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Page No 16:
Question 1:
Match the first column with appropriate entries in the second and third columns and remake the table.
S.no 
Column 1

Column 2  Column 3 
1  Negative acceleration  the velocity of the object remains constant  A car,initially at rest reaches a velocity of 50 km/hr in 10 seconds 
2  Positive acceleration  the velocity of object decreases  A vehicle is moving with a velocity of 25 m/s 
3  Zero acceleration  the velocity of object increases  A vehicle moving with the velocity of 10 m/s, stops after 5 seconds. 
Answer:
S.no  Column 1  Column 2  Column 3 
1  Negative acceleration  the velocity of object decreases  A vehicle moving with the velocity of 10 m/s, stops after 5 seconds. 
2  Positive acceleration  the velocity of object increases  A car initially at rest, reaches a velocity of 50 km/hr in 10 seconds. 
3  Zero acceleration  the velocity of the object remains constant  A vehicle is moving with a velocity of 25 m/s. 
Page No 16:
Question 2:
Answer:
A.
Distance

Displacement

It is defined as the length of the actual path travelled by an object in motion.

It is defined as the shortest distance between the starting and finishing points.

It is a scalar quantity.

It is a vector quantity.

The values of distance are always
positive. 
The values of displacement can be positive, negative or zero.

Uniform motion  Nonuniform motion 
When an object covers equal distance in equal time interval, it is said to be in uniform motion.  When an object covers unequal distance in equal time interval, it is said to be in nonuniform motion. 
The distancetime graph obtained in case of uniform motion is a straight line graph.  The distancetime graph obtained in case of non uniform motion can be of any shape depending on how the acceleration changes with time. 
eg: A car running on a straight road without any change in its speed.  eg: A car running on a road with different speeds in different interval of time. 
Page No 17:
Question 3:
Complete the following table.
u (m/s)  a (m/s^{2})  t (sec)  v = u + at (m/s) 
2  4  3   
  5  2  20 
u (m/s)  a (m/s^{2})  t (sec)  $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^{2}\left(\mathrm{m}\right)$ 
5  12  3   
7    4  92 
u (m/s)  a (m/s^{2})  s (m)  ${\mathrm{v}}^{2}={\mathrm{u}}^{2}+2\mathrm{as}{\left(\mathrm{m}/\mathrm{s}\right)}^{2}$ 
4  3    8 
  5  8.4  10 
Answer:
u (m/s)  a (m/s^{2})  t (sec)  v = u + at (m/s) 
2  4  3  14 
10  5  2  20 
u (m/s)  a (m/s^{2})  t (sec)  $s=ut+\frac{1}{2}a{t}^{2}\left(\mathrm{m}\right)$ 
5  12  3  69 
7  8  4  92 
u (m/s)  a (m/s^{2})  s (m)  ${v}^{2}={u}^{2}+2as{\left(\mathrm{m}/\mathrm{s}\right)}^{2}$ 
4  3  8  8 
4  5  8.4  10 
Page No 17:
Question 4:
Answer:
a. The minimum distance between the start and finish points of the motion of an object is called the displacement of the object.
The shortest distance between the initial and final position of an object is displacement.
b. Deceleration is negative acceleration.
When a body is decelerating, it means its speed is decreasing speed. An acceleration with a negative also shows that the speed of the body is decreasing.
c. When an object is in uniform circular motion, its velocity changes at every point.
In a uniform circular motion, the direction of motion of the body changes at every instant of time. This continuous change in the direction of motion of the body at every instant accounts for the change in its velocity at every point ot instant.
d. During collision momentum of the system remains constant.
When two bodies collide with each other as shown, action and reaction force comes into action i.e. two bodies exerts force on each other.
Now, momentum of the ball system is given as:
Before collision: m_{1}u_{1} + m_{2}u_{2}
After collision: m_{1}v_{1} + m_{2}v_{2}
Rate of change of momentum of ball 1 is given as:
Where, t → Collision time
Rate of change of momentum of ball 2 is given as:
Using Newton’s third law of motion, we can relate the forces F_{12} and F_{21}_{ }as:
F_{12} = −F_{21}
∴ Momentum of the system before collision = Momentum of the system after collision
e. The working of a rocket depends on Newton’s third law of motion.
The burning fuel inside the rocket creates a forward push on the rocket. This forward push creates an equal and opposite push on the exhaust gases. Thus, the working of a rocket depends on Newton’s third law of motion.
Page No 17:
Question 5:
b. Even though the magnitudes of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled.
Answer:
a. When an object falls freely to the ground, it is under the effect of a constant force known as force of gravity. No other forces act on it. Hence, from Newton's second law of motion, we can say that this constant force of gravity accelerates the the freely falling object uniformly. This uniform acceleration is known as acceleration due to gravity which acts towards the centre and is denoted by g.
b. When two bodies interact, the action and reaction forces come into action. Even though their magnitude are same and their direction are opposite, their effects do not get cancelled because these action and reaction forces do not act on the same body.
c. Momentum of a body is the product of its mass and velocity i.e.
$p=mv$
and the rate of change of momentum of a body is equal to the force applied i.e.
$F=\frac{m(vu)}{t}$
In the given information, both the balls move with same velocity and finally stops, thus force applied to stop the ball will be directly proportional to the mass of the ball.
$F\propto m$
As the mass of cricket ball is greater than the tennis ball, its momentum will be more than the tennis ball. Thus, greater force will be required to stop the cricket ball as compared to that of tennis ball.
d. Velocity of an object is said to be uniform when its speed and direction does not change. The object at rest has uniform speed of zero all the time and does not even change the direction. Hence, the velocity of an object at rest is considered to be uniform.
Page No 17:
Question 6:
Take 5 examples from your surroundings and give explanation based on Newtons laws of motion.
Answer:
 If we try to push an empty cart and a cart full of bricks, it would be easier to push an empty cart than the cart full of bricks. This happens because the inertia of the cart full of bricks is more than the empty cart and by Newton's first law of motion we know that more the inertia, more force is required to change the state of rest of a body.
 If we push a bicycle and a car with the same force, the bicycle will have greater acceleration than the car because the bicycle has less mass compared to the car.
 When we place a book on table, the book does not fall. This is because the same amount of force is applied by the table on the book as is applied by the book on the table.
 Our walking is an example of Newton's third law of motion. When we walk we push the ground in backward direction with some force. The ground in reaction pushes us forward with the same force.
 When we shake a tree, the leaves of the tree fall to the ground. This is because the leaves were in the inertia of rest before shaking of tree. So when the tree was shook, they fall down to attain their inertia of rest again.
Page No 17:
Question 7:
b) An object of mass 16 kg is moving with an acceleration of 3 m/s^{2} . Calculate the applied force. If the same force is applied on an object of mass 24 kg, how much will be the acceleration?
c) A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 90 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.
d) A person swims 100 m in the first 40 s, 80 m in the next 40 s and 45 m in the last 20 s. What is the average speed?
Answer:
a.
$\text{Averagespeed=}\frac{\mathrm{Total}\mathrm{distance}}{\mathrm{Total}\mathrm{time}}=\frac{18+22+14}{3+3+3}\phantom{\rule{0ex}{0ex}}=\frac{54}{9}=\text{6m/s}$
b.
(i) m = 16 kg, a = 3 m/s^{2 }
F = m $\times $ a
F = 16 $\times $ 3 = 48 N
Now, F = 48 N, m =24 kg
F = m $\times $ a
48 = 24 $\times $ a
a = 2 m/s^{2}
c. Given,
Mass of bullet, m_{1 }= 10 g
Initial speed of bullet, u_{1 }= 1.5 m/s
Mass of plank, m_{2 }= 90 g
Initial speed of plank, u_{2 }= 0
From law of conservation of momentum,
m_{1}u_{1 + }m_{2}u_{2} = m_{1}v_{1 }+ m_{2}v_{2}
Since, the bullet gets embedded in the palnk and both move with the same speed, so v_{1 }= v_{2 }= v
So, we can rewrite the equation as
m_{1}u_{1 + }m_{2}u_{2} = (m_{1}_{ }+ m_{2})v
10 $\times $1.5 + 90 $\times $ 0 = (10 + 90)v
15 = 100v
$\Rightarrow $v = 0.15 m/s
d.
$\text{Averagespeed=}\frac{\mathrm{Average}\mathrm{distance}}{\mathrm{Average}\mathrm{time}}\phantom{\rule{0ex}{0ex}}=\frac{100+80+45}{40+40+20}=\frac{225}{100}=2.25\text{m/s}$
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