Impulse in Collisions
Impulse is the change in momentum of an object. When objects collide, they exert forces on each other over a certain period, resulting in changes to the momentum of each object. The effect of these changes depends on the initial momentum of each object, which is a product of its mass and velocity. This understanding helps us analyse how forces interact during a collision, enabling better designs that reduce impact and improve safety.
Use this page to revise the following concepts within impulse in collisions:
Impulse
Momentum , \(p\), is the vector quantity defined by the product of an object's mass and velocity.
\[ p = mv \]
Where:
- \(p\) = momentum \((\mathrm{kgms}^{-1})\)
- \(m\) = mass \((\mathrm{kg})\)
- \(v\) = velocity \((\mathrm{ms}^{-1})\)
The impulse, \(J\), is the change in momentum of an object:
\[ \begin{aligned} J &= \Delta p \\ J &= p_{\text{final}} - p_{\text{initial}} \\ J &= m(v - u) \end{aligned} \]
Where:
- \(J\) = impulse \((\mathrm{Ns})\)
- \(\Delta p\) = change in momentum \((\mathrm{kgms}^{-1})\)
- \(u\) = initial velocity \((\mathrm{ms}^{-1})\)
As Newton's second law states:
\[ \begin{aligned} F &= ma \\ F &= m\left(\frac{v - u}{\Delta t}\right) \end{aligned} \]
Where:
- \(F\) = force \((\mathrm{N})\)
- \(a\) = acceleration \((\mathrm{ms}^{-2})\)
- \(v\) = final velocity \((\mathrm{ms}^{-1})\)
- \(u\) = initial velocity \((\mathrm{ms}^{-1})\)
- \(\Delta t\) = change in time \((\mathrm{s})\)
We can describe impulse in terms of a force applied over a period of time:
This means that the force applied to an object to change its momentum depends on the time taken. The relationship is expressed as:
\[F= \frac{J}{\Delta t}\]
Since the time component \((\Delta t)\) is in the denominator, an increase in time reduces the force, and a decrease in time increases the force. For example, the longer it takes for an object to change its momentum, the less force is required to achieve that change.
Worked example
A \(70 \mathrm{~kg}\) person gets onto a train at a platform. The train is stationary and accelerates to a speed of \(22 \mathrm{~ms}^{-1}\). What is the impulse experienced by the person?
\[ \begin{aligned} m &= 70 \mathrm{~kg} \\ v &= 22 \mathrm{~ms}^{-1} \\ u &= 0 \mathrm{~ms}^{-1} \end{aligned} \]
\[ \begin{aligned} J &= m(v - u) \\ &= 70(22 - 0) \\ &= 1540 \mathrm{~Ns} \end{aligned} \]
Check your understanding
View
In summary:
- Impulse is a measure of the change in an object's momentum.
-
To stop an object, the force applied depends on the time taken to create the change in momentum:
- The longer the duration over which the force is applied, the smaller the average force.
- The shorter the duration over which the force is applied, the greater the average force.
Force in a collision
Impulse is described as the product of force applied and the period of time in which it is applied. Therefore, the force applied can be determined by the impulse divided by the period of time.
A key feature of this relationship is that, for a given change in momentum , the force applied is inversely proportional to the duration of the force. In other words, the longer the time period over which the force is applied, the smaller the magnitude of the force.
This principle is crucial in understanding how to reduce the impact of forces during a collision. For example, airbags in cars increase the time over which the force is applied to a person during a crash, thereby reducing the force experienced and minimising injury.
Check your understanding
View
Worked example
A \(70 \mathrm{~kg}\) person jumps from a height of \(1 \mathrm{~m}\) onto a soft mat, and it takes \(0.10 \mathrm{~s}\) for their momentum to change to zero upon landing. What is the average force applied to the person when they land?
The speed of the person just before they land can be calculated using:
\[ v = \sqrt{2gh} \]
Where:
\[ \begin{aligned} g &= \text{acceleration due to gravity at Earth's surface} \approx 9.8 \mathrm{~ms}^{-2} \\ h &= \text{height from the reference point} \; (\mathrm{~m}) \end{aligned} \]
Substitute the values:
\[ v = \sqrt{2(9.8)(1)} \approx 4.43 \mathrm{~ms}^{-1} \]
The change in momentum is:
\[ \Delta p = m(v_f - v_i) \]
Taking upwards as positive, the velocity just before landing is \(v_i = -4.43 \mathrm{~ms}^{-1}\), and the final velocity is \(v_f = 0 \mathrm{~ms}^{-1}\). Therefore:
\[ \Delta p = 70(0 - (-4.43)) = 310.10 \mathrm{~kgms}^{-1} \]
The average net force can be calculated using:
\[ F_{\text{net}} = \frac{\Delta p}{\Delta t} \]
Substitute the values:
\[ F_{\text{net}} = \frac{310.10}{0.10} = 3101 \mathrm{~N} \]
Therefore, the average net force on the person during landing is \(3101 \mathrm{~N}\) upwards.
If the average force applied by the mat is required, then:
\[ F_{\text{mat}} - mg = F_{\text{net}} \]
\[ F_{\text{mat}} = 3101 + 70(9.8) = 3787 \mathrm{~N} \]
If the stopping time were \(\Delta t = 0.01 \mathrm{~s}\), typical for landing on a hard surface, the average net force would be:
\[ F_{\text{net}} = \frac{310.10}{0.01} = 31010 \mathrm{~N} \]
This is \(10\) times the magnitude of the average net force experienced during landing on the soft mat.
Check your understanding
View
Straight-line collision
In a straight-line collision without external forces, momentum in the system is conserved. This means that the total momentum before the collision is equal to the total momentum after the collision.
|
\[ \Sigma p_\text{before} = \Sigma p_\text{after} \] |
This is known as the law of conservation of momentum.
There are several types of collisions:
- Objects collide and move separately
- Objects collide and then combine to move together as one
- Objects separate into multiple objects from a single object
Each of these scenarios requires an analysis of the number of objects before and after the collision to determine the total momentum in the system.
For two objects, where \( u_n \) is the initial velocity of object \( n \); and \( v_n \) is the final velocity of object \( n \):
- Object 1 collides into object 2 and they move separately:
- Object 1 collides into object 2 and then combine to move together as one object 3:
- Object 1 separates into object 2 and object 3:
\[ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \]
\[ m_1u_1 + m_2u_2 = m_3v_3 \]
\[ m_1u_1 = m_2v_2 + m_3v_3 \]
In all of these cases, the conservation of momentum can be written as
Total momentum before collision = Total momentum after collision.
When in doubt, this general expression will always work, no matter the type of collision you’re considering.
Worked example
A child is playing with a \(50 \mathrm{~g}\) magnetic train, which is pushed with an initial velocity of \(u_1 = 0.2 \mathrm{~ms}^{-1}\) into another stationary \(50 \mathrm{~g}\) magnetic train. After they collide and combine, they move together in the same direction. What is their final velocity?
As there are two objects that collide and then move together, use the principle of conservation of momentum:
\[ m_1u_1 + m_2u_2 = (m_1 + m_2)v \]
Here, \(m_1 + m_2\) is the combined mass of the two objects:
\[ m_1 + m_2 = 100 \mathrm{~g} = 0.1 \mathrm{~kg} \]
Substitute the given values into the equation:
\[ (0.05)(0.2) + (0.05)(0) = (0.1)v \]
\[ 0.01 = 0.1v \]
Rearrange for \(v\):
\[ v = \frac{0.01}{0.1} = 0.1 \mathrm{~ms}^{-1} \]
Therefore, the final velocity of the combined train is \(0.1 \mathrm{~ms}^{-1}\).
Check your understanding
View
Energy in collisions
Although momentum is always conserved in collisions within a closed system, kinetic energy is not necessarily conserved. Collisions can be classified as elastic or inelastic based on whether kinetic energy is retained or transformed into other forms of energy.
Elastic collisions are where kinetic energy is conserved. These occur only in ideal conditions and are not truly observed on a macroscopic scale, but many collisions in real life are approximately elastic, such as the collision of billiard balls. This is important to understand especially when modelling situations where the losses are minimal or assumed negligible.
Inelastic collisions occur when kinetic energy is not conserved. Instead, it is partially transformed into other types of energy such as heat or sound.
These collisions can be further classified as two types: standard inelastic collisions, where the objects collide and separate with energy losses; and perfectly inelastic collisions, where the objects stick together and move as a single entity after the collision.
An example of a standard inelastic collision is where a car crashes into another vehicle. The front of the car crumples and deforms, and a loud sound and heat are generated.
