Matric Part 1/9th Class Physics Chapter 3 Long Question With Answer for Unit 3 Dynamics

Question # 1

Two masses 26 Kg and 24 kgare attachedto the ends ofstringwhich passes over a frictionless pulley. 26 kg islayingover a smooth horizontal table. 24 N mass is

Ans 1: Solution
₁= 24kg
₂=26 kg
= 10ms-²
Solution;
a= m₁g/m₁ +m₂
a= 24x10/24+26
a =240/50
a =4.8ms-²
T =m₁ m₂ g/m₁ +m₂
T= 24x26x10/24+26
T= 6240/50
T = 124.8N
T =125 N

Result:
a=4.8ms-²
T= 125N

Question # 2

Explain Newtons first law of motion

Ans 1: Statement: A body continues its state of rest or of uniform motion in a straight line provided no net force acts on it.
Ans 2: Explanation: first law of motion deals with bodies which are either at rest or moving with uniform speed in a straight line. According to Newtons first law of motion, a body at rest remains at rest provided no net force acts on it.
Ans 3: At rest: this part of the law is true as we observe that object do no move by themselves unless someone move them. For example a book lying on a table remains at rest as long as no force acts on it
Ans 4: At uniform speed: similarly a moving object does not stop move by itself. A ball rolled on a rough ground stops earlier than that rolled on a smooth ground. It is because rough surfaces offer greater friction. If there would be no force to oppose the motion of a body then the body would never stop
Ans 5: Law of inertia: Since Newtons first law of motion deals with the inertial property of matter , therefore , Newtons first law of motions is also known as law of inertia
Example: We have observed that the passengers standing in a buss fall forward when its driver applied breaks suddenly. it is because the upper parts of their bodied tend to continue their motion , while lower parts o their bodies in contact with the bus stop with it. Hence they fall forward

Question # 3

Explain law of conservation of momentum

Ans 1: Statement: The momentum of an isolated system of two or more than two interacting bodies remains constant
Ans 2: Isolated system: An isolated system is a group of interacting bodies on with no external force is acting
Ans 3: Explanation: Consider an isolated system of two spheres of masses m1 and m2. They are moving in a straight line with initial velocities u1 and u2 respectively such that u1 is greater than u2 sphere of mass m1 approaches the sphere m2 as they moveinitial momentum of mass m1= m1u1initial momentum of mass m2= m2u2Total initial momentum of the system before collision =m1u1+m2u2After sometime mass m1 hits m2 with some force. According to Newton's third law of motion, m2 exerts an equal and opposite reaction force on m1. Let their velocities become v1 and v2 respectively after collision. Thenfinal momentum of mass m1=m1v1final momentum of mass m2= m2v2Total final momentum of the system after collision=m1v1+m2v2................by comparing we get Total initial momentum of the system before collision= total momentum of the system after collisionm1u1+m2u2=m1v1+m2v2..............AEquation A shows that the momentum of an isolated system before and after collision remains same which is the law of conservation of momentum. Law of conservation of momentum is an important law and has vast applications

Question # 4

Explain Newtons second law of motion

Ans 1: 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
Ans 2: Newton: One newton is the force that produces an acceleration of 1 ms-2 in a body of mass 1 kg1N= 1kg * 1ms-2or 1 N = 1kgms-2

Question # 5

Define rolling friction. Also explain why rolling friction is less than sliding friction

Ans 1: Rolling friction: force of friction between body and surface when a body rolls over a surface e.g. when a wheel rolls on ground
Ans 2: Sliding friction: force of friction between body and surface when a body slides over a surface. when a wooden block moves over asurface of table then the force friction between block and surface of table is sliding friction
Ans 3: Comparison of rolling and sliding friction: rolling friction is less than sliding friction because in rolling friction a body rolls without rupturing the cold welds , while in sliding friction a body slides' by rupturing the cold welds so sliding friction is more than rolling friction

Question # 6

Write differences between mass and weight

Ans 1: 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
Ans 2: Weight: weight of a body is the force of gravity acting on itit is a vector quantityit is measured by using a spring balanceit varies depending upon the value of acceleration due to gravity gthe SI unit of weight is newton(N)it is a derived quantity

Question # 7

how much centripetal force is needed to make a body of mass 0.5 kg to move to a circle of radius 50cm with a speed 3 ms-¹

Ans 1:
m= 0.5kg
r = 50cm = 0.5m
v = 3ms-¹
Fc=?
As we know that
Fc= mv²/r
Fc = 0.5 x(3)²/0.5
Fc = 4N

Question # 8

Explain braking and skidding

Ans 1: 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 # 9

Derive relation for tension and acceleration when two bodies moving along vertically

Ans 1: 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 # 10

Explain Centripetal force and Drive F_c= mv²/r.

Ans 1: A force that keeps a bodyto move in a circle is known as centripetal force.
Explanation:
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 to move in a circle by pulling the body towards the center of the circle. The sting pulls the center of the circle .the string pulls the body perpendicular to its motion. The pulling force continuously changesthe direction of motion and remains towards the center seeking force is called the centripetal force.
Equation:
Fc = ma,
So ma,+ mv²/r
a,=v²/r

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Matric Part 1/9th Class Physics Chapter 3 Long Question With Answer for Unit 3 Dynamics

9th Class English Medium Subjects

9th Class Urdu Medium Subjects

Online Long Questions For Chapter 3 "9th Class Physics Chapter 3 Long Question Preparation "

Physics (New Book) - 9th Class Physics Chapter 3 Long Question Preparation

Two masses 26 Kg and 24 kgare attachedto the ends ofstringwhich passes over a frictionless pulley. 26 kg islayingover a smooth horizontal table. 24 N mass is

Explain Newtons first law of motion

Explain law of conservation of momentum

Explain Newtons second law of motion

Define rolling friction. Also explain why rolling friction is less than sliding friction

Write differences between mass and weight

how much centripetal force is needed to make a body of mass 0.5 kg to move to a circle of radius 50cm with a speed 3 ms-¹

Explain braking and skidding

Derive relation for tension and acceleration when two bodies moving along vertically

Explain Centripetal force and Drive Fc= mv2/r.

Prepare Complete Set Wise Questions For Chapter 3 "9th Class Physics Chapter 3 Long Question Preparation "

Set 1

Set 2

9th Class Physics Chapter 3 Long Question Preparation - Set 1

Two masses 26 Kg and 24 kgare attachedto the ends ofstringwhich passes over a frictionless pulley. 26 kg islayingover a smooth horizontal table. 24 N mass is

Explain Newtons first law of motion

Explain law of conservation of momentum

Explain Newtons second law of motion

Define rolling friction. Also explain why rolling friction is less than sliding friction

Write differences between mass and weight

how much centripetal force is needed to make a body of mass 0.5 kg to move to a circle of radius 50cm with a speed 3 ms-¹

Explain braking and skidding

Derive relation for tension and acceleration when two bodies moving along vertically

Explain Centripetal force and Drive Fc= mv2/r.

9th Class Physics Chapter 3 Long Question Preparation - Set 2

Explain law of conservation of momentum

Write note on application of centripetal force

Explain Newtons first law of motion

State the Newton's first law of motion & Second law of motion.

Two masses 26 Kg and 24 kgare attachedto the ends ofstringwhich passes over a frictionless pulley. 26 kg islayingover a smooth horizontal table. 24 N mass is

How you can relate a force with the change in momentum of a body

Explain Newtons second law of motion

Write differences between mass and weight

Define rolling friction. Also explain why rolling friction is less than sliding friction

Explain Centripetal force and Drive Fc= mv2/r.

9th class Physics Chapter 3 Long Q's (English Medium)

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Explain Centripetal force and Drive F_c= mv²/r.

Explain Centripetal force and Drive F_c= mv²/r.

Explain Centripetal force and Drive F_c= mv²/r.