Impulse and Momentum

Impulse and Momentum

Impulse and Momentum

Impulse

It’s defined as the quantity of describing the net effect of forces which act on any object upon a specific period of time. Impulse is expressed by the symbol J and takes the unit Newton*Second (N.S) according to its equation.

Impulse can be calculated mathematically according to the following equation:

J = F* Δt

Where,

-        J: is the impulse.

-        F: is the net force acting on the object.

-        Δt: is the duration of force effect.

Problem: A ball of 0.5 kg hits the ground when travelling downwards. The ball hits the ground in time of 0.02 seconds before go back upwards. The ground affects the ball by a force of 10 N. Calculate the impulse.

Given Data:

-        F = 10N.

-        Δt = 0.02 s.

Required:

-        J = ??

Solution

J = F* Δt

J = 10*0.02

J = 0.2 N. s

Momentum

It simply refers to the resistance of any moving object to stop. This expression is widely used in the sports field. In order to stop any moving object, the application of a force against the motion direction is required for a given time. Momentum is a vector quantity and is expressed physically by the symbol P.

It’s defined physically as the measurement of mass in motion. Momentum is described by the unit kg.m/s which is the multiplication of the unit of mass and unit of velocity. Momentum can be calculated mathematically by the following equation:

P = m*v

The change in momentum of a moving object which is expressed by the symbol ΔP can be calculated mathematically as follows:

ΔP = m.v_f – m.v_i = m(v_f - v_i)

Problem: A basketball ball having 1.5kg mass and 8m/s velocity moves to the west. Calculate the momentum of this ball.

Given Data:

-        m = 1.5 kg.

-        v = 8 m/s

Required:

-        P = ??

Solution

P = m*v

P = 1.5*8 = 12 kg.m/s, west.

Impulse and Momentum Theorem

It was found that the impulse can be described by the change in momentum. Since v = a * Δt so, the equation of momentum can be written as follows:

ΔP = m * a * Δt

By using Newton’s second equation: F = m*a, the momentum equation will be written as follows: ΔP = F * Δt, which is the same equation of impulse J. So, in impulse and momentum theorem, there is a direct connection between the impulse and momentum. Here is a solved problem on the impulse and momentum theorem.

Problem: A 0.25 kg ball free fall from a height of h₁ = 3.6 meters and reflected a height of h₂ = 1.6 meters. Acceleration due to gravity (g) = 10 m/s². Determine impulse.

Given Data:

-        m = 0.25 kg.

-        h₁ = 3.6 m.

-        h₂ = 1.6 m.

-        g = 10 m/s².

Required:

-        Impulse (J).

Solution

1.    We have to get the velocity before and after the collision, by using the equations of free-falling bodies.

v_t² = v_o² + 2 g h

Before collision:

v_t² = 0 + 2*10*3.6 = 72

v_t = 8.5 m/s

After collision:

0 = v_o² + 2*(-10)*1.6

v_o = 5.66 m/s

So,

-        v_i = -8.5 m/s (opposite direction)

-        v_f = 5.66 m/s

J = ΔP

J = m (v_f - v_i)

J = 0.25 (5.66 – (-8.5)) = 3.54 N.s