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Dear students, prepare for physics class 9th chapter 3 long questions. These important long questions are carefully added to get you best preparation for your 9th class physics ch. 3 exams.
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Our database contains a total of 0 questions for physics Short Questions. You’ll prepare using this huge databank.

Question: 1
Explain law of conservation of momentum? Also drive its formula?
Answer: 1
1-23
The momentum of an isolated system of two or more than two or more than two interacting bodies remain constant.<div>An isolated system is a group of interacting bodies on which no external force is acting<gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856591353921922309943"> .</gwmw> IF no unbalanced on net force acts on a <gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856591870407530247526">system then</gwmw> its momentum remains constant.</div><div>Consider a system of gun and a bullet. Before firing, the Velocity of the bullets as well as that of <gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856593326757021521887">gun</gwmw>was zero. <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856593889829120562564">therefore</gwmw> the total momentum of both the objects was also zero.</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856593945041009595596">we</gwmw> can write it as,</div><div><gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856594189233096819997">Total momentum</gwmw> of gun and bullet before fixing =0</div><div>When the gun is fired bullet shoots out of guns and acquire momentum<gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856595275934442500742">.</gwmw>To converse momentum the <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856594876513304236821">qun</gwmw>recoils backward. <gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856597024157513523626">Now according</gwmw> to the law of conversation of momentum, the total momentum of the gun and <gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856597024158289211235">bullet</gwmw> will also be zero after the gun is fired.</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856597816053245211755">let</gwmw> <gwmw class="ginger-module-highlighter-mistake-type-2" id="gwmw-15856597816050048326423">m</gwmw> be the mass of the bullet and v be its velocity on firing the gun; M be the mass of the gun and V is the Velocity with which it <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856597816059063836156">recoids</gwmw>. <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856598699918591684447">thus</gwmw> the total momentum of the gun is fired will be: The momentum of the gun and the bullet after the gun is fired = Mv+ <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856598699914708670096">mv</gwmw></div><div>According to the law of conversation of momentum.</div><div><gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856599864793239791545">Total momentum</gwmw> before firing = Total momentum after firing</div><div> MV + <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856600037038816583236">mv</gwmw>=0</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856600095157063118982">Mv</gwmw> =-<gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856600198523688183966">mv</gwmw></div><div><gwmw class="ginger-module-highlighter-mistake-anim ginger-module-highlighter-mistake-type-1" id="gwmw-15856600453515943532301">or</gwmw> v=-m/M v</div>
Question: 2
Two masses 26 Kg and 24 kg<gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856603890924975919179">are attached</gwmw>to the ends of<gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856603890921330649628">string</gwmw>which passes over a frictionless pulley. 26 kg is<gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856603896464221407055">laying</gwmw>over a smooth horizontal table. 24 N mass is
Answer: 2
2-23
Solution<div> <sub>1</sub>= 24kg</div><div> <sub>2</sub>=26 kg</div><div> = 10ms-²</div><div>Solution;</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856607517106209739515">a</gwmw>= m<sub>1</sub>g/m<sub>1</sub> +m<sub>2</sub></div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856609284918583311988">a</gwmw>= 24x10/24+26</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856609459738634686564">a</gwmw> =240/50</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856610400770384465222">a</gwmw> =4.8ms-²</div><div> T =m<sub>1</sub> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856610771317988133651">m</gwmw><sub>2</sub> g/m<sub>1</sub> +m<sub>2</sub></div><div> T= 24x26x10/24+26</div><div> T= 6240/50</div><div> T = 124.8N</div><div> T =125 N</div><div> </div><div>Result:</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856612792918048889869">a</gwmw>=4.8ms-²</div><div> T= 125N</div>
Question: 3
State the Newton's first law of motion &amp; Second law of motion.
Answer: 3
3-23
Newton first law:<div> A body continues in its state of rest or of uniform motion in a straight line provided no net force acts on it.</div><div>Explanation for rest:</div><div> Similarly Newton's first law of motion deals with bodies which are either at rest or moving with uniform speed in straight lineAccording to first law of motion, a body at rest remains at rest provided no net force on it<gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856619772405772300476"> .</gwmw><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856620258809566517971">this</gwmw>part of the law is true as we observe that objects do not move by themselves unless some moves it.</div><div>Explanation for motion:</div><div> Similarly a moving object does not stop moving by itself. A ball rolled on a rough ground stops earlier than that rolled on smoothground. It is because <gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856622096051795670970">rough surface</gwmw> <gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856622096056175670128">offer</gwmw> greater friction. If there would be no force of <gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856623094769127626675">oppose</gwmw> the motion of the body would never stop</div><div>Newton's Second Law of Motion:</div><div> When a net force acts upon a body, it produces as acceleration <gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856625212533353998561">in</gwmw> the body in the direction of force and the magnitude of acceleration is directlyproportional to the net force and is inversely proportional to the mass of the body.</div><div>Mathematical From:</div><div> If the force f is acting on the body of mass m then we can write this in the mathematical form is as</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856626505957826016251">a</gwmw> <gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856626505955661917573">x f</gwmw> _______&gt;(1)</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856626768166442823048">a</gwmw> x 1/m _____&gt; (2)</div><div>From (1) and (2) we have</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856627114091314534993">a</gwmw> x F/m</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856627433970894134091">a</gwmw> = Constant x F/m</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856627556648396031372">a</gwmw> = k x F/m</div><div>In the above equation according to the international system units if m= 1kg</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856629408014811703695">a</gwmw> = 1ms-²<gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856629408010533415692">,</gwmw>F= <gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856629408011492319683">1N then</gwmw> the value of k will be 1. So the equation can be written as</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856630029666868695236">a</gwmw>=1x F/m</div><div>F= <gwmw class="ginger-module-highlighter-mistake-anim ginger-module-highlighter-mistake-type-1" id="gwmw-15856630171058516231019">ma</gwmw></div>
Question: 4
A force of 20 N moves a body with an acceleration of 2ms-² What is the mass.
Answer: 4
4-23
F = 20 N<div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856631780746585946730">a</gwmw>= 2ms-²</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856631941898921762325">m</gwmw>=?</div><div>Solution:</div><div> F=<gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856632120422568892386">ma</gwmw></div><div>So m= 20/2</div><div> <gwmw class="ginger-module-highlighter-mistake-anim ginger-module-highlighter-mistake-type-1" id="gwmw-15856632948970963306653">m</gwmw> = 10 kg</div>
Question: 5
Explain Centripetal force and Drive F<sub>c</sub>= <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856633719363121477100">mv<sup>2</sup></gwmw>/<gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856633908860274704462">r</gwmw>.<div></div>
Answer: 5
5-23
A force that keeps a body<gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856634604145487918247">to move</gwmw> in a circle is known as centripetal force.<div>Explanation:</div><div> Consider a body tied at the end of a string moving with uniform speed in a circular path. A body has the tendency to move in a straight line due to inertia. The string to which body is tied keeps it <gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856636783403067828187">to move</gwmw> in a circle by pulling the body towards the center of the circle. The sting pulls the center of the circle<gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856637123153373405279"> .</gwmw><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856637345910395231702">the</gwmw> string pulls the body perpendicular to its motion. The pulling force <gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856638245888682694378">continuously changes</gwmw>the direction of motion and remains towards the center seeking force is called the centripetal force.</div><div>Equation:</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856638639821479570627">Fc</gwmw> = <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856638639821003913352">ma</gwmw>,</div><div>So <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856638796053081275088">ma</gwmw><gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856638796057440154241">,</gwmw>+ mv²/r</div><div><gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856638958336744813161">a</gwmw><gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856638958334975283763">,</gwmw>=v²/r</div><div><br></div>
Question: 6
<gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856688040949382175928">how</gwmw> much centripetal force is needed to make a body of mass 0.5 kg to move <gwmw class="ginger-module-highlighter-mistake-type-3" id="gwmw-15856688040946401839963">to</gwmw> a circle of radius 50cm with a speed 3 ms-¹
Answer: 6
6-23
<div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856688766847751444529">m</gwmw>= 0.5kg</div><div> <gwmw class="ginger-module-highlighter-mistake-type-2" id="gwmw-15856689395655474330905">r</gwmw> = 50cm = 0.5m</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856689459003266656254">v</gwmw> = 3ms-¹</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856689802587094340141">Fc</gwmw>=?</div><div>As we know that</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856690412027001690296">Fc</gwmw>= mv²/r</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856690516843570157448">Fc</gwmw> = 0.5 x<gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856690562698976655313">(</gwmw>3<gwmw class="ginger-module-highlighter-mistake-type-6" id="gwmw-15856690589796360661078">)</gwmw>²/0.5</div><div> <gwmw class="ginger-module-highlighter-mistake-type-1" id="gwmw-15856690816524535762575">Fc</gwmw> = 4N</div>
Question: 7
Define dynamics, also briefly explain force , inertia and momentum<div><br></div>
Answer: 7
7-23
Dynamics: the branch of mechanics that deals with the study of motion fo an object and cause of its motion is called dynamics
Question: 8
Explain Newtons first law of motion
Answer: 8
8-23
Statement: A body continues its state of rest or of uniform motion in a straight line provided no net force acts on it.<div><br></div>
Question: 9
Explain Newtons second law of motion
Answer: 9
9-23
When a net force acts on a body, its produces acceleration in the body in the direction of the force. The magnitude of this acceleration is directly proportional to the net force acting on the body and inversely proportional to its mass
Question: 10
Write differences between mass and weight
Answer: 10
10-23
Mass: Mass o a body is the quantity of matter possessed by the bodyits a scaler quantityit is measured by using a common balance( beam balance)it remains same everywherethe SI unit of mas is kilogram(kg)it is a base quantity
Question: 11
Write third law of motoin. explain it with examples
Answer: 11
11-23
Statement: to every action there is always an equal but opposite reaction
Question: 12
Derive relation for tension and acceleration when two bodies moving along vertically
Answer: 12
12-23
Consider two bodies A and B of masses m1 and m2 respectively. Let m1 is greater than m2. The bodies are attached to the opposite ends of an inextensible string. The string passes over a frictionless pulley. The body A being heavier must be moving downward with some acceleration. Let this acceleration be a. At the same time, the body B attached to the other end of the string moves up with the same acceleration a. As the pulley is frictionless, hence tension will be the same throughout the string. Let the tension in the string be T.
Question: 13
Derive tension and acceleration when one body move vertically and other body move horizontally
Answer: 13
13-23
Consider two bodies A and B of masses m1 and m2 respectively attached to the ends of an inextensible string. Let the body A moves downward with an acceleration a. Since string is inextensible , therefore, body B also moves over the horizontal surface with the the same acceleration. As the pulley is frictionless, hence tension T will be the same throughout the string.
Question: 14
How you can relate a force with the change in momentum of a body
Answer: 14
14-23
Consider a body of mass m moving with initial velocity vi. Let a force F acts on the body which produces an acceleration a in it. The changes the velocity of the body. Lets its final velocity after time t become vf. Ig pi and pf be the initial momentum and final momentum of the body related to initial and final velocities arepi=mvipf=mvfchange in momentum= final momentum- initial momentumpf-pi= mvf-mvi
Question: 15
Explain law of conservation of momentum
Answer: 15
15-23
Statement: The momentum of an isolated system of two or more than two interacting bodies remains constant
Question: 16
Explain gun and bullet system using law of conservation of momentum
Answer: 16
16-23
consider a gun and bullet system before firing the gun, both gun and bullet are at rest so the total momentum of system is zero.As the gun is fired bullet shoots out of the gun and acquires momentum. To conserve momentum of system the gun recoils. According to law of conservation of momentum, the total momentum of gun and bullet will also be zero after gun is fired . Let m be the mas of bullet and v be its velocity on firing the gun. M be the mass of gun and V be the velocity of gun with which it recoils'Total momentum of the gun and the bullet after the gun is fired = MV+mvaccording to law of conservation of momentumTotal momentum of gun and bullet after fire= Total momentum of gun and bullet before fireMV+mv=0MV=-mv
Question: 17
define and explain friction
Answer: 17
17-23
Friction: the force that opposes the motion of moving objects is called frictionFactors: in case of solid, forces of friction between two surfaces depends upon many factors such as:nature of two surfaces in contactpressing force between them
Question: 18
Define rolling friction. Also explain why rolling friction is less than sliding friction
Answer: 18
18-23
Rolling friction: force of friction between body and surface when a body rolls over a surface e.g. when a wheel rolls on ground
Question: 19
Explain braking and skidding
Answer: 19
19-23
The wheels of moving vehicles have two velocity componentsmotion of wheels along the roadrotation of wheels about their axisto move a vehicle on the road as well as to stop a moving vehicle requires friction between its tyres and the road. For example, if the road is slippery or the tyre are worn out then the typre instead of rolling, slip over the road,The vehicle will not move if the wheels starts slipping at the same point on the slipper ty road. Thus for the wheels to roll, the force of friction9gripping force 0 between the tyres and the road must be enough that prevents them from slipping.if we want to stop a car quickly, a large force of friction between the tyres and the road is needed. But there is a limit to this force of friction that tyres can provide. If the breaks are applied too strongly, The wheels of th car will lock up ( stop turning) and the car will skid due to its large momentum. If we want t o reduce the skidding, it is necessary not to apply breaks too hard that lock up their rolling motion. It is unsafe to derive a vehicles wit worn out tyres .
Question: 20
Write the advantages and disadvantages of friction, also give methods to reduce friction
Answer: 20
20-23
Advantage; sometimes friction is most desirablewe cannot write if there is no friction between paper and pencilfriction enables us to walk on road. We cannot run on a slippery ground because a slippery ground offers very little frictionbirds cannot fly if there is no air resistance
Question: 21
Define and explain centripetal force
Answer: 21
21-23
Centripetal force: a force that keeps a body to move in a circular path is called centripetal force
Question: 22
Define centrifugal force and explain it briefly
Answer: 22
22-23
Centrifugal force :A force that pulls outward the body moving in circular path us called centrifugal forceORThe reaction of centripetal force is known as centrifugal force
Question: 23
Write note on application of centripetal force
Answer: 23
23-23
There are many application of centripetal force but some are following