NCERT Solutions for Science Class 9 Chapter 8
Chapter 8: Force and Laws of Motion
Intext Exercise 1:
Question 1: Which of the following has more inertia:
(a) a rubber ball and a stone of the same size?
Answer: The inertia of an object is measured by its mass. Heavier or more massive objects offer larger inertia.
Stone is heavier than the rubber ball of the same size. e. Hence, the inertia of the stone is greater than that of a rubber ball.
(b) a bicycle and a train?
Answer: Train is heavier than bicycle. Hence, inertia of the train is greater than that of the bicycle
(c) a five-rupees coin and a one-rupee coin?
Answer: A five rupee coin is heavier than a one rupee coin. Hence, inertia of the five-rupee coin is greater than that of the 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.
Answer: The ball’s velocity changes four times.
First change: The ball’s speed changes from 0 to a specific amount as the football player kicks it. value. As a result, the ball’s velocity is altered.
Second change: Another player is kicking the ball to the goal post in the second change. As a result of this, the direction of the ball is changed. As a result, its speed varies. In this case, the player used force. to change the velocity of the ball.
Third change: The ball is being collected by the goalie in the third change. The ball finally comes to a halt. As a result, its speed is lowered to zero from a specific value. The pace of the ball has changed. In this situation, the goalie utilised a counterforce to slow down or modify the pace of the ball.
Question 3: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer: Because of the inertia of rest, when the branch is quickly moved, the leaves attached to it tend to stay in their resting position. The leaves and branch junctions are put under a lot of stress as a result of this. This strain causes some leaves to detach off the branch.
Fourth change-The goalkeeper kicks the ball to his teammates. As a result, the ball’s velocity increases from zero to a certain number. As a result, its velocity shifts once more. In this case, the goalkeeper used force to change the ball’s velocity.
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?
Answer: We move in the forward direction when a moving bus is braking because our upper portion of the body and the bus are both in motion when the bus is moving, and when the bus is breaking our body is trying to be in motion due to inertia of motion and thereby we experience a forward push. Similarly, when the bus accelerates from the rest, the passenger tends to fall backwards. This is because the passenger’s inertia tends to oppose the bus’s forward motion when the bus accelerates. Therefore, when the bus accelerates, the passenger tends to fall backwards.
INTEXT EXERCISE 2
Question 1: If action is always equal to the reaction, explain how a horse can pull a cart.
Answer: With his foot, a horse pushes the earth in a rearward way. According to Newton’s third law of motion, the Earth exerts a reaction force on the horse in the forward direction. As a result, the cart advances.
Question 2: Explain why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer: When a significant volume of water is discharged from a hose at a high velocity, Newton’s Third Law of Motion states that the water pushes the hose backwards with the same
force. As a result, gripping a hose that ejects a significant volume of water at a rapid rate is difficult for a firefighter.
Question 3: From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an intial velocity of 35 \(ms^{-1}\). Calculate the intial recoil velocity of the rifle.
Given:
Mass of the rifle, \(m_1=4\space kg\)
Mass of the bullet, \(m_2=50\spaceg=0.05\space kg\)
Recoil velocity of the rifle \(=v_1\)
Intial velocity of bullet, \(v_2=35\space m/s\)
Answer:
As, the riffle is at rest, its initial velocity, \(v=0\)
Total initial momentum of the rifle and bullet system \(=(m_1+m_2)v=0\)
Total momentum of the rifle and bullet system after firing:
\(=m_1v_+m_2v_2\)
\(=4(v_1)+0.05\times35\)
\(=4v_1+1.75\)
According to the law of conservation of momentum,
Total momentum after the firing = Total momentum before the firing
\(4v_1+1.75=0\)
\(4v_1=-1.75\)
\9v_1=\frac{-1.75}{4}\)
\(v_1=-0.4375\space m/s\)
The negative sign indicate that the rifle recoils beckwards with a velocity \(v_1=-0.4375\space m/s\)
Question 4: Two objects of masses 100 g and 200 g are moving alont the same line and direction with velocity of 2 \(ms^{-1}\) and 1 \(ms^{-1}\), respectively. They collide asn after the collision, the first object moves at a velocity of 1.67 \(ms^{-1}\). Determine the velocity of the second object.
Given:
Mass of one of the objects, \(m_1=100\space g=0.1\space kg\)
Mass of the other object, \(m_2=200\space g=0.2\space kg\)
Velocity of \(m_1\) before collision, \(v_1=2\space m/s\)
Velocity of \(m_2\) before collision, \(v_2=1\space m/s\)
Velocity of \(m_1\) after collision, \(v_3=1.67\space m/s\)
Answer:
Velocity of \(m_2\) after collision \(=v_4\)
According to the law of conservation of momentum:
Total momentum before collision= Total momentum after collision
\(m_1v_1+m_2v_2=m_3v_3+m_4v_4\)
\((0.1)2+(0.2)1=(0.1)1.67+(0.2)v_4\)
\(0.2+0.2=0.167+0.2v_4\)
\(0.4=0.167+0.2v_4\)
\(0.4-0.167=0.2v_4\)
\(v_4=\frac{0.233}{0.2}\)
\(v_4=1.165\space m/s\)
Hence, the velocity of the second object becomes 1.165 m/s after the collision.
NCERT Excercise
Question 1: An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling at a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If not, provide a reason.
Answer: Yes. An object can travel at a non-zero velocity even if it has a net zero external unbalanced force. This is only possible if the item moves at a consistent speed in a specified
direction. As a result, the body is not subjected to any net imbalanced forces. The item will continue to travel at a velocity greater than zero. A net non-zero external unbalanced force
must be supplied to the item to change its state of motion.
Question 2: When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer: Using a stick to beat a carpet; causing the carpet to move quickly, while dust particles trapped in the carpet’s pores prefer to stay still, since inertia of an item resists any change in its state of rest or motion. The dust particles, according to Newton’s first rule of motion, remain at rest as the carpet moves. As a result, dust particles emerge from the carpet.
Question 3: Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer: According to Newton’s First Law of Motion, luggage on a bus’ roof tends to maintain its condition of rest when the bus is at rest and retain its state of motion when the bus is in motion. When the bus starts moving again after a period of rest, luggage on the roof may fall down to maintain the resting spot. Similarly, owing to inertia of motion, luggage on the roof top of a moving bus will tumble forward when it arrives in the rest state. To avoid this, any luggage kept on a bus’s roof should be tied with a rope.
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.
Answer: Option(c). When the ball moves on the ground, the force of friction opposes its movement and after some time ball comes to a state of rest.
Question 5: A truck starts from rest and rolls down a hill with 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).
Given:
Initial velocity of the truck, \(u=0\) (since the truck is initially at rest)
Distanece travelled, \(s=400\space m\)
Time taken, \(t=20\space s\)
Answer: According to the second equation of motion:
\(s=ut+\frac{1}{2}at^2\)
\(400=0+\frac{1}{2}a(20)^2\)
\(400=a(200)\)
\(a=\frac{400}{200}\)
\(a=2\space m/s^2\)
1 metric tonne=100 kg
∴ 7 metric tonnes=7000 kg
From Newton’s second law of motion:
Force, \(F=Mass\times Acceleration\)
\(F=ma\)
\(F=7000\times2\)
\(F=14000N\)
Hence, the acceleration of truck is \(2\space m/s^2\) and the force acting on the truck \(F=14000\space N\)
Question 6: A stone of 1 kg thrown with a velocity of 20 m \(s^{-1}\) across the frozen surface of lake and comes to rest after travelling 50 m. What is the force of friction between the stone and the ice?
Initial velocity of the stone, \(u=20m/s\)
Final velocity of the stone, \(v=0\) (finally the stone comes to rest_
Diatance covered by the stone, \(s=50\space m\)
Answer:
According to the third equation of motion:
\(v^2=u^2+2as\)
\(0^2(20)^2+2\times a\times50\)
\(0=400+100a\)
\(100a=-400\)
\(a=-\frac{400}{100}a=-4\)
\(a=-4\frac{m}{s^2}\)
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, \(m=1\space kg\)
From Newton’s second law of motion:
Force, \)F=Mass\times Acceleration\)
\(F=ma\)
\(F=1\times-4\)
\(F=-4\space N\)
Hence, the force of friction between the stone and the ice \(F=-4\space N\)
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 5000N, then calculate:
(a) The net accelerating force:
Given:
Force exerted by the engine, \(F=40000\space N\)
Frictional force offered by the track, \(F=5000\space N\)
Answer:
Net accelerating force,
\(F_{net}=F-F_{friction}\)
\(F_{net}=40000-5000\)
\(F_{ne}=35000\space N\)
Hence, the net accelerating force \(F_{net}=350000\space N\)
(b) The acceleration of the train; and
Given:
The engine exerts a force of \(40000\space N\) on all the five wagons.
Net accelerating force on the wagons, \(F_{net}=35000\space N\)
Mass of a wagon \(=2000\space kg\)
Number of wagons \(=5\)
Formula:
Total Mass of the wagons,
\(m=Mass\space of \space a \space wagons\times Number\space of\space wagons\)
Answer:
Total Mass of the wagons,
\(m=2000\times5m=100000\space kg\)
Mass of the engine, \(m’=8000\space kg\)
Total mass, \(M=m+m’=10000+8000=18000\space kg\)
From newton’s second law of motion:
\(Fa=Ma\)
\(a=\frac{Fa}{m}\)
\(a=\frac{35000}{18000}\)
\(a=1.944\space m/s^2\)
Hence, the acceleration of the wagons and the train \(a=1.944\space m/s^2\)
(c) The force of wagon 1 on wagon 2.
Answer: The force of wagon 1 on wagon 2= mass of four wagons \(\times\) acceleration
Madd of 4 wagons \(=4\times 2000=8000\space kg\)
\(F=8000\space kg\times1.944\space m/s^2\)
\(F=1552\space N\)
Question 8: An automobile vehicle has a mass of 1500 kg. What be the froce between the vehicle and road if the vehicle is to be stopped with a negative acceleration of \(1.7\space m^{-2}\)?
Given:
Mass of the automobile vehicle, \(m=1500\space kg\)
Final velocity, \(v=0\)
Acceleration of the automobole, \(a=-1.7\space ms^{-2}\)
Answer:
From Newton’s second law of motion
Force = Mass \(\times\) Acceleration \(=1500\times(1.7)=-2550\) N
Hence, the force between the automobile and the road \(=-2550\) N
Negative sign shows that the force is acting in the opposite direction of the vehicle.
Question 9: What is the momentum of an object of mass m, moving with a velocity v?
(a) \((mv)^2\)
(b) \(mv^2\)
(c) \(1/2\space mv^2\)
(d) \(mv\)
Answer: \(mv\)
Mass of the object \(=m\)
Velocity \(=v\)
Momentum = Mass\(\times\) Velocity
Momentum = 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?
Answer: The same amount of force will act in the opposite direction, according to Newton’s third law of motion. Friction is the name of this force. As a result, the cabinet is subjected to a 200 N frictional force.
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.
Answer: The static friction force is quite strong due to the truck’s massive bulk. Because the student’s effort is insufficient to overcome the static friction, the truck cannot be moved. In this circumstance, the net imbalanced force in either direction is zero, which explains why there is no movement. The force exerted by the learner and the force exerted owing to static friction cancel each other out.
As a result, the student is correct in claiming that the two equal and opposing forces cancel each other out.
Question 12: A hockey ball of mass 200 g travelling at \(10\space ms^{-1}\) is struck by a hockey stick so as to return it along its original path with a velocity at \(5\space ms^{-1}\). Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer: Mass of the hockey ball, \(m=200\space g=0.2\space kg\)
velocity of the ball, \(v_1=10\space m/s\)
Intial momentum \(=mv_1\)
velocity of the ball after struck by the stick, \(v_2=-5m/s\)
Final momentum \(=mv_2\)
Change in momentum \(=mv_1-mv_2=m(v_1-v_2)=0.2(10-(-5))=0.2\times15=3kg\space ms^{-1}\)
Hence, the change in momentum of the hockey ball \(=3kg\space ms^{-1}\)
Question 13: A bullet of mass \(10\space g\) travelling horizontally with a velocity of \(150\space ms^{-1}\) strikes a stationary wooden block and comes to rest in \(0.03\space s\). 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.
Given:
Intial velocity of the bullet, \(u=150\space m/s\)
Final velocity, \(v=0\)
Time, \(t=0.03\space s\)
Answer:
Accroding to the first equation of motion, \(v=u+at\)
Acceleration of the bullet, a
\(0=150+(a\times0.03s)a=-\frac{150}{0.03}a=-5000\space m/s^2\)
(Negative sign indicates that the velocity of the bullet is decreasing)
According to the third equation of motion:
\(v^2=u^2+2as\)
\(0^2=(150)^2+2(-5000)s\)
\(0=22,500-10000s\)
\(10000s=22,500s=\frac{22,500}{10000}s=2.25\space m\)
Hence, the distance of penetration of the bullet into the blcok \(s=2.25\space m\)
From Newton’s second law of motion:
Force, \(F=Mass\times Acceleration\)
Mass of the bulle, \(m=10\space g=0.01\space kg\)
Acceleration of the bullet, \(a=-5000\frac{m}{s^2}F=ma=0.01\times-5000=-50\)
Hence, the magnitude of force exrted by the wooden block on the bullet \(=-50\space N\) but it acts in opposite direction.
Question 14: An object of mass 1 kg travelling in a straight line with a velocity of \(10\space ms^{-1}\) 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.
Given:
Mass of the object,
\(m_1=1\space kg\)
Velocity if the object before collision, \(v_1=10m/s\)
Mass of the wooden block, \(m_2=5\space kg\)
Velocity of the wooden block before collision, \(v_2=0\space m/s\)
Answer:
