In this post we are going to show the detailed solution for class 9th science chapter 9. All the solution are prepared by our esteemed who are very well experienced in the teaching.
Chapter 9 Force and Laws of Motion
Question 1. An object experience 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 – When a net zero external force is applied on the body, it is possible for the object to be travelling with a non- zero velocity. In fact, once an object comes into motion and there is a condition in which its motion is unopposed by any external force; the object will continue to remain in motion. It is necessary that the object moves at a constant velocity and in a particular direction.
Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Solution – Beating of a carpet with a stick; makes the carpet come in motion suddenly, while dust particles trapped within the pores of carpet have tendency to remain in rest, and in order to maintain the position of rest they come out of carpet. This happens because of the application of Newton’s First Law of Motion. Which states that any object remains in its state unless any external force is applied over it.
Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope ?
Solution – Luggage kept on the roof of a bus has the tendency to maintain its state of rest when bus is in rest and to maintain the state of motion when bus is in motion according to Newton’s First law of Motion. When bus will come in motion from its state of rest, in order to maintain the position of rest, luggage kept over its roof may fall down. Similarly, when a moving bus will come in the state of rest or there is any sudden change in velocity because of applying of brakes, luggage may fall down because of its tendency to remain in the state of motion. This is the cause that it is advised to tie any luggage kept on the roof a bus with a rope so that luggage can be prevented from falling down.
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. Explanation: When ball moves on the ground, the force of friction opposes its movement and after some time ball comes to the state of rest.
Question 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne – 1000 kg.)
Solution – Given –
Initial velocity of truck (u) = 0 (Since, truck starts from rest) Distance travelled, s = 400m
Time (t) = 20 Acceleration (a) = ?
We know that,
⇒ s = ut + ½ at²
⇒ 400m = 0 × 20s + ½ × a × (20)²
⇒ 400m = 0 + ½ × a × 400 s²
⇒ 400m = a × 200 s²
⇒ a = 400 ÷ 200 m/s²
⇒ a = 2 m/s¯²
Force acting upon truck:
Given – Mass of truck = 7 tonne = 7 × 1000 kg ⇒ 7000 kg
We know that ,
Force , P = m × a
Therefore, P = 7000 kg × 2ms¯²
Or, P = 14000 Newton
Thus, Acceleration = 2ms¯² and force acting upon a truck in the given condition = 14000 N
Question 6. A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice ?
Solution – Given –
Mass of stone = 1 kg
Initial velocity, u = 20 m/s
Final velocity, v = 0 (as stone comes to rest)
Distance covered, s = 50 m
Force of friction = ?
We know that,
⇒ v² – u² = 2 as
⇒ 0² – 20² = 2a × 50m
⇒ – 400 = 100a m
⇒ a = -400/100 m
⇒ a = -4ms‾²
Now, we know that,
Force, F = mass × acceleration
Therefore, F = 1 kg × – 4ms¯²
Or, F = – 4ms¯²
Thus, force of friction acting upon stone = -4ms¯². Here negative sing shows that force is being applied in the opposite direction of the movement of the stone.
Question 7. A 8000 kg engine pull a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the accelerating of the train; and
(c) the force of wagon 1 on wagon 2.
Solution -Given,
force of engine = 40,000 N ‘and ‘ Force of friction = 5000N
Mass of engine = 8000 kg
Total weight of wagons = 5 × 2000 kg = 10,000 kg
(a) The net accelerating force:
⇒ Force exerted by engine – Force of friction
⇒ 40,000 N – 5000 N = 35,000 N
(b) The acceleration of the train:
We know that, F = m × a
⇒ Or, 35,000 N = (mass of engine + mass)
⇒ 35,000 N = 8000 kg + 10,000 kg) × a
⇒ 35,000 N = 18,000 kg × a
⇒ a = 35,000 / 18,000
⇒ a = 1.944 m/s²
Thus, acceleration of train is 1.944 m/s²
(c) The force of wagon 1 on wagon 2
⇒ Since, net accelerating force = 35,000 N
⇒ Mass of all 5 wagons = 10,000 kg
⇒ We know that, F = m × a
⇒ a = 35000N / 10,000 kg
⇒ 3.5 ms¯²
Therefore, force wagon 1 on 2 = mass of four wagons × acceleration
F = 4 × 2,000 kg × 3.5 ms¯²
F = 8,000 kg × 3.5 ms‾²
F = 28,000 N
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 wit ha negative acceleration of 1.7 ms‾² ?
Solution – Given –
Mass of the vehicle m = 1500 kg
Acceleration, a = 1.7 ms‾²
We know that force ,
⇒ F = mass × acceleration
⇒ F = 1500 × -1.7
⇒ F = – 2550 N
Negative sign shows that force is in opposite direction of motion.
Question 9. What is the momentum of an object of mass m, moving with a velocity v ?
(a) (mv)² (b) mv² (c) ½ mv² (d) mv
Answer – We know that,
⇒ Momentum = mass × velocity
⇒ Thus,
⇒ Momentum = mv
So, option (d) is correct.
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 – Since we move the cabinet with a constant velocity, its acceleration will be = 0
Since acceleration is 0 no net force applied on it.
So total force on body = 0
Now, total for on body will be force applied and friction.
⇒ Force on body + Friction = 0
⇒ 200 N + Friction = 0
⇒ Friction = 0 – 200 N
⇒ Friction = – 200 N
Negative sign shows that force is in opposite direction of motion.
Question 11. Two object, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms¯¹ before the collision during which they stick together. What will be the velocity of the combined object after collision ?
Solution – Mass of object 1 = m1 = 1.5 kg
Mass of object 2 = m2 = 1.5 kg
Velocity of object 1 = u1 = 2.5 ms¯¹
Velocity of object 2 = u2 = -2.5ms¯¹
Note: u2 is negative since it is in opposite direction of u1.
After collision, the object stick together
Combined mass = m1 + m2
= 1.5 + 1.5
= 3 kg
Velocity of combined object = v
Finding momentum before and after
Momentum before:
Momentum = momentum of object 1 + momentum of object 2
= 1.5 × 2.5 + 1.5 × -2.5
= 1.5 × 2.5 – 1.5 × 2.5
= 0
Momentum after:
Momentum = Total mass × velocity after
= 3 × v
= 3v
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
⇒ 0 = 3v
⇒ v = 0/3
⇒ v = 0 m/s
Question 12. 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 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 – The logic of the student is wrong.
According to the third law of motion,
For every action, there is an equal and opposite reaction
But, they act on two different bodies
Action and reaction do not cancel each other as they are on two different bodies
Not the same body. Hence action and reaction can never cancel each other.
In case of truck, when we puss the truck
The logic that the two forces cancel each other is not correct since action and reaction force act on two different bodies and hence cannot cancel each other.
The reason that why the truck does not move even through a force is applied on it because, due to the large mass of the truck the force required to produce acceleration would be large. (F = m × a)
The force applied on truck must have not be enough to overcome the frictional force between the wheels of the truck and the ground and hence the truck does not move.
Question 13. A hockey ball of mass 200 g travelling at 10 m/s is stuck 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 hockey ball, m = 200 g = 200/1000 kg = 0.2 kg
Initial velocity of hockey ball, u = 10 m/s
Final velocity of hockey ball, v = -5 m/s (because direction becomes opposite)
Change in momentum = ?
We know that,
Momentum = mass × velocity
Therefore, Momentum of ball before getting struck
⇒ 0.2 kg × 10 m/s = 2 kg m/s
Momentum of ball after getting struck = 0.2 kg × -5 m/s = -1 kg m/s
Thus,
Initial momentum = 2 kg m/s
Final momentum = -1 kg m/s
Change in momentum = Initial momentum – final momentum
= 2 – (-1)
= -3 kg m/s
Note: (Negative sign denotes change in momentum in the direction opposite to the direction of initial momentum of the ball.)
Question 14. 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 bullet = m = 10 g
= 10 / 100
= 0.01 kg
Initial velocity of the bullet = u = 150 m/s
Time taken = t = 0.03 s
Since the bullet come to rest,
Final velocity = v = 0 m/s
We need to find distance of penetration and force
We cannot find distance before finding acceleration
Finding acceleration
We know v, u and t
Finding a using 1st equation of motion
⇒ v = u + at
⇒ 0 = 150 + 0.03 × a
⇒ 0.03 a = – 150
⇒ a = -150 × 100 / 3
⇒ a = -5000 m/s²
Finding distance
We know v , u and a
Finding distance (s) using 3rd equation of motion
⇒ v² – u² = 2as
⇒ 0² – 150² = 2 × -5000 × s
⇒ -22500 = 10,000s
⇒ s = -22500 / 10,000
⇒ s = 2.25 m
Distance of penetration of bullet is 2.25 m
Finding Force
We know that,
⇒ Force = mass × acceleration
⇒ F = 0.01 × -5000
⇒ F = -50 N
Magnitude of force is 50 N.
Note: Magnitude is only the value without the sign.
Question 15. 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 – Given, mass of moving object, m1 = 1 kg
Mass of the wooden block, m2 = 5kg
Initial velocity of object, u1 = 10 m/s
Since, the wooden block is stationary,
Initial velocity of wooden block, u2 = 0
After, collision they move together
Mass of the system = m1 + m2
= 1 + 5
= 6 kg
Let the velocity after collision = v
Finding momentum before and after collision
Momentum Before
Momentum = momentum of object + momentum of block
= 1 × 10 + 5 × 0
= 10 + 0
= 10 kg m/s
Momentum After
Momentum = total mass × velocity after
= (m1 + m2) × v
= 6v
Since momentum is conserved,
Total momentum before collision = Total momentum after collision
⇒ 10 = 6v
⇒ v = 10/ 6
⇒ v= 5/3 m/s
Therefore,
Momentum before collision = 10 kg m/s
Since Momentum before collision = momentum after collision
Momentum after collision = 10 kg m/s
Velocity of combined object = 5/3 m/s
Question 16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 ss. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Solution – Given,
Mass of the object = m = 100 kg
Initial velocity of the object , u = 5 m/s
Final velocity of the object, v = 8 m/s
Time taken, t = 6 s
Initial momentum
⇒ m × u
⇒ 100 × 5
⇒ 500 kg m/s
Final momentum
⇒ m × v
⇒ 100 × 8
⇒ 800 kg m/s
Finding magnitude of force
By newton’s 2nd law of motion,
Force = Change in momentum / time taken
= Final momentum – Initial momentum
= (800 – 500) / 6
= 300 /6
= 50 N
Question 17. 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 the 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 – Let’s look what all 3 people said,
- Kiran – Due to greater change in momentum of insect, as compared to the car and insect died.
- Akhtar – Since motorcar was moving with a larger velocity, it exerted a larger force on the insect. And the insect died.
- Rahul – Both motorcar and insect experienced the same force and change in their momentum.
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
When the insect hits the cars,
it experiences larger change in momentum
(Because initial momentum will be less because it’s mass is less and final momentum will be much more because mass of car + mass of insect)
[Hence, Kiran Observation is correct ]
If the car was moving slowly, the insect might have not died.
Since the car was moving with high velocity, it exerted higher force on the insect.
[So, Akhtar observation is correct]
Now, Rahul said that both motorcar and insect experienced the same force and change in their momentum.
[Rahul is partially correct] because –
- According to the third law of motion – For every action, there is an equal and opposite reaction. So, the force exerted by car on insect will be same as force exerted by insect on car.
- But, change in momentum of insect will be much higher than change in momentum of car.
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 m/s¯².
Solution – Given –
Mass of dumb-bell = 10 kg
Acceleration, a = 10 m/s ²
Since we drop the dumb-bell
Initial velocity of dumb-bell, u = 0
Distance, s = 80 cm
= 80/100 = 0.8
Let, final velocity of dumb-bell = v
Momentum = mass × velocity
= m × v
We have u, a and s,
Finding v using 3rd equation of motion:
⇒ v² – u² = 2as
⇒ v² – (0)² = 2 × 10 × 0.8
⇒ v² = 16
⇒ v² = ± 16
⇒ v = ± 4 m/s
Since the acceleration is positive,
Therefore, Velocity increases
So, we take positive value of v
Thus, v = 4 m/s
Now,
Momentum = m × v
= 10 × 4
= 40 kg m/s
You can see other posts here:
- NCERT Class 9th Science Chapter 8 – Motion Summary
- NCERT Science Class 9th Chapter 8 Motion
- NCERT Science Class 9th Chapter 9 Force and Laws of Motion Summary
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