State the Law of Conservation of Momentum
The momentum of an object is the product of the velocity and mass of an object. It is a vector quantity. Conservation of momentum is a fundamental law of physics, which states that the total momentum of an isolated system is conserved in the absence of an external force. In other words, the total momentum of a system of objects remains constant during any interaction if no external force acts on the system. The total momentum is the vector sum of individual momenta. Therefore, the component of the total momentum along any direction remains constant (whether the objects interact or not). Momentum remains conserved in any physical process.
Overview of the Law of Conservation of Momentum
Conservation of momentum states that the total momentum of an isolated system remains the same in the absence of an external force, i.e., the momentum can neither be created nor be destroyed, however, it can be changed through the action of forces as described by Newton's laws of motion.
Momentum is the product of the mass of the object and the velocity at which it is travelling and is also equal to the total force required to bring the object to rest.
One of the real-life aspects of the conservation of momentum is collision problems in which the momentum remains conserved and the net external force remains zero.
Additionally, there are several applications of momentum conservation in our day-to-day life that we will cover on this page. Along with this, we will understand the logic behind this concept and the proof of the conservation of momentum.
Illustration in One-Dimension
Conservation of momentum can be explained through a one-dimensional collision of two objects. Two objects of masses m1 and m2 collide with each other while moving along a straight line with velocities u1 and u2, respectively. After the collision, they acquire velocities v1 and v2 in the same direction.
Total momentum before collision pi=m1u1+m2u2
Total momentum after collision pf=m1v1+m2v2
If no other force acts on the system of the two objects, total momentum remains conserved. Therefore,
Pi = pf
m1u1+m2u2= m1v1+m2v2
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Derivation of Conservation of Momentum
If no external force is exerted on the system of two colliding objects, the objects apply impulse on each other for a short interval of time at the point of contact. According to Newton’s third law of motion, the impulsive force applied by the first object on the second one is equal and opposite to the impulsive force applied by the second object on the first object.
During the one-dimensional collision of two objects of masses m1 and m2, which have velocities u1 and u2 before collision and velocities v2 and v2 after the collision, the impulsive force on the first object is F21 (applied by the second object) and the impulsive force on the second object is F12 (applied by the first object). Applying Newton’s third law, these two impulsive forces are equal and opposite, i.e.,
F21 = − F12
If the time of contact is t, the impulse of the force F21 is equal to the change in momentum of the first object.
F21. t = m1v1 − m1u1
The impulse of force F12 is equal to the change in momentum of the second object.
F12. t = m2v2 − m2u2
From F21 = − F12
F21. t = − F12. t
m1v1 − m1u1 = − (m2v2 − m2u2)
m1u1+m2u2= m1v1+m2v2
This relation suggests that momentum is conserved during the collision.
Collision in Two - Dimensions
Before the collision, the total momentum is pix = p1 = m1v1, along the X - axis and piy = p2 = m2v2 along the Y - axis. After the collision, the total momentum is pfx = (m + M) ucosθ, along X-axis and pfy = (m+M)usinθ
Applying conservation of momentum,
pix = pfx
m1v1 = (m + M) ucosθ….(1)
piy = pfy
m2v2 = (m+M) usinθ…..(2)
Therefore, squaring and adding equations (1) and (2),
\[(m_{1}v_{1})^{2} + (m_{2}v_{2})^{2} = (m+M)^{2} u^{2} (Cos^{2}\Theta + Sin^{2}\Theta )\]
\[u = \frac{\sqrt{m_{1}^{2}v_{1}^{2} + m_{2}^{2}v_{2}^{2}}}{(m + M)}\]
It is the speed of the combined object.
Dividing equation (2) by (1),
\[tan \Theta = \frac{m_{2}v_{2}}{m_{1}v_{1}}\]
θ gives the direction of the velocity.
Conservation of Momentum Examples
Recoil of a Gun: If a bullet is fired from a gun, both the bullet and the gun are initially at rest i.e. the total momentum before firing is zero. The bullet acquires a forward momentum when it gets fired. According to the conservation of momentum, the gun receives a backward momentum. The bullet of mass m is fired with forward velocity v. The gun of mass M acquires a backward velocity u. Before firing, the total momentum is zero so that the total momentum after firing is also zero.
0 = mv + Mu
u = -\[\frac{m}{M}\]v
u is the recoil velocity of the gun. The mass of the bullet is much less than that of the gun i.e. m ≪ M. The backward velocity of the gun is very small,
u ≪ v
Rocket Propulsion: Rockets have a gas chamber at one end, from which gas is ejected with enormous velocity. Before the ejection, the total momentum is zero. Due to the ejection of gas, the rocket gains a recoil velocity and acceleration in the opposite direction. This is a consequence of the conservation of momentum
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If a rocket of mass m ejects the propellent of small mass dm with an exhaust velocity ve such that the residual rocket of mass m - dm acquires a velocity dv in the opposite direction, the momenta of the propellant and the residual rocket are equal in magnitude and opposite in direction.
vedm = − (m − dm)dv
Since both dm and dv are small, the equation can be approximated as
dv = − ve\[\frac{dm}{m}\]
If the mass of the rocket reduces from mo to m’ as its velocity increases from 0 to v’, integrating the above equation
\[\int_{0}^{v^{1}}dv = -ve\int_{mo}^{m^{1}} dm/m\]
v’ = ve ln\[(\frac{m}{mo})\]
Solved Examples
I. A bullet of mass 6 g is fired with a speed of 500 m/s from a gun of mass 4 kg. What would be the recoil velocity of the gun?
Solution: The initial momenta of the bullet and the gun are zero such that the total initial momentum is zero. The bullet of mass m = 6g is fired with forward velocity v = 500 m/s. The gun of mass M = 4kg acquires a backward velocity V.
m = 6 g =\[\frac{6}{1000}\]kg
According to the conservation of momentum formula,
0 = mv + MV
0 =\[\frac{6}{1000}\]kg(500m/s) + (4kg) v
v = - 0.75 m/s
The recoil speed of the gun is 0.75 m/s. The negative sign implies that the recoil velocity is opposite to the velocity of the bullet.
FAQs on Law of Conservation of Momentum
1. What is the law of conservation of momentum?
The law of conservation of momentum states that for an isolated system, the total linear momentum remains constant. An isolated system is one where no net external force acts upon it. In simpler terms, if two or more objects interact only with each other, the total momentum they had before the interaction is equal to the total momentum they have after the interaction. Momentum, a vector quantity, is the product of an object's mass and its velocity (p = mv).
2. What is the mathematical formula for the law of conservation of momentum?
For a system involving two colliding objects, the law of conservation of momentum is expressed by the formula: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. In this equation:
- m₁ and m₂ are the masses of the first and second objects, respectively.
- u₁ and u₂ are their initial velocities before the collision.
- v₁ and v₂ are their final velocities after the collision.
3. How can the law of conservation of momentum be explained with a simple real-world example?
A classic example is the recoil of a gun. Before firing, the gun and the bullet are at rest, so the total momentum of the system is zero. When the gun is fired, the bullet moves forward with a certain momentum. To keep the total momentum of the system at zero (as per the law), the gun must move backward with an equal and opposite momentum. This backward movement is the recoil. This demonstrates that momentum is transferred within the system, but the total amount remains conserved.
4. How is the law of conservation of momentum derived from Newton's laws of motion?
The law of conservation of momentum is a direct consequence of Newton's Second and Third Laws of Motion. According to Newton's Third Law, when two objects collide, the force exerted by the first object on the second (F₂₁) is equal and opposite to the force exerted by the second on the first (F₁₂), so F₂₁ = -F₁₂. According to Newton's Second Law, force is the rate of change of momentum (F = dp/dt). Therefore, the change in momentum of the first object is equal and opposite to the change in momentum of the second object. This means the total change in momentum for the entire system is zero, proving that momentum is conserved.
5. Under what conditions is the law of conservation of momentum applicable?
The law of conservation of momentum is strictly applicable only under a specific, crucial condition: the system must be isolated. An isolated system is one where the net external force acting on it is zero. Internal forces, such as the forces between two colliding billiard balls, are allowed. However, external forces like friction, air resistance, or a non-zero gravitational pull from an outside source would change the system's total momentum, and the law would not hold true.
6. Is momentum always conserved in every real-world collision?
No, momentum is not conserved in every real-world collision because most real-world systems are not perfectly isolated. For the law to apply, the net external force must be zero. In many practical scenarios, forces like friction and air resistance act on the colliding objects. These external forces cause a change in the total momentum of the system. However, for many quick collisions (like a bat hitting a ball), these external forces are negligible compared to the large internal forces of the collision, so momentum is considered approximately conserved.
7. How does rocket propulsion demonstrate the principle of conservation of momentum?
Rocket propulsion is a prime example of momentum conservation. The rocket and its fuel constitute a single system, which is initially at rest (total momentum is zero). When the rocket engine fires, it expels hot gases (propellant) downwards at a very high velocity. This stream of gas has a significant downward momentum. To conserve the total momentum of the system (which must remain zero), the rocket itself gains an equal and opposite upward momentum. This upward momentum results in the rocket accelerating into the sky.
8. What is the relationship between impulse and the law of conservation of momentum?
Impulse is defined as the change in an object's momentum (Impulse = Δp). The law of conservation of momentum and impulse are fundamentally linked. In an isolated system with two interacting objects, the impulse on one object is equal in magnitude and opposite in direction to the impulse on the other. This means their changes in momentum cancel each other out (Δp₁ = -Δp₂). Consequently, the total change in the system's momentum is zero (Δp₁ + Δp₂ = 0), which is the very definition of the conservation of momentum.
