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I Multiple Choice Questions:
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i) The law of motion given by Newton which gives the definition of force is :
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a) 1st law b) 2nd law c) 3rd law d) none of these
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ii) Inertia of a body is:
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a) Directly proportional to its mass b) Inversely proportional to mass
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c) Independent of its mass d) none of these
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iii) The physical quantity which is a product of mass and velocity of a body is known as:
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a) Inertia b) Momentum c) Force d) Change in momentum.
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iv) Action and reaction are equal and opposite and they act on:
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a) Same bodies b) Different bodies c) Both a and b d) None
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v) A gun recoils after firing to conserve:
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a) Velocity b) Momentum c) Force d) Speed
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II Give one word for the following.
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i) Law stating action and reaction are equal and opposite
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ii) Tendency of a body to oppose any change in its state of rest or uniform motion.
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iii) The quantity of motion possessed by a moving body.
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iv) Product of mass and acceleration.
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v) Physical quantity which causes or tends to cause motion in an object at rest or changes or tends to change the direction of motion of a moving object or the shape or size of the object.
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III Answer the following questions.
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i) 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 ball 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.
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ii) It is difficult to balance our body to balance our body when we accidently step on a peel of banana. Explain why.
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iii) An object of mass 100kg is accelerated uniformly from a velocity of 5m/s to 8m/s in 6 seconds. Calculate the initial and final momentum of the object. Also; find the magnitude of the force exerted on the object?
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iv) Explain why a gun recoils after firing a bullet.
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v) Why do we fall in a forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
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IV High Order Thinking Questions.
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i) The force of friction between the surface of a floor and the surface of a box in contact with the floor is 200N. We wish to move the box on the floor with constant velocity. How much force has to be applied on the box?
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ii) A force of 100N acts on a body moving with a constant velocity of 20m/s on the floor in a straight line. What is the force of friction between the floor and the body? Give reasons for your answer?
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V Project Work.
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Take a big rubber balloon and inflate it fully. Tie its neck with a thread. Also using adhesive tape. fix a straw on the surface of the balloon. Pass a thread through the straw and hold one end of the thread in your hand or fix it on the wall .Fix the other end of the thread into another wall at some distance. Now remove the thread tied onto the neck of the balloon. Let the air escape from the mouth of the balloon. In which direction will the straw move. Explain why will that happen?
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Showing posts with label 9th Force and law of Motion. Show all posts
Showing posts with label 9th Force and law of Motion. Show all posts
27 August 2012
9th Force and Laws of Motion - Assignments and Worksheet Term-1
17 August 2012
IX Physics force motion problems Numerals On Google Search
| Physics Homework Help: Laws of Motion, Net Force |
| PHYSICS FORM 4 [ FORCE AND MOTION-CHAPTER 2] |
| guruthong.files.wordpress.com/2011/05/linear_motion1.pdf |
| Graphs of Motion - Problems - The Physics Hypertextbook |
| physics.info/motion-graphs/problems.shtml |
| Force & Motion - Science-class.net - Science Resources for the . |
| www.science-class.net/Physics/force_motion.htm |
| SPM PHYSICS FORM 4 forces and motion - Upload & Share … |
| www.slideshare.net/MaxWong1/spm-physics-form-4-forces-and-motion |
| Force - definition of force - force in physics |
| physics.about.com/od/glossary/g/force.htm |
| CBSE PHYSICS: Solved Numerical Problems in Force and Laws of motion |
| physicsadda.blogspot.com/2011/05/solved-numerical-problems-in-force-and.html |
| Physics4Kids.com: Motion: Forces - Rader's PHYSICS 4 KIDS.COM |
| www.physics4kids.com/files/motion_force.html |
| Equations of Motion - Problems - The Physics Hypertextbook |
| physics.info/motion-equations/problems.shtml |
| physics motion problem - Physics Help and Math Help – Physics |
| www.physicsforums.com/showthread.php?t=4176 |
| Urdu Physics Lecture About Force And Motion – YouTube |
| QUESTION BANK Class 9th Physics (Forces and Laws of motion) |
| physicsadda.blogspot.com/2011/06/question-bank-class-9th-physics- |
| Force and Motion - Links to all IU HEP DOE Tasks |
| hep.physics.indiana.edu/~rickv/force_and_motion.html |
| The Physics Classroom |
| www.physicsclassroom.com/ |
| Physics Tutorial Lesson: Planetary Motion Gravity Centripetal |
| www.youtube.com/watch?v=_S0o8yAVcsA |
| une 25, 2012 05:55:0424physics force motion problems |
| Simple harmonic motion problems – Physics Help and Math Help … |
| www.physicsforums.com/showthread.php?t=99092 |
12 August 2012
IX Term -1 Solved numerical Gravitation and Laws of motion
Q.A rifle of mass 3 kg fires a bullet of mass 0.03 kg. The bullet leaves the barrel of the rifle at a velocity of 100 m/s. If the bullet takes 0.003 s to move through its barrel, calculate the force experienced by the rifle due to the recoil.
Ans: Mass of the rifle (m1) = 3 kg Mass of the bullet (m2) = 0.03 kg
Initial velocity of the rifle (u1) = u2 = 0, Final velocity of the rifle (v1) = ?
Final velocity of the bullet (v2) = 100 m s-1
According to law of conservation of momentum = m1u1 + m2u2 = m1v1 + m2v2
Using the law of conservation of momentum, 0 + 0 = 3v1 + 0.03 x 100 =-1m/s
Negative sign shows that the rifle moves in a direction opposite to that of the bullet,
Force experienced by the rifle due to its recoil= m1v1/t =(3x-1)/0.003=-1000N
Therefore, the person would experience a force of 1000 N in the backward direction due to the recoil of the rifle.
Initial velocity of the rifle (u1) = u2 = 0, Final velocity of the rifle (v1) = ?
Final velocity of the bullet (v2) = 100 m s-1
According to law of conservation of momentum = m1u1 + m2u2 = m1v1 + m2v2
Using the law of conservation of momentum, 0 + 0 = 3v1 + 0.03 x 100 =-1m/s
Negative sign shows that the rifle moves in a direction opposite to that of the bullet,
Force experienced by the rifle due to its recoil= m1v1/t =(3x-1)/0.003=-1000N
Therefore, the person would experience a force of 1000 N in the backward direction due to the recoil of the rifle.
Q. Action and reaction forces is never equal to zero. Why?
Ans: The action and reaction force are equal and opposite but their resultant is never equal to zero or cancel because The action and reaction force are acting on two different bodies.
Q. A bullet leaves a rifle with a velocity of 100m/s and the rifle of mass 2.5 kg recoils with a velocity of 1m/s. Find the mass of the bullet?
Ans: According to law of conservation of momentum = m1u1 + m2u2 = m1v1 + m2v2
Þ 0=2.5x1 +m2 x100 Þ 2.5/100= the mass of the bullet Þ 25gm= the mass of the bullet
Q. A body of mass 5 kg undergoes a change in speed from 30 to 40m/s . Calculate its increase in momentum?
Ans: Increase in momentum =m(v-u) =5 kg(40-30)m/s=50kgm/s
Q. Two bodies of mass 10 kg and 12 kg are falling freely. What is the acceleration produced in the bodies due to force of gravity?
Ans: 9.8 m/s2.ast he acceleration due to gravity produced in both the bodies is the same as it is independent of the mass of the body.
Q. What will happen to the force of gravitation between two objects A and B if the distance between them is reduced to half?
Ans: F=1/r2
F’ = 1/(1/2r)2 =4(1/r2) =4(F)
if the distance between two objects A and B is reduced to half the force of gravitation between two objects A and B increases 4 times
Q. What will happen to the force of gravitation between two objects A and B If the mass of the object A is doubled.
Ans: F = m M
F’= 2mxM=2(mM)=2F
the force of gravitation between two objects A and B increases 2 times
Q. Show that universal gravitational constant is nothing but force of gravitation between two unit masses separated by unit distance.
Ans: F=GmM/r2
if m=M=1kg and r =1m
then, F = G i.e., gravitational constant is equal to the force of gravitation.
Q. A boy drops a stone from a cliff, which reaches the ground in 20 seconds. Calculate the height of the cliff.
Ans: the height of the cliff.=S=ut+1/2at2 =0+1/2x10x20x20=2000m=2km
23 July 2012
Newton's second and Third Laws of motion :In Book Principia
“When an external, unbalanced force acts on an object, the object will accelerate in the same direction of the force. Then,
The acceleration varies directly as the force,
=> a a F
=> a a F
This just means that if the force increases, the acceleration will increase. If the force decreases,
the acceleration decreases.
The acceleration varies inversely the mass,
=> a a 1/m
=> a a 1/m
This means that if the mass is bigger, the acceleration is less. If the mass is less, the acceleration is more.
=> a a F/m
=> F = k ma Here k is a constant
=> F = ma
Hence, Force is equal to product of mass and acceleration
F = force (Newton, N)
m = mass (kg)
a = acceleration (m/s2)
By definition a one kilogram mass will be accelerated at 1 m/s2 if a 1 Newton force is applied to it.
Don't forget to turn the mass into kilograms when you solve numerical problems.
Newton's Third Law (Action-Reaction)
What happens when you have objects interacting, affecting each other?
“For every action force there is an equal and opposite reaction force.”
Action: the tires on a car push on the road… Reaction: the road pushes on the tires.
Action: while swimming, you push the water backwards... Reaction: the water pushes you forward.
Post related to 9th Force and law of Motion.
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22 July 2012
Newton's First Law :In Book Principia
Newton’s Laws of Motion, as written in his book his book the Principia.On July 5, 1687
The First Law (The Law of Inertia) :
“Every body continues in a state of rest or uniform velocity in a straight line, unless an external force acts on it.”
An object resisting a change in its “state of motion” (stopped or moving in a straight line) is something that Newton called inertia. That’s why this law is sometimes called the Law of Inertia.
The idea of inertia is confirm us if an object is moving right now, it will keep on moving in the direction of force.
It is not necessarily you see this because of the effects of friction.
If you roll a ball across the floor it will slow down and eventually stop.
This is not because it is violating inertia, but because there is an external force acting on it--- friction.
Why people are injured in car accidents when they do not wear seat belts?
In the car the person try to be in inertia of motion, so the person will continue to move forward at the same speed as the car was originally traveling and hit car .
Newton’s First Law goes against what Aristotle said, but is basically what Galileo had said a few years earlier.
The First Law (The Law of Inertia) :
“Every body continues in a state of rest or uniform velocity in a straight line, unless an external force acts on it.”
An object resisting a change in its “state of motion” (stopped or moving in a straight line) is something that Newton called inertia. That’s why this law is sometimes called the Law of Inertia.
The idea of inertia is confirm us if an object is moving right now, it will keep on moving in the direction of force.
It is not necessarily you see this because of the effects of friction.
If you roll a ball across the floor it will slow down and eventually stop.
This is not because it is violating inertia, but because there is an external force acting on it--- friction.
Why people are injured in car accidents when they do not wear seat belts?
In the car the person try to be in inertia of motion, so the person will continue to move forward at the same speed as the car was originally traveling and hit car .
Newton’s First Law goes against what Aristotle said, but is basically what Galileo had said a few years earlier.
Aristotle had said that if you stop pushing an object, it will come to rest. He believed that “at rest” was the natural state for any object.
Galileo told us to ignore friction and basically came up with Newton’s First Law. It is called Newton’s First Law because he was the one that formally published it and had the mathematical proofs to back it up.
Google search for PPT
[PPT]
Galileo told us to ignore friction and basically came up with Newton’s First Law. It is called Newton’s First Law because he was the one that formally published it and had the mathematical proofs to back it up.
Google search for PPT
[PPT]
Newton's Three Laws of Motion
iws.collin.edu/.../Newtons%20Three%20Laws/Newtons... - United States
Newton's First law of motion. Comments. This means that if you leave a book on a bench over night, when you return in the morning, unless an outside force ...07 July 2012
Challenging Questions:IX Physics: Force and laws of Motion
CBSE PHYSICS : Challenging Questions
1. A machine gun can fire bullet of 50gm at speed of 2000m/s ,the man holding the gun can exert an average force of 200N against the gun . Calculate the maximum number of bullet which he can fire per minute ( Ans: 120 )2. A machine gun can fire 20 bullet per second ,if the speed of the bullet is 100m/s and if each bullet is of weight 100gms , then calculate force require to hold the gun in its position. ( Ans: 200N )
3. A man in a circus show jumps from a height of 10m and is caught by a net spread below him. The net sags below by 2m . Find the force exerted by the net on the man to stop him. Mass is 60kg and acceleration is 10m/s2. (Ans: 300N)
4. A machine gun is fitted on a trolley of mass 2000kg on a horizontal frictionless surface. If the gun fires bullet each of mass 10gm with a velocity of 500m/s with respect to the trolly and if the number of bullet fired per second is 10 then calculate the average thrust exerted by ejected bullet on the system (Ans: 50N)
5. Two identical blocks A and B each of mass M are connected to each other through a light string. The system is placed on a smooth horizontal floor. When constant force F is applied horizontally on the block A, find the tension in the string. (Ans: F/2)
6. A hunter fires 50 g bullets from a machine gun. If each bullet moves with a velocity of 150ms−1 how many such bullets are to be fired into a tiger of mass 60 kg coming towards him with a velocity of 10ms−1 in order to stop the tiger? (Ans: 80)
7. A machine gun fires 240 bullets per minute with a velocity of 500m/s . If the mass of each of the bullet is 5×10−2 kg , find the power of the gun.( Ans: 2500W)
8. The momentum of a body is double. By what percentage does its kinetic energy increase?
(Ans: 300%)
(Ans: 300%)
9. A machine gun is mounted on a flat rail road car. The gun is firing bullets at the rate of 10 bullets per second each of mass 10 g. The bullets come out with velocity 500 m/s relative to the car. Calculate the acceleration of the car at the instant when its mass is 200 kg. Also calculate the force at the instant. (Ans: F = 50N)
10. A machine gun fires 240 bullets per minute with certain velocity . If the mass of each bullets 10-2 kg and the power of the gun is 7.2 k.W, find the velocity which each bullet is fire. (Ans: 600 m/s)
"Post Solution in comment or mail jsuniltutorial@gmail.com"
Enrich your study: IX:Force and laws of motion solved Problems
CBSE PHYSICS: IX:Force and laws of motion solved Problems
1. An athlete runs a certain rest before taking a long jump. Why?
Solution: An athlete runs a certain distance to accelerate himself and gain enough momentum so that he can jump through the maximum possible length.
2. Springs are provided in car seats. Why?
Solution: The springs in the car seats absorb shock (sudden jumps) due to the roughness of the road. Thus, making the ride more comfortable.
3. A gun of mass 1500 kg fires a shell of mass 15kg with velocity 150 m/s. calculate velocity of recoil of the gun.
Solution: Before firing, total momentum of the gun and shell is = 0 (they were all at rest)
After firing, the momentum of the shell = 15 X 150 = 2250 Ns
The momentum of the gun is = 1500v Ns
By conservation of momentum,
Total momentum before firing = Total momentum after firing
0 = 2250 + 1500v
v = -2250/1500 = -1.5 m/s
4. Why cricketer pulls his hands backwards while catching the ball
6. What is impulse of Force?
I = F x t= change in p
Like force, impulse of force is also a vector quantity. Direction of impulse of force is same as the direction of force. Unit of impulse of force is kg m/s or Ns.Like force, impulse of force is also a vector quantity. Direction of impulse of force is same as the direction of force. Unit of impulse of force is kg m/s or Ns.
1. An athlete runs a certain rest before taking a long jump. Why?
Solution: An athlete runs a certain distance to accelerate himself and gain enough momentum so that he can jump through the maximum possible length.
2. Springs are provided in car seats. Why?
Solution: The springs in the car seats absorb shock (sudden jumps) due to the roughness of the road. Thus, making the ride more comfortable.
3. A gun of mass 1500 kg fires a shell of mass 15kg with velocity 150 m/s. calculate velocity of recoil of the gun.
Solution: Before firing, total momentum of the gun and shell is = 0 (they were all at rest)
After firing, the momentum of the shell = 15 X 150 = 2250 Ns
The momentum of the gun is = 1500v Ns
By conservation of momentum,
Total momentum before firing = Total momentum after firing
0 = 2250 + 1500v
v = -2250/1500 = -1.5 m/s
4. Why cricketer pulls his hands backwards while catching the ball
Solution: cricketer pulls his hands backwards while catching the ball. When he does so, momentum of the ball reduces slowly, and time t required for this increases. As per F =DP/t , as t increase, magnitude of F decreases. As a result cricketer can catch the ball, easily, without any injury.
5. Why A karate player in order to break a brick, hits it quickly
Solution: A karate player in order to break a brick, hitsit quickly, so in a short time there is a large change
in momentum and as per F =change in p/t as large force acts on the brick and it breaks.in momentum and as per F =change in p/t as large force acts on the brick and it breaks.in momentum and as per F =change in p/t as large force acts on the brick and it breaks.
Solution: A karate player in order to break a brick, hitsit quickly, so in a short time there is a large change
in momentum and as per F =change in p/t as large force acts on the brick and it breaks.in momentum and as per F =change in p/t as large force acts on the brick and it breaks.in momentum and as per F =change in p/t as large force acts on the brick and it breaks.
6. What is impulse of Force?
I = F x t= change in p
Like force, impulse of force is also a vector quantity. Direction of impulse of force is same as the direction of force. Unit of impulse of force is kg m/s or Ns.Like force, impulse of force is also a vector quantity. Direction of impulse of force is same as the direction of force. Unit of impulse of force is kg m/s or Ns.
7. A ball of 150 g is thrown at 20 m/s towards the batsman. He hits the ball in the direction oppsite to intial direction of motion with velocity 25 m/s. If the ball is hit in 0.01 s, calculate change in momentum of the ball and force applied by the batsman on the ball.
Solution: m = 150 g = 150/1000kg= 0.15 kg
u = 20 m/s, v = 25 m/s, t = 0.01 s
Initial momentum of the ball
pi = mu = (0.15)(20) = 3 kgm/s
Final momentum of the ball after being hit
pf = mv = (0.15)( - 25) = - 3.75 kgm/s
Here negative sign indicates that direction of motion of the ball after hitting is opposite to initial
direction of motion of the balls.
Change in momentum of the ball Dp = pf - pi = ( - 3.75 - 3) = - 6.75 kgm/s
So change in momentum of the bat = 6.75 kgm/s
Force applied by the bat,
F =Change in p/t= 6.75/ 0.01 = 675 kg m/s2 = 675 N
8.A bullet of 20 g is fired horizontally from a pistol of 2 kg mass with velocity 150 ms-1. How much would be the velocity of pistol in backward direction after firing the bullet ?
Solution : Mass of bullet m1 = 20 g = 0.02 kg
Mass of pistol m2 = 2 kg
Initial velocity of bullet u1 = 0 m/s
Initial velocity of pistol u2 = 0 m/s
Final velocity of the bullet = v1 = 150 ms-1
Final velocity of the pistol v2 = ?
According to law of conservation of momentum
m1v1 + m2v2 = m1u1 + m2u2
(0.02)(150) + (2)(v2) = (0.02)(0) + (2)(0)
v2 = - (0.02)(150)/2 = - 1.5 ms–1
Here, negative sign indicates that motions ofpistol and bullet are in opposite direction.
9. A bullet of mass 20 g is fired from a gun of mass 10kg with a velocity of 180m/s. Find the velocity of the recoil of the gun . Find the force required to stop the gun before it moves 20cm. Ans: Here we shall use the conservation of momentum to calculate the recoil velocity.
Momentum of the bullet + gun before firing = 0
Momentum of the bullet + gun after firing = (20 × 10 -3) (180) + (10)(v)
‘v’ is the recoil velocity of the gun.
So, by momentum conservation,
(20 × 10 -3 )(180) + (10)(u g ) = 0
=> u g = -0.36 m/s
The negative sign indicates that the gun’s velocity is directed opposite to the velocity of the bullet.
Now,
Distance travelled before coming to rest, S = 20 cm = 0.2 m
Let ‘a’ be the acceleration with which it stops.
Using, v 2 = u 2 + 2as
=> 0 = (180) 2 + 2a × 0.2
=> a = -8100 m/s 2
So, force on the bullet, F = ma = (20 × 10 -3 ) × (-8100) = -162 N
Negative sign indicates the force is against the motion of the bullet.
10. Two billiard balls each of mass 50 g moving in opposite directions with a speed of 10m/s colloid and rebound with the same velocity .What is the impulse imparted to each ball due to other?
Ans: Impulse = change is momentum.
Initial velocity of one billiard ball, u = 10m/s Final velocity after collision, v = -10m/s
Change in momentum in one billiard ball, Δ p = m(v-u) = -1kgm/s
Similarly, change in momentum of the other billiard ball, Δ p' = +0.1kgm/s
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Solution: m = 150 g = 150/1000kg= 0.15 kg
u = 20 m/s, v = 25 m/s, t = 0.01 s
Initial momentum of the ball
pi = mu = (0.15)(20) = 3 kgm/s
Final momentum of the ball after being hit
pf = mv = (0.15)( - 25) = - 3.75 kgm/s
Here negative sign indicates that direction of motion of the ball after hitting is opposite to initial
direction of motion of the balls.
Change in momentum of the ball Dp = pf - pi = ( - 3.75 - 3) = - 6.75 kgm/s
So change in momentum of the bat = 6.75 kgm/s
Force applied by the bat,
F =Change in p/t= 6.75/ 0.01 = 675 kg m/s2 = 675 N
8.A bullet of 20 g is fired horizontally from a pistol of 2 kg mass with velocity 150 ms-1. How much would be the velocity of pistol in backward direction after firing the bullet ?
Solution : Mass of bullet m1 = 20 g = 0.02 kg
Mass of pistol m2 = 2 kg
Initial velocity of bullet u1 = 0 m/s
Initial velocity of pistol u2 = 0 m/s
Final velocity of the bullet = v1 = 150 ms-1
Final velocity of the pistol v2 = ?
According to law of conservation of momentum
m1v1 + m2v2 = m1u1 + m2u2
(0.02)(150) + (2)(v2) = (0.02)(0) + (2)(0)
v2 = - (0.02)(150)/2 = - 1.5 ms–1
Here, negative sign indicates that motions ofpistol and bullet are in opposite direction.
9. A bullet of mass 20 g is fired from a gun of mass 10kg with a velocity of 180m/s. Find the velocity of the recoil of the gun . Find the force required to stop the gun before it moves 20cm. Ans: Here we shall use the conservation of momentum to calculate the recoil velocity.
Momentum of the bullet + gun before firing = 0
Momentum of the bullet + gun after firing = (20 × 10 -3) (180) + (10)(v)
‘v’ is the recoil velocity of the gun.
So, by momentum conservation,
(20 × 10 -3 )(180) + (10)(u g ) = 0
=> u g = -0.36 m/s
The negative sign indicates that the gun’s velocity is directed opposite to the velocity of the bullet.
Now,
Distance travelled before coming to rest, S = 20 cm = 0.2 m
Let ‘a’ be the acceleration with which it stops.
Using, v 2 = u 2 + 2as
=> 0 = (180) 2 + 2a × 0.2
=> a = -8100 m/s 2
So, force on the bullet, F = ma = (20 × 10 -3 ) × (-8100) = -162 N
Negative sign indicates the force is against the motion of the bullet.
10. Two billiard balls each of mass 50 g moving in opposite directions with a speed of 10m/s colloid and rebound with the same velocity .What is the impulse imparted to each ball due to other?
Ans: Impulse = change is momentum.
Initial velocity of one billiard ball, u = 10m/s Final velocity after collision, v = -10m/s
Change in momentum in one billiard ball, Δ p = m(v-u) = -1kgm/s
Similarly, change in momentum of the other billiard ball, Δ p' = +0.1kgm/s
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