IIT JEE Linear Momentum | JEE Mechanics Momentum

Table of Content

Quantitative Definition of Linear Momentum
Non-Relativistic Momentum
Relativistic Momentum
Newton’s Second Law of Motion and Momentum
Momentum of a System of Particles
Related Resources

Momentum of a Man From Its Rest Position Momentum of a body is defined as the amount of motion contained in a body.

Quantity of motion or the momentum of the body depends upon,

(a) mass of the body.

(b) velocity of the body.

Therefore momentum of a body of mass ‘m’ and velocity ‘v’ will be,

$\vec{p} = m\vec{v}$

Quantitative Definition of Linear Momentum

Momentum of a body is equal to the product of its mass and velocity. Momentum is a vector quantity and possesses the direction of velocity.

Non-Relativistic Momentum of Guys

Units:-

S.I:- kg m s^-1

C.G.S:- g cm s^-1

Momentum can be put into following two categories.

Dimension:-

[MLT^-1]

(a) Non-Relativistic Momentum

According to classical physics (or non-relativistic physics) which is based upon the concepts of Newton’s laws of motion, mass of a body is considered to be a constant quantity, independent of the velocity of body. In that case momentum $\vec{p}$ is given by,

$\vec{p} = m\vec{v}$ .

Thus, momentum of a body is a linear function of its velocity.

(b) Relativistic Momentum

Relativistic Momentum

In accordance to Einstein’s special theory of relativity, mass of a body depends upon the relative velocity ‘v’ of the body with respect to the observer. If ‘m₀’ is the mass of body observed by an observer at rest with respect to body, its relativistic mass ‘m’ is given by,

$m = \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{^{2}}}}}$

Therefore, momentum of a body according to the concepts of theory of relativity is given by,

Thus, relativistic momentum is not a linear function of v.

Newton’s Second Law of Motion and Momentum

Movement of the Block Due to The Force Applied By the Man

The rate of change of momentum of a body is proportional to the resultant external force acting on the body and is the direction of that force.

Newton’s second law provides a quantitative definition of the force.

Let $\vec{v}$ be the instantaneous velocity of the body. Momentum $\vec{p}$ of the body is given by,

$\vec{p} = m\vec{v}$

According to second law,

$\vec{F}$ ∝ (rate of change of momentum)

Or,

$\vec{F}\alpha \frac{d\vec{p}}{dt}$

Or,

$\vec{F}\alpha \frac{d(m\vec{v})}{dt}$

Or,

$\vec{F}\ =k \frac{d(m\vec{v})}{dt}$

Here ‘k’ is the constant of proportionality. Mass ‘m’ of a body is considered to be a constant quantity.

$\vec{F}\ =k m\frac{d(\vec{v})}{dt}$

or,

$\vec{F}\ =k m\vec{a}$

The units of force are also selected that ‘k’ becomes one.

Thus, if a unit force is chosen to be the force which produces a unit acceleration in a unit mass,

i.e., F = 1, m = 1 and a = 1.

Then, k = 1

So, Newton’s second law can be written , in mathematical form, as

$\vec{F}\ =m\vec{a}$

i.e., Force = (mass) (acceleration)

Momentum of a System of Particles

Momentum Of a System of Particle

Consider a system of n particles with masses m₁,m₂ etc. Each with its own velocity, and linear momentum. The particle may interact with each other, and external forces may act on them as well. The system as a whole has a total linear momentum P, which is defined to be the vector sum of the individual particle linear momenta.

Here we assume that total mass M = ∑_i m_i of the system remains constant with time. If we assume that each particle will have some momentum then the system as a whole will have momentum P given by,

P = p₁ + p₂ + p₃ + . . . + p_n

If we compare this equation with

v_cm= (m₁v₁ + m₂v₂+ . . . + m_nv_n)/ M,

We see that,

P = Mv_cm

Which gives us another way to define the linear momentum of a system of particles.

The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass.

Taking the time derivative of this equation and comparing with $\Sigma$ F = Ma_cm, we can write Newton's second law for a system of particles in the equivalent form

$\sum F_{ext} = \sum \frac{dP}{dt}$

Table below displays the important relations that we have derived for single particles and for systems of particles.

Problem 1:-

A 4.88-kg object with a speed of 31.4 m/s strikes a steel plate at an angle of 42.0º and rebounds at the same speed and angle (Below Figure). What is the change (magnitude and direction) of the linear momentum of the object?

A Object Strikes with a Steel Plate at an Angle 42.0º

Concept:-

Momentum of the object (p) is defined as the mass of the object (m) time’s velocity of the object (v).

p = mv …… (1)

Solution:-

The below figure shows a 4.88-kg object with a speed of 31.4 m/s strikes a steel plate at an angle of 42.0^° and rebounds at the same speed with same angle which is shown geometrically.

The initial momentum of the object is p_i when it strikes the steel plate.

To find initial momentum p_i, substitute p_i for p, 4.88-kg for mass m of the object and 31.4 m/s for velocity of the object in the equation p = mv,

p_i = mv

= (4.88 kg) (31.4 m/s)

= 153 kg. m/s …… (2)

Since the final momentum p_f of the object is equal to the initial momentum p_i of the object, therefore,

p_f = 153 kg. m/s …… (3)

In the above figure the angle θ will be,

θ = 42º + 42º

= 84º

Using cosine law, the magnitude of change of the linear momentum of the object Δp will be,

Δp = √( p_i)² + (p_f)² + 2 (-p_i)(p_f)cos θ …… (4)

To obtain magnitude of linear momentum Δp, substitute 153 kg. m/s for p_i, 153 kg. m/s for p_f and 84° for the angle θ in the equation Δp = √( p_i)² + (p_f)² + 2 (-p_i)(p_f) cos θ,

Δp = √( p_i)² + (p_f)² + 2 (-p_i)(p_f) cos θ

= √(153 kg. m/s)² + (153 kg. m/s)² - 2 (153 kg. m/s)( 153 kg. m/s) cos 84°

= 205 kg. m/s

From the above observation we conclude that, the change of the linear momentum of the object will be, 205 kg. m/s and also from the figure the direction of change of the linear momentum of the object is perpendicular to the steel plate.

Problem 2:-

A 2000-kg truck travelling north at 40.0 km/h turns east and accelerates to 50.0 km/h. What is the magnitude and direction of the cahnge of the truck’s momentum?

Concept:-

Momentum of the body p is equal to the mass of the body m times velocity of the body v.

So, p = mv

Initial momentum p_i of the truck having initial velocity v_i will be,

p_i = mv_i

and the final momentum p_f of the truck having final velocity v_f will be,

p_f = mv_f

Momentum is a vector quantity. If Δp_x is the x-component of momentum p and Δp_y is the y-component of momentum p, then the angle θ between Δp_x and Δp_y will be,

θ = tan^-1(Δp_y/ Δp_x)

Solution:-

To obtain the Initial momentum p_i of the truck having initial velocity v_i, substitute 2000 kg for m, 40 km/h for v_i in the equation p_i = mv_i,

p_i = mv_i

= (2000 kg) (40 km/h)

= 8.00×10⁴ kg.km/h

So, p_i = 8.00×10⁴ kg.km/h j (j for north direction)

To obtain the final momentum p_f of the truck having final velocity v_f, substitute 2000 kg for m, 50 km/h for v_i in the equation p_f = mv_f,

p_f = mv_f

= (2000 kg) (50 km/h)

= 1.00×10⁵ kg.km/h

So, p_f = 1.00×10⁵ kg.km/h i (i for east direction)

Thus, Δp = p_f - p_i

= (1.00×10⁵ kg.km/h i) – (8.00×10⁴ kg.km/h j)

So the magnitude of Δp will be,

Δp = √(Δp_x)² +(Δp_y)²

= √ (1.00×10⁵ kg.km/h)² +(8.00×10⁴ kg.km/h)²

= 1.28 ×10⁵ kg.km/h

To obtain the direction, substitute 8.00×10⁴ kg.km/h for Δp_y and 1.00×10⁵ kg.km/h for Δp_x in the equation θ = tan^-1(Δp_y/ Δp_x),

θ = tan^-1(Δp_y/ Δp_x)

= tan^-1[(8.00×10⁴ kg.km/h) /( 1.00×10⁵ kg.km/h)]

=38.7 ° (south of east)

From the above observation we conclude that, the magnitude of the change of the truck’s momentum will be 1.28 ×10⁵ kg.km/h and the direction will be 38.7 ° (south of east).

Problem 3:-

The block in below figure slides without friction. what is the velocity v of the 1.6-kg block after the collision?

Two Blocks (Slides Without Friction) Collides with Each Other

Concept:-

Momentum of the body p is equal to the mass of the body m times velocity of the body v.

So, p = mv

In accordance to the principle of conservation of energy, the final momentum of the system is equal to the initial momentum of the system.

Consider the initial momentum of the block 1 is p_i,1, initial momentum of the block 2 is p_i,2, final momentum of the block 1 is p_f,1 and final momentum of the block 2 is p_f,2.

To obtain the initial momentum p_i,1 of the block 1, substitute p_i,1 for p, m₁ for the mass of the block 1 and v_1,i for the initial velocity of the block 1in the equation p = mv,

p = mv

p_i,1 = m₁ v_1,i

To obtain the initial momentum p_i,2 of the block 2, substitute p_i,2 for p, m₂ for the mass of the block 2 and v_2,i for the initial velocity of the block 2 in the equation p = mv,

p = mv

p_i,2 = m₂ v_2,i

To obtain the final momentum p_f,1 of the block 1, substitute p_f,1 for p, m₁ for the mass of the block 1 and v_1,f for the final velocity of the block 1in the equation p = mv,

p = mv

p_f,1 = m₁ v_1,f

To obtain the final momentum p_f,2 of the block 2, substitute p_f,2 for p, m₂ for the mass of the block 2 and v_2,f for the final velocity of the block 2 in the equation p = mv,

p = mv

p_f,2 = m₂ v_2,f

Solution:-

So applying conservation of momentum to this system, the sum of the initial momentum of the block 1 and block 2 will be equal to the sum of the final momentum of the block 1 and block 2.

p_f,1 + p_f,2 = p_i,1 + p_i,2

Substitute, m₁ v_1,f for p_f,1, m₂ v_2,f for p_f,2, m₁ v_1,i for p_i,1 and m₂ v_2,i for p_i,2 in the equation p_f,1 + p_f,2 = p_i,1 + p_i,2,

p_f,1 + p_f,2 = p_i,1 + p_i,2

m₁ v_1,f+ m₂ v_2,f = m₁ v_1,i + m₂ v_2,i

m₁ v_1,f = m₁ v_1,i + m₂ (v_2,i - v_2,f)

So,

v_1,f = v_1,i + m₂/ m₁ (v_2,i- v_2,f)

To obtain the velocity v of the 1.6-kg block after the collision (v_1,f), substitute 5.5 m/s for v_1,i, 2.4 kg for m₂ and 1.6 kg for m₁, 2.5 m/s for v_2,i and 4.9 m/s for v_2,f in the equation v_1,f = v_1,i + m₂/ m₁ (v_2,i- v_2,f),

v_1,f = v_1,i + m₂/ m₁ (v_2,i- v_2,f)

= (5.5 m/s) + (2.4 kg)/(1.6 kg) [(2.5 m/s) – (4.9 m/s)]

= 1.9 m/s

From the above observation we conclude that, the velocity v of the 1.6-kg block after the collision (v_1,f) would be 1.9 m/s.

Problem 4:-

A golfer hits a golf ball, imparting to it an initial velocity of magnitude 52.2 m/s directed 30º above the horizontal. Assuming that the mass of the ball is 46.0 g and the club and ball are in contact for 1.20 ms, find (a) the impulse imparted to the ball, (b) the impulse imparted to the club, and (c) the average force exerted on the ball by the club.

Concept:-

The impulse of the net force J acting on a particle during a given time interval is equal to the change in momentum Δp of the particle during that interval.

So, J = Δp

= p_i – p_f

Here p_f is the final momentum and p_i is the initial momentum of the body.

Force (F) is equal to the rate of change of momentum (Δp) body.

Therefore the average force will be,

F = Δp/ Δt

Solution:-

The initial momentum p_i of the ball will be,

p_i = mv_i

To obtain the initial momentum p_i of the ball, substitute 46.0 g for the mass of the ball and 52.2 m/s for v_i in the equation p_i = mv_i,

p_i = mv_i

= (46.0 g) (52.2 m/s)

= (46.0 g×10^-3 kg/1 g) (52.2 m/s)

= 2.4 kg.m/s

= (2.4 kg.m/s) (1 N/1 kg.m/s²)

= 2.4 N.s

As the final speed of the ball is zero, thus the final momentum p_f of the ball will be zero.

So, p_f = 0

To find out the impulse J imparted to the ball, substitute 2.4 N.s for p_i and 0 for p_f in the equation J = p_i – p_f,

J = Δp

= p_i – p_f

= 2.4 N.s -0

= 2.4 N.s

Therefore the impulse J imparted to the ball would be 2.4 N.s.

b) The impulse J imparted to the club is just opposite that imparted to the ball. Therefore the impulse J imparted to the club will be 2.4 N.s.

c) To obtain the average force F exerted on the ball by the club, substitute 2.4 N.s for Δp and 1.20 ms for t in the equation F = Δp/ Δt,

F = Δp/ Δt

= 2.4 N.s/1.20 ms

= (2.4 N.s)/((1.20 ms)(10^-3 s/1ms))

= 2000 N

From the above observation we conclude that, the average force F exerted on the ball by the club would be 2000 N.

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