When two cars collide, the total momentum before and after the collision is the same, even though the cars may change speed or direction. This case can be better understood by studying the Law of Conservation of Linear Momentum in Physics. The below article will help us to understand the Law of Conservation of Linear Momentum with their examples in detail.
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The law of conservation of linear momentum works on the principle that when no external forces act on a system of two colliding objects, the vector sum of their linear momentum remains constant and is unaffected by their mutual interaction.
"The law of conservation of linear momentum states that if a system's net external force is zero, the system's total momentum remains constant."
S.I Unit of linear momentum is $\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ and the direction of linear momentum remains same as direction of velocity.
$$
\mathbf{p}=m \cdot \mathbf{v}
$$
Where:
Linear momentum of a body is the quantity of motion contained in the body.
It is measured in terms of the force required to stop the body in a unit time.
If a body of mass $m$ is moving with velocity $\vec{v}$, then its linear momentum $\vec{p}$ is given by $\vec{p}=m \vec{v}$.
It is a vector quantity and its direction is the same as the direction of the velocity of the body.
S.I. Unit: kg-m/sec
Dimension $-\mathrm{MLT}^{-1}$
If two objects of different masses have the same momentum, the lighter body possesses greater velocity.
$\begin{aligned} & \text { As } \mathrm{p}=\mathrm{m}_1 \mathrm{v}_1=\mathrm{m}_2 \mathrm{v}_2=\text { constant } \\ & \therefore \frac{V_1}{V_2}=\frac{m_2}{m_1} \Rightarrow V \alpha \frac{1}{m}\end{aligned}$
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When object A exerts a force on object B, object B reacts with a force of equal magnitude but in the opposite direction, according to Newton's third law.
The principle of conservation of momentum is deduced from Newton's third law of motion
Consider two colliding particles A and B with masses of $m_1$ and $m_2$ with starting and ultimate velocities of $u_1$ and $v_1$ for A and $u_2$ and $v_2$ for B, respectively. The contact time between two particles is denoted by the letter t.
As shown in the figure:
The change in the momentum of particle A is:
$
\Delta P_A = m_1 (v_1 - u_1)
$
The change in the momentum of particle B is:
$
\Delta P_B = m_2 (v_2 - u_2)
$
Using Newton's Third Law of Motion, we can write:
$
F_{BA} = -F_{AB}
$
Substituting for the forces:
$
F_{BA} = m_2 a_2 = \frac{m_2 (v_2 - u_2)}{t}
$
$
F_{AB} = m_1 a_1 = \frac{m_1 (v_1 - u_1)}{t}
$
Equating the two:
$
\frac{m_2 (v_2 - u_2)}{t} = -\frac{m_1 (v_1 - u_1)}{t}
$
Simplifying the equation:
$
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
$
As a result, if no external force is exerted on the system, the momentum after the collision is equal to the momentum before the collision.
As a result, the equation of the law of conservation of momentum is as follows: $\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2$ represents the total momentum of particles A and B before the collision, and $\mathrm{m}_2 \mathrm{v}_2+\mathrm{m}_2 \mathrm{v}_2$ represents the total momentum of particles A and B after the collision.
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These are some of the examples of the conservation of linear momentum that we see in our day-to-day lives. Now we will see some solved examples of the law of conservation of linear momentum.
Question 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.
(a) 10
(b) 40
(c) 30
(d) 15
Solution:
Given : Mass, $(m)=1 \mathrm{Kg}$
$ Speed (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 option (a).
Question 2: If the speed and mass of an object are doubled, which of the following is true?
(a) The momentum of the object is doubled
(b) The momentum of the object is quadrupled.
(c) The momentum remains the same.
(d) The momentum is halved.
Solution:
Given :
Speed, $(v)=2 v$
Mass, $(m)=2 m$
We know,
Momentum, $(p)=m v$
According to the question if mass and speed are doubled. Then,
$
p=2 m \times 2 v=4 m v=4 p
$
Hence, the answer is option (b).
Question 3: A ball of mass 0.20 kg hits a wall at an angle of $45^{\circ}$ with a velocity of $25 \mathrm{~m} / \mathrm{s}$. Suppose 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.
(a) $5 \sqrt{2} N s$
(b) 5 Ns
(c) $3 \sqrt{2} N s$
(d) 10 Ns
Solution:
Given-
mass, $\mathrm{m}=0.2 \mathrm{~kg}$
initial and final speed, $v=25 \mathrm{~m} / \mathrm{s}$
Magnitude of Initial linear momentum $\left(\vec{p}_i\right)$ 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}
$
Angle between initial and final momemtum 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 an option (a)
Question 4: An object is thrown vertically upwards. At its maximum height, which of the following quantities becomes zero?
(a) Momentum
(b) Potential Energy
(c) Acceleration
(d) Force
Solution:
At maximum height, its velocity becomes zero. Therefore, the momentum of the object becomes zero
The correct option is (1)
Question 5: A body of mass $M$ at rest explodes into three pieces, in the ratio of masses $1: 1: 2$. Two smaller pieces fly off perpendicular to each other with velocities of $30 \mathrm{~ms}^{-1}$ and $40 \mathrm{~ms}^{-1}$ respectively. The velocity of the third piece will be :
(a) $15 \mathrm{~ms}^{-1}$
(b) $25 \mathrm{~ms}^{-1}$
(c) $35 \mathrm{~ms}^{-1}$
(d) $5 \mathrm{~ms}^{-1}$
Solution:
Conservation of momentum,
$
\begin{aligned}
& \frac{\mathrm{m}}{4} \times 30 \mathrm{i}+\frac{\mathrm{m}}{4} 30 \mathrm{~J}+\frac{\mathrm{m}}{2} \overrightarrow{\mathrm{v}}=0 \\
& \tilde{\mathrm{v}}=-15 \mathrm{i}-20 \mathrm{~J} \\
& |\tilde{\mathrm{v}}|=\sqrt{15^2+20^2}=25 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
Hence, the Correct answer is Option (b).
The importance of conservation of the linear momentum in a system or body in motion is that it maintains total momentum and is equal to the product of mass and vector velocity when an external force is applied.
Linear momentum p is expressed symbolically as:
p=mv
Where,
The mass of the system is m, and its velocity is v.
As a result, the S.I unit of momentum is kg.m/sec.
The net external force is equal to the change in momentum of a system divided by the rate of change of time, according to Newton's 2nd law of motion.
Linear momentum is equal to the product of a system’s mass multiplied by its velocity:
p = mv
Where p=linear momentum
m=mass
v=Velocity
The law of conservation of momentum applies to all physical processes. Here are several examples:
Collision: The conservation of momentum and energy governs the collision of things.
The momentum of the propellant gas causes the rocket to go in the opposite direction, which is due to the law of conservation of momentum.
When a bullet is ejected from a gun, the gun suffers recoil momentum.
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