Types Of Collision

Types Of Collision

Vishal kumarUpdated on 02 Jul 2025, 07:33 PM IST

Linear momentum, a fundamental concept in physics, plays a crucial role in understanding the motion of objects. It is defined as the product of an object's mass and velocity, indicating how much force is needed to change its motion. There are various types of linear momentum, such as translational, angular, and relativistic, each relevant in different scenarios. For instance, when you push a shopping cart, the cart's linear momentum increases as it accelerates, making it harder to stop. Similarly, understanding linear momentum helps explain everyday phenomena like car collisions, where the momentum of each vehicle determines the impact's outcome. By exploring these types, we gain a deeper insight into the laws of motion that govern our daily lives.

This Story also Contains

  1. Linear Momentum
  2. Solved Examples Based on Linear Momentum
  3. Summary
Types Of Collision
Types Of Collision

Linear Momentum

Linear momentum is a core concept in physics that helps us understand how objects move and interact. Defined as the product of an object's mass and its velocity, linear momentum reflects how much force is required to alter the motion of that object.

  1. The linear momentum of a body is the quantity of motion contained in the body.

  2. It is measured in terms of the force required to stop the body in a unit of time.

  3. If a body of mass m is moving with velocity, then its linear momentum $\vec{p}$ is given by $\vec{p}=m \vec{v}$.

  4. It is a vector quantity and its direction is the same as the direction of the velocity of the body.

  5. S.I. Unit : kg-m/sec

  6. Dimension- MLT-1

  7. If two objects of different masses have the same momentum, the lighter body possesses greater velocity.

As $\mathrm{p}=\mathrm{m}_1 \mathrm{v}_1=\mathrm{m}_2 \mathrm{v}_2=$ constant

$\therefore \frac{V_1}{V_2}=\frac{m_2}{m_1} \Rightarrow V \alpha \frac{1}{m}$

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Solved Examples Based on Linear Momentum

Example 1: An object of mass 1 kg travels at a constant speed of 10 m/s. Calculate the linear momentum (in kg-m/s )of the object.

1) 10

2) 40

3) 30

4) 15

Solution:

Given : Mass, $(m)=1 \mathrm{KgSpeed},(v)=10 \mathrm{~m} / \mathrm{sSo}$, Linear momentum, $\vec{p}=m \vec{v}$ $\vec{p}=m \vec{v} \Rightarrow \vec{p}=1 \mathrm{~kg} \times 10 \mathrm{~m} / \mathrm{s}=10 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

Hence the answer is the option (1).

Example 2: If the speed and mass of an object are doubled, which of the following is true?

1) The momentum of the object is doubled

2) The momentum of the object is quadrupled.

3) The momentum remains the same.

4) The momentum is halved.

Solution :

Given :
Speed, $(v)=2 v$
Mass, $(m)=2 m$
We know,
Momentum, $(p)=m v$
According to the question of mass and speed is doubled. Then, $p=2 m \times 2 v=4 m v=4 p$

Hence the answer is the option (2).

Example 3: A ball of mass 0.20 kg hits a wall at an angle of $45^{\circ}$ with a velocity of $25 \mathrm{~m} / \mathrm{s}$. If the ball rebounds at $90^{\circ}$ to the direction of incidence with the same speed. Calculate the magnitude of change in momentum of the ball.

1) $5 \sqrt{2} \mathrm{Ns}$
2) 5 Ns
3) $3 \sqrt{2} \mathrm{~N} \mathrm{~s}$
4) 10 Ns

Solution:

Given
mass, $\mathrm{m}=0.2 \mathrm{~kg}$
initial and final speed, $\mathrm{v}=25 \mathrm{~m} / \mathrm{s}$
The magnitude of Initial linear momentum $\left(\overrightarrow{p_i}\right)$ and final linear momentum $\left(\overrightarrow{p_f}\right)$

$
p=\left|\overrightarrow{p_f}\right|=\left|\overrightarrow{p_i}\right|=m v \Rightarrow p=5 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}
$
The angle between the initial and final momentum vectors is $90^{\circ}$, as shown in the figure below

Change in linear momentum

$\begin{aligned} & \Delta \vec{p}=\overrightarrow{p_f}-\overrightarrow{p_i} \\ & |\Delta \vec{p}|=\left|\overrightarrow{p_f}-\overrightarrow{p_i}\right| \\ & |\Delta \vec{p}|=\sqrt{p^2+p^2-2 p^2 \cos 90^0} \\ & |\Delta \vec{p}|=\sqrt{2} p \\ & \Rightarrow|\Delta \vec{p}|=5 \sqrt{2} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\end{aligned}$

Hence the answer is the option (1).

Example 4: A ball is thrown vertically up (taken as + z-axis) from the ground. The correct momentum-height (p-h) diagram is :

1)

2)

3)

4)

Solution

As we know,

$
\begin{aligned}
& v^2=u^2-2 g h \\
& v=\sqrt{u^2-2 g h} \\
& |P|=m v \\
& |P|=m \sqrt{u^2-2 g h}
\end{aligned}
$
Therefore, the graph between p and h cannot have a straight line.
During the upward journey as h increases, p decreases, and in the downward journey as h decreases p increases.

Hence the answer is the option (4).

Example 5: An object is thrown vertically upwards. At its maximum height, which of the following quantities becomes zero?

1) Momentum

2) Potential Energy

3) Acceleration

4) Force

Solution:



$\rightarrow$ At maximum height, its velocity becomes zero
$\rightarrow$ Therefore, the momentum of the object becomes zero

Hence the answer is the option (1).

Summary

Linear momentum, defined as the product of an object's mass and velocity, is a crucial concept in physics that helps explain the motion of objects. It is a vector quantity, with its direction aligning with the object's velocity. The momentum of an object determines the force required to change its motion. Key examples include calculating momentum changes in collisions and understanding the effects of doubling an object's mass and velocity.

Frequently Asked Questions (FAQs)

Q: How are collisions used in particle physics experiments?
A:
Collisions are central to particle physics experiments in several ways:
Q: How do collisions contribute to pressure in gases?
A:
Collisions play a fundamental role in creating pressure in gases:
Q: What is the relationship between collision dynamics and the laws of thermodynamics?
A:
Collision dynamics and the laws of thermodynamics are closely related, particularly in the context of energy transformations:
Q: How do collisions affect the internal energy of colliding objects?
A:
Collisions can significantly affect the internal energy of colliding objects, especially in inelastic collisions. During an inelastic collision, some of the initial kinetic energy is converted into internal energy of the objects. This can manifest as an increase in temperature (thermal energy), deformation of the objects (elastic potential energy), or even changes in the objects' molecular or atomic structure. In perfectly inelastic collisions, the maximum amount of kinetic energy is converted to internal energy. Understanding these energy transformations is crucial for analyzing real-world collisions and their effects on materials.
Q: How do collisions at different scales compare, from subatomic to astronomical?
A:
Collisions occur at vastly different scales in the universe, from subatomic to astronomical, and while the fundamental principles of conservation of momentum and energy apply at all scales, the nature and effects of these collisions can vary dramatically:
Q: What is the significance of momentum exchange in collisions?
A:
Momentum exchange is a fundamental aspect of all collisions. It refers to the transfer of momentum between colliding objects. The significance of momentum exchange lies in its universality - it occurs in all types of collisions, whether elastic, inelastic, or perfectly inelastic. The principle of conservation of momentum is based on this exchange. Understanding momentum exchange helps predict the motion of objects after collision, which is crucial in various applications, from particle physics to spacecraft design. It also explains phenomena like Newton's cradle and the behavior of billiard balls after collision.
Q: How do collisions relate to the concept of work and energy?
A:
Collisions are intimately related to the concepts of work and energy. During a collision, forces between objects do work, leading to energy transfers and transformations. In elastic collisions, the work done by collision forces is reversible, conserving kinetic energy. In inelastic collisions, some of this work goes into deforming the objects or generating heat, leading to a loss of kinetic energy. The work-energy theorem, which states that the net work done on an object equals its change in kinetic energy, is particularly useful in analyzing collisions and understanding energy transformations during impacts.
Q: What is the role of friction in collisions?
A:
Friction plays a significant role in many real-world collisions, especially those involving surfaces in contact. During a collision, friction can affect the objects' motion in several ways:
Q: What is the significance of the line of impact in collisions?
A:
The line of impact is the line along which the force acts during a collision. It's significant because it determines how the collision affects the objects' motion. In a central collision, where the line of impact passes through both objects' centers of mass, the collision affects only the objects' translational motion. In an oblique collision, where the line of impact doesn't pass through the centers of mass, the collision can induce both translational and rotational motion. Understanding the line of impact is crucial for predicting the behavior of objects after collision, especially in complex scenarios like billiards or vehicle collisions.
Q: Can you explain the concept of coefficient of restitution in more detail?
A:
The coefficient of restitution (e) is a measure of the "bounciness" of a collision, ranging from 0 to 1. It's defined as the ratio of the relative velocity of separation to the relative velocity of approach for two colliding objects. Mathematically, e = -(v2f - v1f) / (v2i - v1i), where v represents velocities, and subscripts 1 and 2 denote the two objects. A coefficient of 1 indicates a perfectly elastic collision, 0 indicates a perfectly inelastic collision, and values between 0 and 1 represent partially inelastic collisions. The coefficient of restitution helps predict the behavior of objects after collision and is crucial in fields like sports physics and materials science.