brought to rest over a distance of 15 m. Find the retarding force developed by the brakes of the car.
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Answer:
Initial velocity (u) = 108 km/hr
= 30 m/s
Final velocity (v) = 0
Distance covered (S) = 15 m
[III equation of motion] Applying v2 - u2 = 2aS
Mass of the car = 800 kg
Retarding force = 24000 N
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Question (35):
The velocity of a car of mass 1800 kg increases by 90 km/h in 15 seconds. Calculate the force applied to increase the speed.
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Answer:
Mass of the car (m) = 1800 kg
= 5 x 5
= 25 m/s
Force applied = ma
Time (t) = 15 seconds
Force = ma
= 600 x 5
= 3000 N
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Question (36):
What is the acceleration produced by a force of 12 Newton exerted on an object of mass 3 kg?
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Answer:
From Newton's second law of motion F = ma
F = 12 N, mass (m) = 3 kg
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Question (37):
What force would be needed to produce an acceleration of 4 m/s2 on a ball of mass 6 kg?
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Answer:
Acceleration (a) = 4 m/s2
Mass of the ball (m) = 6 kg
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Question (38):
Why does a fielder in a cricket match move his hand backwards while making an attempt to catch the ball?
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Answer:
By moving the hand backwards the fielder increases the time of action of the force on his hands. When the time increases the force is reduced and the chances of the fielder getting hurt and dropping the ball is less.
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Question (39):
For how long should a force of 200 N act on a body of 20 kg, so that it acquires a velocity of 100 m/s?
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Answer:
Time required to change the velocity of the body is to be calculated.
Initial Velocity (u) = 0
Final Velocity (v) = 100 m/s
The force should act on the body for 10 seconds.
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Question (40):
A retarding force is applied for 10 seconds to stop 2000 kg heavy car moving with a uniform velocity of 40 m/s. Calculate the retardation and the retarding force.
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Answer:
Time (t) = 10 seconds
Final Velocity (v) = 0 [as the car comes to rest]
Initial velocity (u) = 40 m/s
Mass of the car = 2000 kg
Force required to stop the car = ma
= 2000 x -4
= -8000 N
Retarding Force = +8000 N
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Question (41):
Which would require a greater force accelerating a 10 g mass at 5 m/s2or a 20 g mass at 2 m/s2?
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Answer:
Let force required to accelerate a 10 g mass i.e., 0.01 kg be F1
F1= m1 a1
Let force required to accelerate a 20 g mass i.e., 0.02 kg be F2
F2 = m2a2
\ Force required to accelerate 10 g is more than the force required to accelerate a 20 g mass.
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Question (42):
An object undergoes an acceleration of 8 m/s2 starting from rest. Find the distance travelled in one second.
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Answer:
Time (t) = 1 s
Distance travelled (S) = ?
Initial velocity (u) = 0 [object is starting from rest]
Distance covered = 4 m
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Question (43):
A truck of mass 5000 kg starts from rest and rolls down a hill with constant acceleration. It travels a distance of 800 metres in 20 seconds. Find its acceleration.
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Answer:
F = ma
Mass (m) = 5000 kg
Initial velocity (u) = 0 [ starting from rest]
Final velocity (v) = ?
Time (t) = 20 seconds
Distance travelled (S) = 800 m
Applying II equation of motion, we get
Acceleration = 4 m /s2
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Question (44):
Determine (a) the acceleration (b) the distance travelled in 12 seconds when a 7500 N force accelerates a 1500 kg car, which is initially at rest.
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Answer:
Force (F) = 7500 N
Mass of the car (m) = 1500 kg
Acceleration = 5 m/s2
Initial velocity (u) = 0
Time (t) = 12 s
Applying II equation of motion,
Distance travelled in 12 seconds = 360 m
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Question (45):
The table below gives the variation of velocity of a coin of mass 20 g with time. Plot a v-t graph for the data and also calculate the force exerted by the table to bring the coin to rest.
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Answer:
Force exerted by the table (F) = ma
Force exerted by the table = - 16 x 10-4 N
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Question (46):
A train running at 108 km/h is brought to rest over a distance of 1500 m. Calculate the force developed by the brakes if the mass of the train is 75000 kg.
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Answer:
Initial velocity u = 108 km/h
= 6 x 5 = 30 m/s
Final velocity v = 0 [as the train is brought to rest]
Distance covered S = 1500 m
Mass of the train m = 75000 kg
Force developed by the brakes F = ma
To calculate the acceleration we make use of the third equation of motion
= 75000 x 0.3
= -225000 N
Negative sign shows that the force is a retarding one.
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Question (47):
Determine the magnitude of the linear momentum of a 200 g hockey puck travelling at a speed of 108 km/h.
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Answer:
Linear momentum = mv
Mass of the hockey puck = 200 g.
Velocity of the hockey puck = 108 km/h 
Linear momentum = mv
= 0.2 x 30 = 6 kg m/s
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Question (48):
A bullet of mass 0.04 kg moving with a velocity of 360 km/h is brought to rest by a target in 0.02 second. Calculate the impulsive force.
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Answer:
Mass m = 0.04 kg
Initial velocity (u) = 360 km/h
But 1 km/h =5/18 m/s
Initial velocity (u) = 360 x 5/18
= 20 x 5
= 100 m/s
Time taken to hit the target = 0.02 second.
Final velocity v = 0 (the bullet comes to rest)
Impulse = Ft
= 0.04 x 5000 =-200 N
Impulsive force = -2000 N
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Question (49):
Name the physical quantities, which have N s as their unit and derive the relation between.
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Answer:
Impulse and momentum have N s as their unit. Impulse is the physical quantity which is equal to change in momentum. Consider an object of mass 'm' moving with an initial velocity 'u'. When a force is acting on the object for a certain time 't' its velocity changes to 'v'.
According to Newton's second law of motion,
F = ma .... (1)
Where 'a' is the acceleration produced in the object due to the application of force.
Substituting the value of a in equation (1) we get,
Since 'v' is the final velocity, 'mv' will be the final momentum and 'u' is the initial velocity, 'mu' will be initial momentum.
Ft is defined as impulse
\ Impulse = Change in momentum
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Question (50):
A cricket ball moving with a velocity 25 m/s is brought to rest by a fielder in 0.4 s. If the mass of the ball is, 0.1 Kg find
(a) the impulse
(b) the retarding force.
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Answer:
Mass of the ball (m) = 0.1 Kg
Initial velocity of the ball (u) = 25 m/s
Final velocity of the ball (v) = 0
Time taken (t) = 0.4 s
Impulse = m(v - u)
= 0.1 (0-25)
= 0.1 x -25
Impulse = - 2.5 N s
Impulse = m(v-u) = Ft
Retarding force = - 6.25 N
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Question (51):
Name the physical quantity, which is equal to rate of change of momentum and derive a relation between them. A car weighing 2000 kg moving with a velocity of 90 km/h retards uniformly and comes to rest in 20 s. Calculate the force required to stop the car.
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Answer:
Force is the physical quantity, which is equal to rate of change of momentum. Consider an object of mass m moving with an initial velocity u. When a force is acting on the object for a certain time t its velocity changes to v. According to Newton's second law of motion,
F = ma ...(1)
Where a is the acceleration produced in the object due to the application of force.
Substituting the value of a in equation (1) we get,
Since v is the final velocity, mv will be the final momentum and u is the initial velocity, mu will be the initial momentum.
Force = Rate of change of momentum
Mass of the car = 2000 kg
Initial velocity (u) = 90 km/h
Initial velocity (u) = 25 m/s
Final velocity (v) = 0
Time (t) = 20 s
Force required to stop the car = -2500 N
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Question (52):
Calculate the momentum of an electron of mass 9 x 10-31kg, moving with a velocity of 3 x 107m/s.
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Answer:
Momentum = mv
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Question (53):
Calculate the velocity of a body of mass 1 kg having a momentum of 10 N s.
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Answer:
Momentum (p) = mv
Momentum (p) = 10 N s
Mass (m) = 1 kg
p = mv
10 = 1 x v
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Question (54):
Why is it easier to stop a tennis ball than a cricket ball moving at the same speed?
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Answer:
Mass of the tennis ball is less than that of the cricket ball hence the rate of change of momentum is less in the case of the tennis ball. Therefore the force required to stop the tennis ball is less.
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Question (55):
Jumping on a cemented floor causes more injury than on a heap of sand. Why?
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Answer:
The change in momentum when a person is jumping on a cemented floor or on a heap of sand remains the same. But when a person jumps on a cemented floor, he comes to rest immediately and hence the force applied by the floor is very large. This large force causes serious injuries. But when a person jumps on a heap of sand he does not come to rest immediately and as a result the impact of the force exerted by the floor decreases considerably.
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Question (56):
A body of mass 15 kg moving with a velocity of 15 m/s is brought to rest in 10 seconds. Find the change in momentum and also the retarding force.
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Answer:
Mass of the body (m) = 15 kg
Initial Velocity (u) = 15 m/s
Final Velocity (v) = 0
Time (t) = 10 s
Change in momentum = mv- mu
= -225 kg m/s
Force applied is equal to rate of change in momentum.
Retarding Force = 22.5 N
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Question (57):
The speed of a car weighing 1000 kg increases from 36 km/h to 108 km/h. Calculate the change in momentum.
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Answer:
Mass of the Car (m) = 1000 kg
Final Velocity (v) = 108 km/h
= 30 m/s
Change in momentum = mv - mu
Change in momentum = 20,000 N s
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Question (58):
Calculate the initial and final momentum of a 1000 g ball thrown vertically up with a speed of 10 m/s.
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Answer:
Mass of the ball (m) = 1000 g
Initial Velocity of the ball (u) = 10 m/s
Final Velocity of the ball (v) = 0
[When it attains the maximum height, the final velocity is zero.]
Initial momentum = mu
= 1 x 10
Initial momentum of the ball = 10 kgm/s
Final momentum of the ball = mv = 1 x 0 = 0
Final momentum of the ball = 0
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Question (59):
A car weighing 1500 kg travelling at a speed of 108 km/h collides with a building and is stopped in 0.8 seconds. What is the impulse exerted on the car?
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Answer:
Mass of the car (m) = 1500 kg
Initial velocity (u) = 108 km/h
Final Velocity (v) = 0 [car stops when it collides with a building]
Impulse = (mv - mu)
= m(0 - u)
= - mu
= - 1500 x 30
= -45000 N s
Impulse exerted on the car = - 45000 N s
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Question (60):
A horse develops a momentum of 6000 N s while running at 30 m/s. Calculate the mass of the horse.
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Answer:
Momentum p = mv
Velocity (v) = 30 m/s
Momentum (p) = 6000 N s
Mass of the horse = 200 kg
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Question (61):
Name the physical quantity that is equal to change in momentum and derive a relation between them.
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Answer:
Impulse is the physical quantity, which is equal to change in momentum. Consider an object of mass `m moving with an initial velocity u. When a force is acting on the object for a certain time t its velocity changes to v. According to Newton's second law of motion,
F = ma ... (1)
Where a is the acceleration produced in the object due to the application of force.
Substituting the value of a in eqn. (1) we get,
Since v is the final velocity,mv will be the final momentum and similarly mu will be the initial momentum as u is the initial velocity .
Ft = Change in momentum
Ft is defined as impulse.
\ Impulse = Change in momentum
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Question (62):
Name the physical quantity, which is equal to rate of change of momentum and derive a relation between them.
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Answer:
Force is the physical quantity, which is equal to rate of change of momentum. Consider an object of mass m moving with an initial velocity u. When a force F is acting on the object for a certain time t its velocity changes to v. According to Newton's second law of motion,
F = ma ...(1)
Where a is the acceleration produced in the object due to the application of force.
Substituting the value of a in equation (1) we get,
Since v is the final velocity,mv will be the final momentum and similarly mu will be the initial momentum as u is the initial velocity .
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Question (63):
A bullet of mass 10 g passing through a 10 cm plank, has its velocity reduced from 300 m/s to 200 m/s. Find the average resistance offered to the bullet.
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Answer:
Distance moved by the bullet = Thickness of the plank = S
Initial velocity = 300 m/s
Final velocity = 200 m/s
40000 - 90000 = 0.2 a
- 50000 = 0.2 a
We have to calculate the force F = ma.
= 0.01 x - 250000
= - 2500 N
Average resistance offered to the bullet = -2500 N
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Question (64):
State law of conservation of momentum.
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Answer:
Law of conservation of momentum states that if a group of bodies are exerting force on each other, i.e., interacting with each other, then their total momentum remains conserved before and after the interaction provided there is no external force acting on them.
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Question (65):
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 seconds. Find its acceleration. Find the force acting on it, if its mass is 7 metric tons.
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Answer:
Initial velocity (u) = 0
Distance covered by the truck (s) = 400 m
Time (t) = 20 s
To calculate the acceleration we make use of the II equation of motion.
Mass of the truck (m) = 7 metric ton (1 metric ton = 1000 kg)
Mass = 7000kg
Force acting on the truck (F) = ma
= 7000 x 2
= 14000 Newtons
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Question (66):
A stationary ball weighing 0.25 kg acquires a speed of 10 m/s when hit by a hockey stick. What is the impulse imparted to the ball?
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Answer:
Mass of the ball = 0.25 kg
Initial velocity (u) = 0 (initially it is at rest)
Final velocity (v) = 10 m/s
Impulse = Ft = mv - mu
Ft = m (v - u)
= 0.25 (10 - 0) = 0.25 x 10
Impulse = 2.5 N s
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Question (67):
State Newton's third law of motion.
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Answer:
" To every action there is an equal and opposite reaction.
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Question (68):
A rifle of mass 3 kg fires a bullet of mass 0.03 kg. The bullet leaves the barrel of the rifle at a velocity of 100 m/s. If the bullet takes 0.003 s to move through its barrel, calculate the force experienced by the rifle due to the recoil.
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Answer:
Mass of the rifle (m1) = 3 kg
Mass of the bullet (m2) = 0.03 kg
Initial velocity of the rifle (u1) = u2 = 0,
Final velocity of the rifle (v1) = ?
Final velocity of the bullet (v2) = 100 m s-1
According to law of conservation of momentum = m1u1 + m2u2 = m1v1 + m2v2
Using the law of conservation of momentum,
0 + 0 = 3v1 + 0.03 x 100
or 
Negative sign shows that the rifle moves in a direction opposite to that of the bullet,
\Force experienced by the rifle due to its recoil 
= -1000 kg ms-2
= -1000 N
Therefore, the person would experience a force of 1000 N in the backward direction due to the recoil of the rifle.
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Question (69):
The resultant of action and reaction forces is never equal to zero. Why?
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Answer:
Even though the action and reaction forces are equal and opposite their resultant is never equal to zero because they are acting on two different bodies.
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Question (70):
Define force of friction.
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Answer:
The force of friction is that force which opposes the motion of an object over another object in contact with it.
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Question (71):
Why do we generally sprinkle a small amount talcum powder on a carrom board?
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Answer:
A small amount of talcum powder reduces the friction i.e., the talcum powder fills in the depressions on the carrom board and the surface becomes smooth.
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Question (72):
A longer time and a longer force are required to make a moving ship come to rest than the time and force required to stop a very heavy vehicle on the road. Why so?
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Answer:
The force of friction existing between water and the ship is very small and because of this small friction a ship requires more time and force to stop. Comparatively smalll force is required to stop a heavy vehicle on road as the force of friction between the road and tyres of the vehicle is very large.
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Question (73):
Justify the statement: Friction is a necessary evil.
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Answer:
Friction is considered to be a necessity because of the following advantages:
 The force of friction enables us to walk on the surface of the earth
 The fibres of thread are held together due to force of friction
 Nails and screws hold the wooden boards together due to force of friction.
Friction is considered to be an evil because of the following disadvantages:
 There is wastage of energy when we try to overcome the force of friction
 There is wear and tear of moving parts
 The speed of the automobiles cannot be increased beyond a certain limit.
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