- 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
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
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.
`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 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