U3AOS1 Topic 10: Impulse

Introduction

Impulse is a fundamental concept in physics that describes the change in momentum of an object when a force is applied over a specific period. It plays a crucial role in understanding how forces influence the motion of objects and is essential in fields ranging from sports to engineering. Here’s a closer look at impulse, including its definition, formula, and practical applications.

Definition of Impulse

Impulse is defined as the product of the force applied to an object and the time duration over which the force acts. It effectively quantifies the impact of a force on an object's momentum. Mathematically, impulse JJ can be expressed as:

J=FΔtJ = F \Delta t

where FF is the force applied, and Δt\Delta t is the time interval during which the force acts.

Relationship to Momentum

Impulse is directly related to momentum, which is the product of an object's mass mm and its velocity vv. The change in momentum (Δp\Delta p) of an object is equal to the impulse applied to it. This relationship is expressed by the impulse-momentum theorem:

J=ΔpJ = \Delta p

where Δp=mΔv\Delta p = m \Delta v (change in momentum). Hence, impulse can also be written as:

J=mΔvJ = m \Delta v

Calculation of Impulse

  1. Constant Force:

    • If the force applied is constant over time, impulse is simply the product of this constant force and the time duration. For example, if a force of 10 N10\ \text{N} is applied for 2 s2\ \text{s}, the impulse is: J=FΔt=10×2=20 N⋅sJ = F \Delta t = 10 \times 2 = 20\ \text{N}\cdot\text{s}

  2. Variable Force:

    • If the force varies with time, impulse can be calculated by integrating the force over time. For a force F(t)F(t) that changes with time, the impulse is given by: J=∫t1t2F(t) dtJ = \int_{t_1}^{t_2} F(t) \, dt

Practical Examples

  1. Sports:

    • In sports, such as baseball or golf, players apply a force over a short duration to change the momentum of the ball. For instance, when a baseball bat hits a ball, the impulse imparted by the bat changes the ball’s momentum, determining its speed and direction.

  2. Automobile Safety:

    • In vehicle crashes, crumple zones are designed to extend the time over which the collision occurs, thereby reducing the force experienced by passengers. This increase in collision time reduces the impulse imparted to the occupants, minimizing injury.

  3. Rocket Propulsion:

    • Rockets work on the principle of impulse. The expulsion of gas at high speed generates a large impulse, which results in a reaction force that propels the rocket forward.

Impulse and the Change of Momentum

Impulse directly affects the change in an object's momentum. For instance, if a soccer ball is kicked with an impulse of 50 N⋅s50\ \text{N}\cdot\text{s} and its mass is 0.5 kg0.5\ \text{kg}, the change in velocity of the ball can be calculated by:

Δv=Jm=500.5=100 m/s\Delta v = \frac{J}{m} = \frac{50}{0.5} = 100\ \text{m/s}

Thus, impulse provides a measure of how much the momentum of an object changes due to a force.

Impulse is the change in momentum and follows the equation

\[ \displaystyle \Huge I = \Delta \rho = m \Delta v = m(v-u)\]


I=Δρ=mΔv=m(v−u)I = \Delta\rho = m\Delta v = m(v-u)


Note: Impulse is also a vector.

Therefore the direction matters!!


Impulse can also be calculated by:

\[ \displaystyle \Huge I=F \Delta t\]

I=FΔtI = F\Delta t


Example 1
A soccer player applies a constant force of 150 N150\ \text{N} to a ball for 0.2 s0.2\ \text{s}. Calculate the impulse imparted to the ball.

Impulse JJ is given by the product of the force FF and the time duration Δt\Delta t:

J=FΔtJ = F \Delta t

Substitute F=150 NF = 150\ \text{N} and Δt=0.2 s\Delta t = 0.2\ \text{s}:

J=150×0.2=30 N⋅sJ = 150 \times 0.2 = 30\ \text{N}\cdot\text{s}

So, the impulse imparted to the ball is 30 N⋅s30\ \text{N}\cdot\text{s}.

Example 2
An ice hockey puck with a mass of 0.15 kg0.15\ \text{kg} is initially at rest. A player hits the puck with a force that results in an impulse of 12 N⋅s12\ \text{N}\cdot\text{s}. What is the final velocity of the puck?

Impulse JJ is equal to the change in momentum Δp\Delta p:

J=Δp=mΔvJ = \Delta p = m \Delta v

Rearrange to find the change in velocity Δv\Delta v:

Δv=Jm\Delta v = \frac{J}{m}

Substitute J=12 N⋅sJ = 12\ \text{N}\cdot\text{s} and m=0.15 kgm = 0.15\ \text{kg}:

Δv=120.15=80 m/s\Delta v = \frac{12}{0.15} = 80\ \text{m/s}

Since the puck was initially at rest, this is its final velocity. Therefore, the final velocity of the puck is 80 m/s80\ \text{m/s}.

Example 3
A car experiences a varying braking force described by F(t)=100−5tF(t) = 100 - 5t, where F(t)F(t) is in newtons and tt is in seconds. If the braking force acts from t=0t = 0 to t=5 st = 5\ \text{s}, calculate the total impulse experienced by the car.

To find the impulse, integrate the force over time:

J=∫05F(t) dt=∫05(100−5t) dtJ = \int_{0}^{5} F(t) \, dt = \int_{0}^{5} (100 - 5t) \, dt

Perform the integration:

J=[100t−5t22]05J = \left[ 100t - \frac{5t^2}{2} \right]_{0}^{5} J=(100×5−5×522)−(100×0−5×022)J = \left( 100 \times 5 - \frac{5 \times 5^2}{2} \right) - \left( 100 \times 0 - \frac{5 \times 0^2}{2} \right) J=500−1252J = 500 - \frac{125}{2} J=500−62.5=437.5 N⋅sJ = 500 - 62.5 = 437.5\ \text{N}\cdot\text{s}

So, the total impulse experienced by the car is 437.5 N⋅s437.5\ \text{N}\cdot\text{s}.

Confident?

Select quiz difficulty