Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 4th Lesson Laws of Motion Class 11 Textbook Exercise Questions and Answers.
Laws of Motion Class 11 Questions and Answers AP Inter 1st Year Physics 4th Lesson
I. Laws of Motion Multiple Choice Questions (1 Mark)
Question 1.
When a bus starts suddenly, the passengers are pushed back. This is an example for………..
(1) Newton’s First Law
(2) Newton’s Second Law
(3) Newton’s Third Law
(4) Newton’s law of gravitation
Answer:
(1) Newton’s First Law
Question 2.
Which of the following is an example of centripetal force?
(1) Pushing door
(2) Mud flying off tire
(3) Planets orbiting the sun
(4) Churning of butter in butter milk
Answer:
(3) Planets orbiting the sun
Question 3.
A non zero net force exerted on a body does not change its ………………
(1) direction of motion
(2) momentum
(3) acceleration
(4) inertia
Answer:
(4) inertia
Question 4.
The rate of change of momentum is proportional to…………
(1) acceleration
(2) velocity
(3) force
(4) displacement
Answer:
(3) force
Question 5.
The net force (F) experienced by a body of mass (m) whose displacement is y= ut+\(\frac{1}{2}\) gt2 is
(1) F=m a
(2) F=m g
(3) F=0
(4) F≠0
Answer:
(2) F=m g
Question 6.
The linear momentum of a particle as a function of time t is given by P=a+b t. The force acting on the particle is………..
(1) a
(2) b
(3) a/b
(4) b/a
Answer:
(2) b
Question 7.
The inertia of a moving object depends on ………….
(1) momentum of the object
(2) speed of the object
(3) shape of the object
(4) mass of the object
Answer:
(4) mass of the object
Question 8.
A force accelerates a lighter vehicle more easily than a heavier vehicle. Guess the law behind this.
(1) Newton’s First Law of motion
(2) Newton’s Second Law of motion
(3) Newton’s Third Law of motion
(4) Newton’s Law of gravitation
Answer:
(2) Newton’s Second Law of motion
Question 9.
Calculate the time needed for a net force of 5 N to change the velocity of a 10 kg mass by 2 m/s
(1) 2 sec
(2) 6 sec
(3) 4 sec
(4) 3 sec
Answer:
(3) 4 sec
II. Laws of Motion Fill in the Blanks (1 Mark)
Question 1.
Friction opposes the ………… between the surfaces in contact with each other.
Answer:
Relative motion
Question 2.
………… is required to keep a body in accelerated motion.
Answer:
Non zero net force
Question 3.
Kinetic friction produces ………… energy.
Answer:
heat
Question 4.
Powder is sprinkled on a carom board to ………… friction between the striker and the board surface.
Answer:
reduce
Question 5.
To every action there is always an equal and ………… reaction.
Answer:
opposite
Question 6.
Impulse is the product of force and time which equals ………..
Answer:
change in momentum
Question 7.
………… gives the measure of inertia.
Answer:
mass
Question 8.
The area under F – t graph represents …………
Answer:
impulse
Question 9.
The Newton’s Second Law is consistent with the first law only if ………….
Answer:
F=0
Question 10.
Sliding friction is ………… than the limiting value of static friction.
Answer:
less
III. Laws of Motion One Word Answer Questions (1 Mark)
Question 1.
State Newton’s First Law.
Answer:
“Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled to change that state by a net external force.”
Question 2.
State Newton’s Third Law.
Answer:
For every action, there is an equal and opposite reaction.
Question 3.
Write the mathematical equation of Newton’s Second Law.
Answer:
F=ma
Question 4.
State the SI unit of force.
Answer:
Newton (N)
Question 5.
Define centripetal force.
Answer:
Centripetal force is the force that keeps a body moving in a circular path, directed towards the center of the circle.
Question 6.
According to Newton’s Third Law every force is accompanied by an equal and opposite force. Then how can a movement ever take place?
Answer:
Because action and reaction act on different bodies, not on the same body. Thus, the net force on a single object is not canceled, allowing it to move.
Question 7.
Why does a heavy rifle not recoil as strongly as light rifle using the same cartridges?
Answer:
For the same force, a heavier object undergoes a smaller acceleration than the lighter one because a=F/ma. That is why a heavy rifle does not recoil as strongly as a light rifle using the same cartridges.
IV. Laws of Motion Very Short Answer Questions (2 Marks)
Question 1.
What is inertia ? What gives the measure of inertia ?
Answer:
Inertia is the property of an object to resist changes in its state of motion or rest. Mass of the object gives the measure of its inertia.
Question 2.
When a bullet is fired from a gun, the gun gives a kick in the backward direction. Explain.
Answer:
When a bullet is fired, it gains forward momentum. To conserve momentum, the gun moves backward with equal momentum (Newton’s Third Law).
Question 3.
If a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain.
Answer:
A bomb at rest has zero total momentum. After explosion, the two pieces must move in opposite directions so that total momentum remains zero (conservation of momentum).
Question 4.
Define force. What are the basic forces in nature?
Answer:
Force is a push or pull that changes or tries to change the state of motion of a body.
Basic forces : Gravitational, Electromagnetic, Strong Nuclear, Weak Nuclear.
Question 5.
State the SI units and dimensions of lineæ momentum.
Answer:
SI Unit: kg m/s
Dimensions : M1L1T-1
Question 6.
State the SI units and dimensions of impulse.
Answer:
SI Unit : N.s (Newton – second) or Kgm/s
Dimensions : M1L1T-1 (same as momentum)
Question 7.
Can the coefficient of friction be greater than one ? Give the reason.
Answer:
Yes, the coefficient of friction can be greater than one.
“It is found experimentally than μs > μk, and both can be more than 1 depending on surface roughness.” This means the frictional force can be greater than the normal force in some cases.
Question 8.
Why does the car with a flattened tyre stop sooner than the one with inflated tyres?
Answer:
A flat tyre increases the contact area and friction, which opposes motion more strongly, causing the car to stop sooner.
Question 9.
A horse has to pull harder during the start of the motion than later. Explain.
Answer:
Initially, the horse must overcome static friction, which is greater than kinetic friction encountered during motion.
Question 10.
What happens to the coefficient of friction if the weight of the body is doubled?
Answer:
The coefficient of friction remains unchanged because it is independent of normal force or weight; it depends only on the nature of surfaces.
Question 11.
Write the expressions for maximum possible speed of a car moving on (a) level road (b) banked road.
Answer:
(a) On a level road:
\(\mathrm{v}_{\max }=\sqrt{\mu \mathrm{Rg}}\)
where μ = coefficient of friction, R= radius, g= gravity
(b) On a banked road (no friction):
\(\mathrm{v}_{\max }=\sqrt{\mathrm{Rg} \tan \theta}\)
where θ = banking angle
V. Laws of Motion Short Answer Questions (4 Marks)
Question 1.
A stone of mass 0.1 kg 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.
Answer:
Let the acceleration due to gravity g =9.8m/s2
Mass of stone = 0.1 kg
Weight = m g 0.1 × 9.8=0.98 N
a) Upward motion :
Magnitude =0.98 N, Direction = Downward (only gravity acts, opposing motion).
b) Downward motion :
Magnitude =0.98 N, Direction = Downward (gravity aids motion).
c) At highest point :
Magnitude = 0.98 N, Direction = Downward (velocity is zero but force of gravity acts).
Question 2.
Define the terms momentum and impulse. State and explain the Law of Conservation of linear momentum. Give examples.
Answer:
Momentum : Product of mass and velocity; p = mv, a vector quantity.
Impulse : Change in momentum; Impulse = Force × Time
Impulse =Δp = F Δ t
“The total momentum of an isolated system of interacting particles is conserved.” If two bodies collide :
m1 u1+m2 u2=m1 v1+m2 v2
Examples :
- Recoil of a gun when a bullet is fired.
- Explosion of a bomb at rest into fragments that move in opposite directions.
- A person jumping off a boat causes the boat to move in the opposite direction.
Question 3.
Why are shock absorbers used in motorcycles and cars ?
Answer:
“Shock absorbers reduce the impact of impulsive forces by increasing the time over which the force acts.”
By increasing the time of impact during bumps or collisions, the force experienced is reduced, protecting both the vehicle and passengers.
Question 4.
Explain the terms limiting friction, dynamic friction and rolling friction.
Answer:
- Limiting friction : Maximum static friction just before motion begins.
- Dynamic (Kinetic) friction : Friction acting when two surfaces are in relative motion.
- Rolling Friction : It is the resistance a body faces when rolling over a surface. It is usually much smaller than sliding friction. For example, a ball bearing rolling causes less friction than a block sliding.
Question 5.
Explain advantages and disadvantages of friction.
Answer:
Advantages of Friction :
- Essential for motion : Walking, running, or driving is possibble due to friction between feet/tires and the ground.
- Enables grip: Helps to hold objects, nails to stay in wood, etc.
- Braking systems : Vehicles stop due to friction between brakes and wheels.
- Transfers motion: Belts in machines rely on friction to transmit power.
Disadvantages of Friction :
- Causes wear and tear: Machinery and moving parts wear out over time.
- Wastes energy : Converts mechanical energy into heat, reducing efficiency.
- Requires more force : More effort is needed to move objects due to resistance.
Question 6.
Mention the methods used to decrease friction.
Answer:
- Lubrication : Applying oil or grease reduces contact between surfaces.
- Polishing surfaces : Makes them smoother, reducing resistance.
- Using ball bearings or rollers : Converts sliding friction into rolling friction (which is much lower).
- Streamlining: In vehicles and aircraft to reduce air resistance.
- Using proper materials : Like Teflon, which has a low coefficient of friction.
Question 7.
State the Laws of rolling friction.
Answer:
- Rolling friction is smaller than sliding or static friction.
- It depends on the nature of the surfaces in contact, especially the deformation of the rolling object or the surface.
- It increases with load but remains significantly less than sliding friction for the same surfaces.
Question 8.
Derive an expression for the maximum possible speed of a car moving on a banked road?
Answer:
Expression for the maximum possible speed of a car moving on a banked road : Consider a body of mass ‘m’ on an banked rough surface making an angle ‘θ’ as shown in the figure.
To find max speed from the figure.
V. Laws of Motion Long Answer Questions (8 Marks)
Question 1.
(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?
Answer:
(a) Newton’s Second Law of Motion and Derivation of F=ma:
Newton’s Second Law states that the rate of change of momentum of a body is directly proportional to the applied external force, and this change occurs in the direction of the force.
Let the mass of the body be m, and its velocity change from u to v in time t.
Change in momentum =m v-m u=m(v-u)
Rate of change of momentum = \(\frac{m(v-u)}{t}=m \frac{(v-u)}{t}=m \cdot a\)
So, force F α ma, and with proper units,
F=ma
(b) Yes, a force is ‘required’ even if the speed is constant because the direction of motion continuously changes, meaning the velocity changes (since velocity is a vector). This change in direction requires centripetal acceleration, which is provided by a centripetal force directed towards the center of the circle. Without this force, the body would move in a straight line due to inertia.
From Newton’s Second Law :
\(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}\)
Since there is acceleration (towards the center), there must be a net force directed towards the center. This is known as the centripetal force, and is given by :
F = mv2/R
Where:
- m= mass of the object
- v = constant speed
- R = radius of circular path
VII. Problem (Marks 8)
Question 1.
A force 2i+j-k Newton acts on a body which is initially at rest. At the end of 20 seconds the velocity of the body is 4i+2j-2 kms-1. What is the mass of the body?
Answer:
Given :
Force, F=2i+j-kN
Initial velocity, u=0 (body is at rest)
Final velocity, v=4i+2j-2 km/s
Time, t=20s
Using Newton’s Second Law :
Question 2.
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 acceleration due to gravity, what is the impulse received by the ball due to gravity force during its flight (neglect air resistance) ?
Answer:
Question 3.
A container of mass 200 kg rests on the back of an open truck. If the truck accelerates 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 ?
Answer:
Given m=200kg, a=1-5ms-2, g=9.8 ms-2
Co-efficient of friction μs=\(\frac{\mathrm{a}}{\mathrm{g}}=\frac{1.5}{9.8}\)=0.153
Question 4.
A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically down-wards with an initial speed of 10m/s. If acceleration due to gravity is 10m/s2, what is the separation between the fragments 2 sec after the explosion.
Answer:
Given :
One fragment moves down with u=10 m/s
Other fragment (by conservation of momentum) must move up with u=10m/s Time t=2s
Acceleration due to gravity is g=10m/s2
Using the equation of motion : s=u t+\(\frac{1}{2}\) gt2
Downward fragment : s1=10 × 2+\(\frac{1}{2}\) ×10.4=20+20=40m (downward)
Upward fragment : s2=10 × 2-\(\frac{1}{2}\) × 10.4=20-20=0m (still at explosion point)
So, separation =40m