1© manhattan press (h.k.) ltd. newton’s first law of motion linear momentum and newton’s second...

© Manhattan Press (H.K.) Ltd. 1

• Newton’s First Law of MotionNewton’s First Law of Motion• • Linear momentum and Linear momentum and

Newton’s Second Law of Newton’s Second Law of MotionMotion

3.1 Newton’s Laws of 3.1 Newton’s Laws of MotionMotion

• • Newton’s Third Law of Newton’s Third Law of MotionMotion


Newton’s First Law of Motion

3.1 Newton’s Laws of Motion (SB p. 104)

Newton’s First Law of Motion:An object will remain at rest or move along a straight line with constant speed, unless it is acted upon by a net force.

Object resists changes to its state of rest or motion.

inertia

Inertia of object : its mass (independent of its velocity)

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Common Error


Newton’s First Law of Motion


1. Use seat belt / head-rest

2. Like being thrown away when turning around corner


Linear momentum and Newton’s Second Law of Motion


1. Linear momentum

Momentum = Mass x Velocity

p = mvvector

Unit – kg m s-1 or N s




2. Newton’s Second Law of Motion

Newton’s Second Law of Motion:

The rate of change of momentum of an object is directly proportional to the net force acted on it, and the motion occurs along the direction of the force.

Rate of change of momentum Net force∝





kmat

uvkm

tmumvk

F

Ft

mumv

)(

)(

Force momentum of change of Rate





Unit – newton (N)

F = kma

1 N = k (1 kg) (1 m s-2)

= k kg m s-2

k = 1

F = ma





Force = Rate of change of momentum

momentumin Change )(

)(

)(

21

mumvmvdFdtmvdFdt

mvdtdF

mvmu

tt

impulse

= Area under F-t curve

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Example 1Example 1

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More to Know 1More to Know 1

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Example 2Example 2

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Example 3Example 3


Newton’s Third Law of Motion


E.g water rocket:

water pushed out from the bottle,

opposite force exerted on the bottle

Newton’s Third Law of Motion:When two bodies interact, they exert equal but opposite forces on each other. They are called action and reaction.

push it upwards




E.g. Ball is being hit by bat

Magnitude of F1 = Magnitude of F2 = F

Action and reaction are equal in magnitude but opposite in direction. They act on different bodies.




E.g. Ball is released

Magnitude of F1 = Magnitude of F2

M >> m,

a1 >> a2


End


It is incorrect to think that a force can always cause a motion. Only a net force can do so. The net force is the vector sum of all forces acting on an object. If the forces cancel each other, there is no net force acted on the object.

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TextText



dtdmv

dtdvmdt

)mv(dF

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TextText



Q:Q: A body of mass 3 kg experiences a force F which varies with time t as shown in the figure. What is the increase in momentum of the body after 8 s?

Solution



Solution:Solution:

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TextText

s N 70

10)68(21

graph under the Area

momentumin Increase 80

Fdt



Q:Q: A plumb-line is hung from the roof of a train carriage. The bob is vertically above a markon the wall as shown when the train is stationary. Draw figures to show the position of the bob relative to the mark when(a) the train accelerates uniformly.(b) the train moves with uniform velocity.(c) the train retards uniformly.Explain your figures.

Solution



Solution:Solution:


(a) The net force acting on the bob is the horizontal component of tension T in the string, i.e. T cosθ. ∴ T cosθ = ma

where m is the mass of the bob. Since T cosθ is in the same direction as ma, the bob is swung to the back of the mark.

(b) Since v = constant, acceleration (a) = 0.

From T cosθ = ma

cosθ = (m/T)a

∴ cosθ = 0, θ = 90°

The bob is vertically above the mark.


Solution (cont’d):Solution (cont’d):

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(c) When acceleration = – a (or retardation)

By Force = Mass × Acceleration

T cosθ = m (– a)

T cosθ = –ma

T cosθ is in the opposite direction to the direction of motion. Therefore, the bob is swung to the front of the mark.


Q:Q: A space research rocket stands vertically on its launching pad. Prior to ignition, the mass of the rocket and its fuel is 1.9 × 103 kg. On ignition, gas is ejected from the rocket at a speed of 2.5 × 103 m s−1 relative to the rocket, and fuel is consumed at a constant rate of 7.4 kg s−1. Find the thrust of the rocket and hence explain why there is an interval between ignition and lift-off. Let acceleration of free-fall (g) be 10 m s−2. Solution



Solution :Solution :

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∴Thrust of rocket = (2.5 103) 7.4 = 1.85 104 N

Initial weight of rocket = mg = (1.9 103) 10 = 1.9 104 N

which is greater than the thrust of the rocket. Therefore, the rocket is not lifted immediately at ignition.

As the fuel is burnt, the weight of the rocket decreases. When the weight of the rocket is less than 1.85 104 N, the rocket will be lifted. Hence, there is an interval between ignition and lift-off.

dtdmv

)mv(dtdF

1© manhattan press (h.k.) ltd. newton’s first law of motion linear momentum and newton’s second...

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