NCERT Solutions for Class 9 Science Chapter – 8 Force and Laws of Motion
| Textbook | NCERT |
| Class | 9th |
| Subject | Science |
| Chapter | 8th |
| Chapter Name | Force and Laws of Motion |
| Category | Class 9th Science |
| Medium | English |
| Source | Last Doubt |
| NCERT Solutions Class 9th Science Chapter – 8 Force and Laws of Motion Question & Answer were prepared based on the latest exam pattern. We have Provided Force and to help students understand the concept very well. What is law of motion ?, What is inertia notes?, What is the SI unit of force?,What are the 3 types of Newton law?, What is Newton’s law formula?, What is Newton’s 2nd law called?, What is Newton’s first law called?, What is called inertia?, What is Newton’s second law?, What is velocity?, What is 9.8 newton equal to? |
NCERT Solutions for Class 9 Science Chapter- 8 Force and Laws of Motion
Chapter – 8
Force and Laws of Motion
Question & Answer
Page: 91
| 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-repees coin and a one-rupee coin? Answer- (a) A stone of the same size (b) a train (c) a five-rupees coin As the mass of an object is a measure of its inertia, objects with more mass have more inertia. |
| 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 velocity of football changed four times. |
| Question 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Answer- 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 tend 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? Answer- When a moving bus brakes-to a stop: When the bus is moving, our body is also in motion, but due to sudden brakes, the lower part of our body comes to rest as soon as the bus stops. But the upper part of our body continues to be in motion and hence we fall in forward direction due to inertia of motion. When the bus accelerates from rest we fall backwards. • When the bus’ is stationary our body is at rest but when the bus accelerates, the lower part of our body being in contact with the floor of the bus comes in motion, but the upper part of our body remains at rest due to inertia of rest. Hence we fall in backward direction. |
Page: 97
| 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. Answer- When an object experiences a net zero external unbalanced force, in accordance with second law of motion its acceleration is zero. If the object was initially in a state of motion, then in accordance with the first law of motion, the object will continue to move in same direction with same speed. It means that the object may be travelling with a non-zero velocity but the magnitude as well as direction of velocity must remain unchanged or constant throughout. |
| Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain. Answer- The carpet with dust is in state of rest. When it is beaten with a stick the carpet is set in motion, but the dust particles remain at rest. Due to inertia of rest the dust particles retain their position of rest and falls down due to gravity. |
| Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope? Answer- In moving vehicle like bus, the motion is not uniform, the speed of vehicle varies and it may apply brake suddenly or takes sudden turn. The luggage will resist any change in its state of rest or motion, due to inertia and this luggage has the tendency to fall sideways, forward or backward. To avoid the fall of the luggage, it is tied with the 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- (c) there is a force 6n 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 tonnes (Hint : 1 tonne = 1000 kg). Answer- Initial velocity, u = 0 (since the truck is initially at rest) Distance travelled, s = 400 m Time taken, t = 20 sAccording to the second equation of motion:- s = ut + 1/2 at2 Where, Acceleration = a 400 = 0 + 1/2a(20)2 400 = 1/2a(400) a = 2 m/s2 1 metric tonne = 1000 kg (Given) ∴ 7 metric tonnes = 7000 kg Mass of truck, m = 7000 kg From Newton’s second law of motion: Force, F = Mass × Acceleration F = ma = 7000 × 2 = 14000 N Hence, the acceleration of the truck is 2 m/s2 and the force acting on the truck is 14000 N. |
| Question 6. A stone of lkg is thrown with a velocity of 20 ms-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? Answer- Initial velocity of the stone, u = 20 m/s Final velocity of the stone, v = 0 (finally the stone comes to rest) Distance covered by the stone, s = 50 mAccording to the third equation of motion: v2 = u2 + 2asWhere, Acceleration, a (0)2 = (20)2 + 2 × a × 50 a = −4 m/s2 The negative sign indicates that acceleration is acting against the motion of the stone. Mass of the stone, m = 1 kg From Newton’s second law of motion: Force, F = Mass × Acceleration F = ma F = 1 × (− 4) = −4 N Hence, the force of friction between the stone and the ice is −4 N. |
| Question 7. A 80000 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: Answer- (a) Force exerted by the engine, F = 40000 N Frictional force offered by the track, Ff = 5000 N Net accelerating force, Fa = F − Ff = 40000 − 5000 = 35000 N Hence, the net accelerating force is 35000 N.(b) Acceleration of the train = a The engine exerts a force of 40000 N on all the five wagons. Net accelerating force on the wagons, Fa = 35000 N Mass of the wagons, m = Mass of a wagon × Number of wagons Mass of a wagon = 2000 kg Number of wagons = 5 ∴ m = 2000 × 5 = 10000 kg Total mass, M = m = 10000 kg From Newton’s second law of motion: Fa = Ma a = Fa/m = 35000/10000 ms = 3.5ms-2 Hence, the acceleration of the wagons and the train is 3.5 m/s2 |
| 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 ms-2? Answer- Mass of the automobile vehicle, m = 1500 kg Final velocity, v = 0 (finally the automobile stops) Acceleration of the automobile, a = −1.7 ms−2 From Newton’s second law of motion: Force = Mass × Acceleration = 1500 × (−1.7) = −2550 N Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile. |
| 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 (d) mv Answer- (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? Answer-
As the wooden cabinet moves across the floor at a constant velocity and the force applied is 200 N. Hence the frictional force that will be exerted on the cabinet will be less than 200 N. |
| 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 mass of truck is too large and hence its inertia is too high. The small force exerted on the truck cannot move it and the truck remains at rest. For the truck to attain motion, an external large amount of unbalanced force need to be exerted on it. |
| Question 13. A hockey ball of mass 200 g travelling at 10 m s−1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s−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 g = 0.2 kg Hockey ball travels with velocity, v1 = 10 m/s Initial momentum = mv1 Hockey ball travels in the opposite direction with velocity, v2 = −5 m/s Final momentum = mv2 Change in momentum = mv1 − mv2 = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg m s−1 Hence, the change in momentum of the hockey ball is 3 kg m s−1. |
| Question 13. A bullet of mass 10 p travelling horizontally with a velocity of 150 m-1 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. Answer- Mass of bullet, m = 10 g = 10 / 1000 = 0.01 kg Initial velocity, u = 150 m/s Final velocity, v = 0 Time taken, t = 0.03 s Using the first equation of motion v = u + at 0 = 150 + (a ×0.03 s) a= -150 0.03 = -5000 m\s2 (Negative sign indicates that the velocity of the bullet is decreasing.) Now, using the third equation of motion: v2 = u2 + 2as 0 = (150)2 + 2 (−5000) s s= -(150)2\-2(5000) = 22500 \1000 =2.25 m Hence, the distance of penetration of the bullet is 2.25 m. We know that Force, F = Mass × Acceleration F = ma = 0.01 × 5000 = 50 N Hence, the magnitude of force is 50 N. |
| Question 14. An object of mass 1 kg travelling in a straight line with a velocity of 10 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 before the impact and just after the impact. Also, calculate the velocity of the combined object. Answer- Mass of the object, m1 = 1 kg Velocity of the object before collision, v1 = 10 m/s Mass of the stationary wooden block, m2 = 5 kg Velocity of the wooden block before collision, v2 = 0 m/s ∴ Total momentum before collision = m1 v1 + m2 v2 = 1 (10) + 5 (0) = 10 kg ms−1 It is given that after collision, the object and the wooden block stick together. Total mass of the combined system = m1 + m2 Velocity of the combined object = v According to the law of conservation of momentum: Total momentum before collision = Total momentum after collision m1 v1 + m2 v2 = (m1 + m2) v 1 (10) + 5 (0) = (1 + 5) v v = 10/6 = 5/3 m/s The total momentum after collision is also 10 kg m/s. Total momentum just before the impact = 10 kg ms−1 Total momentum just after the impact kg m/s = (m1 + m2)v = 6×5/3 = 10 kg m/s Hence, velocity of the combined object after collision 5/3 m/s = 1.67 m/s |
| Question 15. An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms-1 to 8 ms-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object. Answer- Initial velocity of the object, u = 5 m/s Final velocity of the object, v = 8 m/s Mass of the object, m = 100 kg Time take by the object to accelerate, t = 6 s Initial momentum = mu = 100 × 5 = 500 kg ms−1 Final momentum = mv = 100 × 8 = 800 kg ms−1 Force exerted on the object, F= mv-μ/t = m(v- u)/t = 800 – 500/6 = 300/6 = 50N Initial momentum of the object is 500 kg ms−1. Final momentum of the object is 800 kg ms−1. Force exerted on the object is 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 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. Answer- Rahul gave the correct reasoning and explanation that both the motorcar and the insect experienced the same force and a change in their momentum. As per the law of conservation of momentum. When 2 bodies collide: Initial momentum before collision = Final momentum after collision m1 u1+ m2 u2 = m1 v1+ m2 v2 The equal force is exerted on both the bodies but, because the mass of insect is very small it will suffer greater change in velocity. |
| Question 18. 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 ms-2. Answer- Mass of the dumbbell, m = 10 kg Distance covered by the dumbbell, s = 80 cm = 0.8 m Acceleration in the downward direction, a = 10 m/s2 Initial velocity of the dumbbell, u = 0 Final velocity of the dumbbell (when it was about to hit the floor) = v According to the third equation of motion: v2 = u2 + 2as v2 = 0 + 2 (10) 0.8 v = 4 m/s Hence, the momentum with which the dumbbell hits the floor is = mv = 10 × 4 = 40 kg ms−1 |
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