Class 9 Science Newton’s Laws of Motion Notes

Chapter 3 — Newton’s Laws of Motion

This chapter extends ordinary Newtonian mechanics into accelerating frames and connects force with circular motion, gravity and torque.

1. Accelerating frames and pseudo force

In an inertial frame, Newton’s laws work directly.

In an accelerating frame, an object may appear to accelerate even when no corresponding real force is visible.

This leads to the idea of pseudo force.

For a frame accelerating with aframea_{\text{frame}}:Fpseudo=maframeF_{\text{pseudo}}=-ma_{\text{frame}}

The negative sign indicates that the pseudo force is directed opposite to the acceleration of the frame.

Example: when a bus suddenly accelerates forward, passengers appear to fall backward. From the road, their bodies are resisting the change in motion. From inside the accelerating bus, a backward pseudo force is introduced to describe the observation.

2. Gravity and orbital motion

The Earth does not simply fly away from the Sun because the Sun’s gravitational attraction provides the required inward, or centripetal, force.

The Earth also possesses tangential velocity. Its inertia tends to keep it moving along a straight path, while gravity continually bends that path towards the Sun.

The combination produces orbital motion.

3. Air resistance

For two objects of equal mass falling through air:Fg=mgF_g=mg

Both experience the same gravitational force, but air resistance depends partly on cross-sectional area.

A larger cross-sectional area generally produces greater drag.

Therefore, under the conditions discussed in the chapter:

  • smaller area → less air resistance → faster fall;
  • larger area → greater air resistance → slower fall.

In a vacuum there is no air resistance, so objects fall together irrespective of their shape or size.

4. Variation of gg with height

As height above Earth’s surface increases, distance from Earth’s centre increases and gravitational acceleration decreases.

The chapter gives:gh=g(RR+h)2g_h=g\left(\frac{R}{R+h}\right)^2

where:

  • RR = Earth’s radius
  • hh = height above the surface
  • gg = surface value of acceleration due to gravity.

For example, using R=6400R=6400 km, h=800h=800 km and g=9.8m/s2g=9.8\,m/s^2, the chapter obtains approximately:gh=7.74m/s2g_h=7.74\,m/s^2

5. Variation of gg with depth

Assuming uniform density:gd=g(1dR)g_d=g\left(1-\frac dR\right)

where dd is the depth below Earth’s surface.

Thus:

  • gg is maximum at the surface;
  • gg decreases as we move upward;
  • gg also decreases as we move downward;
  • at Earth’s centre, g=0g=0.

6. Moment of force / torque

A force can produce rotation about a pivot.

The turning effect is called moment of force or torque:τ=Fdsinθ\tau=Fd\sin\theta

where:

  • FF = force
  • dd = distance from pivot to point of application
  • θ\theta = angle between the force and lever arm.

Torque is maximum when:θ=90\theta=90^\circ

and zero when:θ=0 or 180\theta=0^\circ \text{ or }180^\circ

SI unit:NmN\,m

This explains why door handles are placed far from hinges and why a long spanner makes it easier to loosen a tight bolt.

Chapter 3 — Newton’s Laws of Motion

The questions below cover the chapter’s treatment of accelerating frames, pseudo force, orbital motion, air resistance, variation of gg, and torque.

Questions

A. Multiple Choice Questions

1. Newton’s laws in their usual form apply directly in:
a) Accelerating frames
b) Non-inertial frames
c) Inertial frames
d) Rotating frames only

2. A pseudo force is introduced when describing motion from:
a) An inertial frame
b) An accelerating frame
c) A stationary laboratory only
d) A vacuum

3. The direction of pseudo force is opposite to the:
a) Velocity of the object
b) Displacement of the object
c) Acceleration of the frame
d) Gravitational field

4. In air, an object with a smaller cross-sectional area generally experiences:
a) Greater air resistance
b) Less air resistance
c) No gravity
d) Greater mass

5. At greater altitude above Earth, the value of gg:
a) Increases
b) Decreases
c) Remains exactly unchanged
d) Becomes infinite

6. At the centre of Earth, under the uniform-density model discussed in the chapter, gg is:
a) Maximum
b) Zero
c) 9.8m/s29.8m/s^2
d) Infinite

7. Torque is maximum when the angle between the force and lever arm is:
a) 00^\circ
b) 3030^\circ
c) 6060^\circ
d) 9090^\circ

B. Fill in the Blanks

8. A pseudo force is also called a ______ force.

9. Pseudo force is observed in an ______ frame.

10. Air resistance acts opposite to the direction of ______.

11. The turning effect of a force is called ______.

12. The SI unit of torque is ______.

C. True or False

13. A pseudo force arises from a physical interaction between two bodies.

14. In a vacuum, two objects of equal mass but different shapes experience no air resistance.

15. Gravity becomes exactly zero immediately outside Earth’s atmosphere.

16. Torque is zero when a force acts through the pivot.

D. Assertion–Reason

17. Assertion: A passenger inside an accelerating bus may appear to move backward.
Reason: A pseudo force directed opposite to the frame’s acceleration is introduced when using the bus frame.

18. Assertion: A longer spanner can make loosening a bolt easier.
Reason: Increasing the perpendicular distance from the pivot increases torque for the same force.

E. Short Answer

19. Why does a passenger appear to fall backward when a bus suddenly accelerates forward?

20. Why do objects of equal mass but different cross-sectional areas fall at different rates in air?

21. Explain why Earth remains in orbit around the Sun rather than travelling in a straight line.

22. Why is the turning effect of a force zero when the force acts parallel to the lever arm?

F. Numerical Questions

23. A lift accelerates upward at 4.5m/s24.5\,m/s^2. Calculate the magnitude of pseudo force for a 60kg60\,kg person.

24. A force of 20N20N acts on a door 0.8m0.8m from the hinge at 9090^\circ. Find the torque.

25. Calculate gg at a height of 800km800km above Earth if R=6400kmR=6400km and g=9.8m/s2g=9.8m/s^2.


Answers

A. MCQ Answers

  1. c) Inertial frames
  2. b) An accelerating frame
  3. c) Acceleration of the frame
  4. b) Less air resistance
  5. b) Decreases
  6. b) Zero
  7. d) 90∘90^\circ

B. Fill in the Blanks

  1. fictitious
  2. accelerating/non-inertial
  3. motion
  4. torque/moment of force
  5. newton-metre (NmNm)

C. True/False

  1. False
  2. True
  3. False
  4. True

D. Assertion–Reason

  1. Both statements are true, and the reason correctly explains the assertion.
  2. Both statements are true, and the reason correctly explains the assertion.

E. Answers

  1. The passenger’s body tends to maintain its previous state of motion due to inertia. The bus moves forward beneath the passenger, making the passenger appear to move backward relative to the bus. In the accelerating bus frame, this can be represented using a backward pseudo force.
  2. Both objects experience the same gravitational force if their masses are equal, but the object with the larger cross-sectional area experiences greater air resistance. Hence the smaller-area object has a greater net downward force.
  3. Earth’s tangential motion tends to carry it in a straight line, while the Sun’s gravitational attraction continuously pulls it inward. The combination produces orbital motion.
  4. Torque is

τ=Fdsinθ\tau=Fd\sin\theta

For θ=0\theta=0^\circ or 180180^\circ, sinθ=0\sin\theta=0, so torque is zero.

F. Numerical Answers

Fpseudo=maF_{\text{pseudo}}=ma=60×4.5=60\times4.5270N\boxed{270N}

τ=Fdsinθ\tau=Fd\sin\theta=20(0.8)sin90=20(0.8)\sin90^\circ16Nm\boxed{16Nm}

gh=g(RR+h)2g_h=g\left(\frac{R}{R+h}\right)^2=9.8(64007200)2=9.8\left(\frac{6400}{7200}\right)^27.74m/s2\boxed{7.74m/s^2}