NCERT Solutions for class 9 Science chapter 8 Forces and Laws of Motion is prepared and uploaded for reference by academic team of expert members of Param Himalaya. Get solutions of all chapters of NCERT class 9 Science from Param himalaya. use as a reference of the following NCERT solutions of chapter 8 prepared by paramhimalaya. Read the theory of chapter-8 Forces and Laws of Motion while before going to exam.
Page no : 91 Question 1 to 4 :
Question 1. Which of the following has more inertia :
(a) a rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five rupees coin and a one-rupee coin?
Solution :
(a) a stone of the same size will have more inertia than a rubber ball.
(b) A train will have more inertia than a bicycle.
(c) A five rupees coin will have more inertia than a one-rupee coin.
Question 2. In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.
Solution :
The velocity of football changes four times.
1. First , when a football player kicks a football to another player ( 0 m/s to u m/s) ,
2. second when that player kicks the football to the goalkeeper ( u m/s to v m/s )
3. Third when the goalkeeper stops the football.( v m/s to 0 m/s )
4. Fourth , when the goalkeeper kicks the football towards his team player. ( 0 m/s to x m/s )
Agent supplying the force :
1. The first case is the First Player.
2. The second case is the second player.
3. The third case is goalkeeper.
4. The fourth case is Goalkeeper.
Question 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Solution :
when the tree's branch is shaken vigorously the branch attain motion but the leaves stay at rest. Due to the inertia of rest , the leaves tends to remain in its position and hence detaches from the tree to fall down.
Question 4. 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 :
when a moving bus brakes to a stop we fall in the forward direction because we are also moving with the speed of bus due to the inertia of motion and when suddenly it puts brakes i.e. comes to rest the lower half of our body also comes to rest but the upper half of our body not being in close contact with bus is still in the phase of motion so we fall in the forward direction.
When the bus accelerates from rest, we are also at rest being on the resting seat as the engine applies force in forward direction we fall backwards due to the inertia now.
Exercise Question
Question 1. An object experiences a net zero external unbalanced force. 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. If no, provide a reason.
Solution : By Newton's First law of motion, a body remains in the state of rest or uniform motion unless an external force is applied on it. So, if the body has some initial velocity and net force applied is zero, it will continue to move with the same initial velocity.
Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Solution : When a carpet is beaten with a stick, dust comes out of it because carpet fibres vibrate in forward and backward direction as carpet is beaten but the loosely bound dust particles due to inertia remain at rest and so they come out.
Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Solution : It is advised to tie any luggage kept on the roof of a bus with a rope because when bus moves the luggage also gets moving with the velocity same as that of the bus and in the same direction but when bus changes direction or deaccelerates, due to inertia of motion luggage moves in the same direction and may get thrown away from roof of buses.
Question 4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Solution : (c) there is a force on the ball opposing the motion.
Question 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.)
Solution :
According to the Question,
initial velocity of truck (u) = 0
distance = s= 400 m and time = 20 s
mass of truck = 7metric tones = 7000kg
Forces and Laws of Motion
$s = ut + \frac{1}{2}at^{2}$
$400 = 0 \times 20 + \frac{1}{2} \times a \times (20)^{2}$
$400 = 0+ \frac{1}{2} \times a\times 400$
$400 = 200 a$
$a = \frac{400}{200}$
$a= 2 m / s^{2}$
therefore, F = m × a
F = 7000 × 2
F = 14000 N
Question 6. A stone of 1 kg is thrown with a velocity of 20m s-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution :
$v^{2} = u^{2}+2as$
$0 = 20^{2} + 2a \times 50$
$0 = 400 + 100a$
$100a = -400$
$a =- \frac{400}{100}$
$a= -4 m/s^{2}$
therefore, the force of friction between the stone and the ice
F = ma
F= -4 × 1
F = -4 N
Negative sign indicates that the direction of the force of friction is opposite to the direction of moving stone.
Question 7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. 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;
(b) the acceleration of the train
Solution :
(a) The net accelerating force = Force exerted by engine – frictional force of track
$F_{Net} = F-f$
$F_{Net}= 40000 – 5000 = 35000 N$
(b) Total mass of train = M + n ×m
Total mass of train = 8000 + 5 × 2000
Total mass of train = 18000
So
F = total mass × acceleration
35000 = 18000 × a
a= 35000 / 18000
a= 35/18
$a= 1.94 m/s^{2}$
Question 8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m S-2?
Solution :
since,
F=m × a
F=1500 × -1.7
F= -2550 N
(negative sign symbolizes acceleration in opposite direction)
Question 9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2
(b) mv2
(c) 1/2 mv2
(d) mv
Solution : (d) mv
Question 10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Solution : The cabinet will move with constant velocity only when the net force on it is zero. Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.
Question 11. 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. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Solution : 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 result is the two opposite and equal forces cancel each other but when one of these forces is bigger than inertia so the object moves in the direction of force applied. As this student explains the truck is massive so the force applied cannot overcome force caused by inertia. Therefore, the truck does not move.
Question 12. A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s . Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution :
Given , mass of the ball (m) = 200 g = 0.2 kg
Initial velocity of the ball (u) = 10 m/s
Final velocity of the ball (v) = -5 m/s
initial momentum of hockey ball = mu
0.2 kg × 10 m/s= 2 kg m/s
Final momentum of hockey ball = mv
= 0.2 kg × -5 m/s = -1 kg m/s
Change in momentum of hockey ball = mv - mu = -1-2 = - 3 kg m/s
Question 13 . A bullet of mass 10 g travelling horizontally with a velocity of 150 m / s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Solution :
Mass of the bullet, m = 10 g = 0.01 kg
It is given that the bullet is travelling with a velocity of 150 m/s.
Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s
Final velocity, v = 0 (since the bullet finally comes to rest)
Time taken to come to rest, t = 0.03 s
According to the first equation of motion,
v = u + at
where , a is the acceleration of the bullet
0 = 150 + (a × 0.03)
-150 = a× 0.03
$a = - \frac{150}{0.03}$
$a = -5000 m/s^{2}$
Concept Note:- Negative sign indicates that the velocity of the bullet is decreasing.
According to the third equation of motion:
$v^{2}= u^{2} + 2as$
0 = $(150)^{2}$+ 2×(-5000)×sx
0 = 22500 + 2×(-5000)×x
0=22500 – 10000 x
10000 x = 22500
$x= \frac{22500}{10000}$
$x= 2.25m$
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion: Concept Note:-
Force, F = Mass × acceleration
bullet, m = 0.01 kg
Acceleration of the bullet, $a = -5000 m/s^{2}$
F = ma = 0.01×(-5000) = -50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.
Question 14. An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Solution :
Mass of the object, m1 = 1 kg
The velocity of the object before the collision, $v_{1}$= 10 m/s
Mass of the stationary wooden block, $m_{2} = 5 kg$
The velocity of the wooden block before the collision, $v_{2} = 0 m s^{-1}$
∴ Total momentum before collision = $m_{1}v_{1} + m_{2} v_{2}$ = 1 (10) + 5 (0)
$m_{1}v_{1} + m_{2} v_{2}= 10 kg ms^{-1}$
It is given that after the collision, the object and the wooden block stick together.
The total mass of the combined system = $m_{1} + m_{2}$
The velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
$m_{1}v_{1}+ m_{2}v_{2} = (m_{1}+ m_{2}) v$
1(10) + 5 (0) = (1 + 5)v
10 =6v
$v = \frac{10}{6}$
$v =\frac{5}{3} ms^{-1}$
v= 1.66 m/s
The total momentum after collision = $(m_{1}+ m_{2}) v$
The total momentum after collision = $(1+5) \times \frac{5}{3}$
The total momentum after collision= $6 \times \frac{5}{3}$
The total momentum after collision = 10 kg m/sec
Question 15. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Solution :
Mass of object m=100kg
Initial velocity, u=5m/s
Final velocity, v=8m/s
Time, t=6s
Initial momentum =mu=100×5=500 kg m/s
Final momentum =mv=100×8=800 kg m/s
Force exerted on the object
$F=\frac{(mv−mu)}{t}$
$F=\frac{(800−500)}{6}$
$F= \frac{300}{6}$
$F= 50N$
Hence the magnitude of the force exerted on the object = 50 N
Question 16. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. 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. Comment on these suggestions.
Solution : Since the mass of insect is negligible in comparison to the mass of motorcar therefore there will be no any change in the momentum of motorcar.
Question 17. 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 m / s2 .
Solution :
Mass of dumb bell , m = 10 kg
Initial velocity u = 0 m/s
Final velocity v = ?
Distance, s = 80 cm = 0.8m
Acceleration $a = 10ms^{-2}$
We know
$v^{2} = u^{2} + 2as$
$v^{2} = 0^{2} + 2 \times 10 \times 0.8$
$v^{2} = 16$
$v = {\sqrt{16}}$
v = 4 m/s
momentum transferred,
p = m v = 10 × 4
p= 40 k g m / s