Impulse

Impulse in physics refers to the effect of a force acting on a body for a brief time interval, resulting in a change in momentum. It is commonly observed during collisions, where the momentum of a body changes instantly before and after impact.

Impulse (J) is the product of the average net force and the time it acts on an object. Impulse describes how a force changes an object’s momentum over time."
\boxed{J = F .\Delta t}

Its SI unit is newton-second (N·s), which is equivalent to kg·m/s, and it is a vector quantity. It is a vector quantity.

Momentum

Momentum is the quantity of motion possessed by a body and it is given by:
\boxed{p = m \times v}
where
(m) = mass of the object
(v) = velocity of the object

Momentum depends on both mass and velocity. A heavy object moving slowly can have the same momentum as a light object moving fast. An object at rest has zero momentum.

Impulse-Momentum Theorem

The impulse-momentum theorem provides a direct connection between impulse and momentum. It tells us that the change in an object's momentum is equal to the impulse applied to it over a period of time.

In simpler terms, impulse measures how much an object's momentum changes due to a force acting over time. This helps students understand that impulse is not a separate concept but rather a way to describe how motion is altered.

We can express this relationship using the following equation:

Impulse = Final Momentum − Initial Momentum

According to Newton’s Second Law:
F = m a = m \frac{\Delta v}{\Delta t}

Multiplying both sides by (𝝙t):
F \Delta t = m \Delta v

Since (m \Delta v = \Delta p),

\boxed{J = \Delta p}

Thus, 'impulse is equal to the change in momentum of the body.'

Newton’s Second Law

The relationship between impulse and Newton's Law of Motion is crucial. Newton's second law is very useful for finding the value of the impulse.

We know that force acting on an object is given using,

F= m \, . a

As we know that acceleration (a= \frac{\Delta v}{\Delta t} )

F=m \,. \frac{\Delta v}{\Delta t}

F \Delta t = m \Delta v

F \Delta t = m .(v_f -v_i)

where FΔt implies that Impulse acting on the body and it is given as the change in linear momentum of the body, m(v_f-v_i)

Impulse in Collisions

During a collision, a large force acts for a brief time, producing a significant change in momentum. The damage caused depends on the magnitude of impulse. Increasing the time of impact (such as with airbags or padded surfaces) reduces the force and minimizes damage.

Learn more about, Newton’s Second Law of Motion

Impulse Examples

• When a person falls on a hard floor, the stopping time is tiny, resulting in a large force and more injury. Falling on sand or a soft surface increases the time of impact and reduces the force, causing less damage.
• Nylon ropes are used in rock climbing because they stretch slightly during a fall. This increases the stopping time and reduces the impact force, protecting the climber.
• In bat and racket sports, players are advised to follow through after hitting the ball. This increases the contact time between the bat/racket and the ball, producing a greater change in momentum and hence a more powerful shot.

You may also read,

• Newton's 2nd Law of Motion
• Equation of Motion
• Work-Energy Theorem

Solved Examples on Impulse

Example 1: An item comes to a halt when it collides with a solid wall. Calculate the object's impulse if the object was 2.0 kg in weight and travelled at a speed of 10 m/s before colliding with the wall.

Solution:

Given,
Mass of the object, m = 2.0 kg
Initial velocity of the ball, v_i = 10 m/s
Final velocity of the ball, v_f = 0 m/s
The formula for impulse is:
J = m × (v_f − v_i)
Substitute all the values in the above equation.
J = 2 × (0 - 10) kg m/s
= -20 kg m/s
Hence, the impulse on the object is -20 kg m/s.

Example: 2A kicks a ball rolling at 6 m/s; after the kick, the ball attains a velocity of 36 m/s. Find the impulse applied to the ball if its mass is 1/2 kg.

Solution:

We know that the Impulse formula is,
J = Δp

Given,
mass of ball (m) = 1/2 kg
Initial velocity of Ball (v_i) = 6 m/s
Final velocity of Ball (vi) = 36 m/s
Initial Momentum = mv_i = 1/2×6 = 3 kgm/s
Final Momentum = mv_f = 1/2×36 = 18 kgm/s
Impulse (J) = mv_f - mv_i  = 18 - 3 = 15 kgm/s
Thus, the Impulse applied to the ball is 15 kgm/s

Example 3: A golfer hits a ball of mass 100 g at a speed of 50 m/s. The golf club is in contact with the ball for 2 ms. Compute the average force applied by the club on the ball.

Solution:

Given,
Change in the velocity, Δv = 50 m/s
Mass of the ball, m = 100 g = 0.1 kg
Time of contact, t = 2 ms = 0.002 s
The formula of impulse is:
J = F × Δt = m × Δv
F = m × Δv / Δt
Substitute all the values in the above equation.
F = (0.1) × (50) / 0.002 N
= 2500 N
Hence, the average force applied on the ball is 2500 N.

Example 4: Calculate the impulse on a body hit by a force of 500 N with a time of contact equal to 0.1 s.

Solution: Given,
Force exerted on body, F = 500 N
Time of contact, Δt = 0.1 s
Formula for impulse is,
J = F × Δt
=(500) × (0.1) N s
= 50 N s
Hence, the impulse on body is 50 N s.

Unsolved Questions

Question 1: A force of 50 N acts on a body for 0.2 s. Find the impulse imparted.

Question 2: A constant force produces an impulse of 12 N·s in 3 s. Find the magnitude of the force.

Question 3: The velocity of a particle changes from 4 m/s to 10 m/s in 0.5 s. If the mass is 2 kg, calculate the impulse.

Question 4: A body with a mass of 1 kg experiences a force of F = 5t² N for 2 seconds. Find the impulse imparted.