What is inertia? What gives the measure of inertia? | Physics

LAWS OF MOTION

Very Short Question Answers

Q1. What is inertia? What gives the measure of inertia?
Ans.

Inertia: The inability of a body to change its state by itself is known as inertia. Mass is the measure of inertia.

Q2. According to Newton’s third law, every force is accompanied by an equal and opposite force. How can a movement ever take place?
Ans.

According to newton’s third law of motion action and reaction act on two different bodies. Hence motion is possible

Q3. When a bullet is fired from a gun, the gun gives a kick in the backward direction.Explain?
Ans.

According to the law of conservation of linear momentum.

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$0=m_{1}v_{1}+m_{2}v_{2}$

$ m_{1}v_{1}=\,-\,m_{2}v_{2}$

minus sign indicates gun and bullet move in opposite direction after fired. Hence, gun will
kick backwards.

Q4. Why does a heavy rifle not recoil as strongly as a light rifle using the same cartridges?
Ans.

According to law of conservation of linear momentum

$https://latex.codecogs.com/png.image?\large \dpi{110} m\propto \frac{1}{V}$

Hence heavy rifle recoil less as compare to light rifle.

Q5. If a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain?
Ans.

According to the law of conservation of linear momentum.

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$0=m_{1}v_{1}+m_{2}v_{2}$

$ m_{1}v_{1}=\,-\,m_{2}v_{2}$

minus sign indicates two pieces move in opposite direction after explosion.

Q6. Define force. What are the basic forces in nature?
Ans.

Force: The physical quantity that changes or tries to change the state of rest or of uniform
motion along a straight line is called force.
The basic forces in nature are :

Gravitational force
Electromagnetic force
Strong nuclear force
Weak nuclear force.

Q7. Can the coefficient of friction be greater than one?
Ans.

Yes. The coefficient of friction is greater than one.

Q8. Why does the car with flattened tyre stop sooner than the one with inflated tyres?
Ans.

A flattened tyre has got more surface area of contact due to deformation than an inflated tyre. Hence friction is more in the case of a flattened tyre.

Q9. A horse has to pull harder during the start of the motion than later. Explain?
Ans.

Static friction is more than kinetic friction. To overcome static friction initially horse
should apply more force than later.

Q10. What happen to the coefficient of friction, if the weight of the body is doubled?
Ans.

Coefficient of friction (𝜇) is independent of mass of the body. Therefore, coefficient of
friction does not change.

Short Question Answers

Q1. A stone of mass 0.1kg is thrown vertically upwards. Give the magnitude and direction of the net force on the stone a) During Its Upward Motion. b) During Its Downward Motion. c) At the highest point, where it momentarily comes to rest?

Ans.
stone mass $m=0.1$ kg acceleration due to gravity $g=9.8$ $m/s^2$
Force acting on stone in three conditions is same
That is $F=ma=0.1×9.8=0.98 N$

Q2. Define the terms momentum and impulse. State and explain the law of conservation of linear momentum. Give its exampls?
Ans.

Momentum : The product of mass and velocity is known as momentum.

$$P=mv$$

Impulse: The product of force and time is known as impulse it is leads to the change in its momentum.

$$J=Ft=∆p$$

The law of conservation of linear momentum: The total momentum of a system remains constant if there is no force acting on the system.

$$P=constant $$(in the absence of external force )

Examples:

Considering the system of gun and bullet, both are at rest before the gun fires the bullet. After firing the bullet, their momentum is equal in magnitude and opposite in direction and thus the total momentum after firing is also zero.
Similarly when a bomb at rest explodes into two unequal pieces, their momentum is equal and opposite.

Q3. Why are shock absorbers used in motor cycles and cars?

Ans. Shock absorbers are used in motor cycles and cars to increase time interval of jerk. As a result the momentum rate slows down thus less force act during the jerk.

Q4. Explain the terms limiting friction, dynamic friction and rolling friction?

Ans. Limiting friction: The maximum value of static friction is known as Limiting friction.
$f_{L}=μ_L N$
Dynamic Friction: The force which is resisting the motion of sliding between the two surfaces is called as Dynamic friction (O r Kinetic friction).
$f_k=μ_k N$
Rolling friction: The force which is resist the motion of rolling of a object on surface is called as Rolling friction.
$f_r=μ_r N$

Q5. Explain advantages and disadvantages of Friction?
Ans.

Advantages	Disadvantages
Walking is not possible without friction	Due to friction wear and tear of machine parts increases
Breaks of vehicles working without friction	Due to friction life time of machine decrease
Nuts and Bolts holdings the surface due to friction	Due to friction energy is lost and hence reduces its efficiency

Q6. Mention the methods used to decrease friction?

Ans. Polishing: polishing will reduce the static friction.
Lubricants: lubricating the surfaces can reduce kinetic friction.
Ball bearings: Use of ball bearing will reduce the rolling friction
Streamlining: streamlining design of vehicles will reduce air resistance.

Q7. Sate the laws of rolling Friction?
Ans.

Rolling friction is directly proportional to the Normal. $ f_r∝N$
Rolling friction is directly proportional to area of contact. $f_r∝A$
Rolling friction is inversely proportional to the radius of wheel.$ f_r∝1/r$

Q8. Why is pulling the lawn roller is preferred to pushing it?

Ans. As shown in the diagram a lawn roller is pulling by a force F making an angle θ with respect to horizontal. Then the normal reaction is

pulling

$N+F sin⁡θ=mg$
$N=mg-F sin⁡θ$
Rolling friction $ f_r=μ_r N$
$f_r=μ_r (mg-F sin⁡θ )$
Pushing: As shown in the diagram a lawn roller is pushing by a force F, making an angle θ with respect to horizontal. Then the normal reaction is

pushing

$ N=F sin⁡θ+mg$
Rolling friction $ f_r^{'}=μ_r N $
$ f_r^{'}=μ_r (mg+F sin⁡θ ) $
Therefore, $ f_r<f_r^{'} $ hence pulling is preferred than pushing.

Q9. State Newton’s second law of motion. Hence derive the equation of motion F = ma from it.
Ans.

Newton’s second law of motion: The rate of change of momentum of a body is directly proportional to the net force acting on it.

Derivation of $F=ma$:

Consider a body of mass $m $moving with velocity $v$ by applied a external force $F$.

Then , according Newton’s second law of motion

$F∝ \frac{\mathrm{d}P }{\mathrm{d} t}$

$F=k \frac{\mathrm{d}P }{\mathrm{d} t}$

$F=k \frac{\mathrm{d}(mv )}{\mathrm{d} t}$

$F=km \frac{\mathrm{d}v }{\mathrm{d} t}$

$F=kma$

Where$k=1$, therefore net force acting on body is $$F=ma$$

Essay Question Answers

Q1. a) State Newton’s second law of motion. Hence derive the equation of motion F = ma from it.b) A body is moving along a circular path such that its speed always remains constant Should there be force acting on the body?

Ans. LAWS OF MOTION

Q2. Define angle of friction and angle of repose.Show that angle of friction is equal to angle of repose for a rough inclined plane.A block of mass 4kg is resting on a rough horizontal plane and is about to move when ahorizontal force of 30 N is applied on it. If g = 10 m/s² find the total contact force exerted by the plane on the block?

Ans. LAWS OF MOTION

Problems Question Answers

Q1. The linear momentum of a particle as a function of time t is given by p= a+bt a and b are positive constants. What is the force acting on the particle?

Ans. LAWS OF MOTION

Q2. Calculate the time needed for a net force of 5N to change the velocity of a 10kg mass by 2 m/s?

Ans. LAWS OF MOTION

Q3. A ball of mass m is thrown vertically upward from the ground and reaches a height h before momentarily coming to rest. If g is the acceleration due to gravity. What is the impulse received by the all due to gravity force during its flight? (Neglect air resistance)

Ans. LAWS OF MOTION

Q4. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m/s to 3.5 m/s in 25s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?

Ans. LAWS OF MOTION

Q5. A man in a lift feels an apparent weight W, when the lift if moving up with a uniform acceleration of 1/3rd of the acceleration due to gravity. If the same man were in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then what is the apparent weight?

Ans. LAWS OF MOTION

Q6. A container of mass 200kg rests on the ack of an open truck. If the truck accelerate at 1.5 m/s² , what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from sliding off the back of the truck?

Ans. LAWS OF MOTION

Q7. A bomb initially at rest at a height of 40m above the ground suddenly explodes in to two identical fragments. One of them starts moving vertically downwards with an initial speed of 10m/s. If acceleration due to gravity is 10 m/s² . What is the separation between the fragments 2s after the explosion?

Ans. LAWS OF MOTION

Q8. A fixed pulley with a smooth grove has a light string passing over it with a 4 kg attached on one side and a 3kg on the other side. Another 3kg is hung from the other 3kg as shown with another light string. If the system is released from rest, find the common acceleration?

Ans. LAWS OF MOTION

Q9. A block of mass of 2kg slides on an inclined plane that makes an angle of 30° with the horizontal. The coefficient of fricti n between the block and the surface is 3 /2 .a) What force should be applied to the block so that it moves down without any acceleration? b) What force should be applied to the block so that it moves up without any acceleration?

Ans. LAWS OF MOTION

Q10. A block is placed on a ramp of parabolic shape given by the equation y = x²/20 If ?=0.5 , what is the maximum height above the ground at which the block can be placed without slipping?

Ans. LAWS OF MOTION

Q11. subjected to a horizontal force by allowing the 0.45kg mass to fall. The coefficient of sliding friction between the block and table is 0.2. Calculate a) The Initial acceleration b)The Tension in the string c) The Distance the block would continue to move if, after 2s of motion the string should break

Ans. LAWS OF MOTION

Q12. On a smooth horizontal surface a block A of mass 10kg is kept. On this block a second block B of mass 5kg is kept. The coefficient of friction between the two blocks is 0.4. A horizontal force of 30N is applied on the lower block as shown. The force of friction between the blocks is (Taking g= 10m /s² ).

Ans. LAWS OF MOTION