# Newton’s Second Law of Motion

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• Momentum and Newton's Second Law of Motion
• Momentum, Impulse

## Newton’s Second Law

• The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

• Alternatively, the relationship between an object's mass m, its acceleration a, and the applied force F is F = ma; the direction of the force vector is the same as the direction of acceleration vector.
F ∝ (dp)/dt [Greater the change in momentum, greater is force]
F = k (dp)/dt
F = (dp)/ dt
F = d/ dt (mv)
Let, m: mass of the body be constant
F = m (dv)/dt
F = ma

• Newton’s Second law is consistent with the First law
F = ma
If F = 0, then a = 0
According to First law, if a = 0, Then F = 0
Thus, both the laws are in sync.

• Vector form of Newton’s Second law
bar F=barF_x hat i + bar F_y hat j+ bar F_z hat k

F_x = "dp"_x/ dt = ma_x

F_y = "dp"_y/ dt = ma_y

F_z = "dp"_z/dt = ma_z

• Newton’s Second Law was defined for point objects. For larger bodies,

a: acceleration of the centre of mass of the system.

F: total external force on the system

Problem 1: A car of mass 2 × 103 kg travelling at 36 km/hr on a horizontal road is brought to rest in a distance of 50m by the action of brakes and frictional forces. Calculate: (a) average stopping force (b) Time taken to stop the car

Solution.
m = 2 × 103
u = 36 km/hr = 10m/s
s = 50m
v = 0

To find: a, F, t
Third equation of motion:
v2 = u2 + 2as
0 = 100 + 2a × 50
Or,
a = -1 m/s2

Therefore, F = ma = (2 × 103) × 1 = 2 × 103 N

V = u + at
0 = 10 – t
t = 10 sec

Problem 2: The only force acting on a 5kg object has components Fx = 15N and Fy = 25N. Find the acceleration of the object.
Solution.
m= 5kg
F_x = 15N, F_y = 25N
F = F_x hat i + F_y hat j
| bar F|=sqrt(F_x^2 + F_y^2)= sqrt( 225 + 625) = sqrt 850

F = ma
Or, a = F/ m = sqrt 850/5= 5 (sqrt34)/5= sqrt34= 5.83 m/s^2

Impulse

• Impulse is defined as a force multiplied by time it acts over.

• For example: Tennis racket strikes a ball, an impulse is applied to the ball. The racket puts a force on the ball for a short time period.

F ∆t = ∆p

F = (∆p)/( ∆t) = Rate of Change of momentum

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