Conservation of Momentum

Claimed by

The conservation of momentum is one of the fundamental laws of physics. Within the definitions of a problem, the total momentum of the system stays constant. Much like the conservation of mass or the conservation of energy, the momentum of the objects before the collision is the same as the momentum of the objects after the collision. The momentum is changed through the action of forces as in Newton’s law of motion. This is a powerful idea in solving problems.

The Main Idea

Conservation refers to something that doesn’t change. Conservation of momentum is the idea that momentum is the same before and after an event i.e. that it is not created nor destroyed. Two common events that obey the conservation of momentum are elastic and inelastic collisions. In an elastic collision, the Kinetic Energy is conserved whereas in an inelastic collision, it is not. This is true in an isolated system, meaning that the system is not acted on by an external force. An elastic collision is one in which the objects collide and do not stick together, and an inelastic collision is when the objects that collided stick together in their final state. During an inelastic collision, the momentum lost by the first object is equal to the momentum gained by the second object. However, this does not mean that other aspects of the two objects do not change. In inelastic collisions, the momenta of the two objects in their initial states is equal to the momentum of both objects in their final state. If the two objects are of equal mass, the velocity of the combined objects will be halved, while the momentum remains conserved because the mass has doubled. The main difference between the conservation of momentum and the conservation of mass or the conservation of energy is that the momentum is a vector quantity, meaning the momentum is conserved in the x, y, and z directions.

The law of conservation of momentum can be logically derived from Newton’s Third Law. When 2 objects collide, the force of object 1 on object 2 ([math]\displaystyle{ \begin{align} F_1 \end{align} }[/math]) is equal in magnitude and opposite in direction to the force of object 2 on object 1 ([math]\displaystyle{ \begin{align} F_1 \end{align} }[/math]). So: [math]\displaystyle{ \begin{align} F_1=-F_2 \end{align} }[/math]

The objects collide during a certain time period ([math]\displaystyle{ \begin{align} \Delta t \end{align} }[/math]) so that the force acting on object 1 and the force acting on object 2 act over the this time period (Δt). So: [math]\displaystyle{ \begin{align} F_1* \Delta 1 =-F_2* \Delta 1 \end{align} }[/math] We know that [math]\displaystyle{ \begin{align} F*t \end{align} }[/math] is the formula for impulse and since the change in impulse is equal to the change in momentum: [math]\displaystyle{ \begin{align} m_1 * \Delta v_1 = -m_2 * \Delta v_2 \end{align} }[/math] (The Law of Conservation of Momentum).

A Mathematical Model

[math]\displaystyle{ \begin{align} \Delta p_1 = m_1 * \Delta v_1 = -m_2 * \Delta v_2 = \Delta p_2 \,. \end{align} }[/math]

[math]\displaystyle{ \begin{align} p_1 \end{align} }[/math] is the momentum of the first object, [math]\displaystyle{ \begin{align} m_1 \end{align} }[/math] is the mass of the first object, [math]\displaystyle{ \begin{align} v_1 \end{align} }[/math] is the velocity of the first object. [math]\displaystyle{ \begin{align} p_2 \end{align} }[/math] is the momentum of the second object, [math]\displaystyle{ \begin{align} m_2 \end{align} }[/math] is the mass of the second object, [math]\displaystyle{ \begin{align} v_2 \end{align} }[/math] is the velocity of the second object.

A Computational Model

The link below shows a simulation of an elastic collision where one object transfers all of its momentum to the second object after the collision.

[1]

Examples

Simple

Question

You are playing pool with your friends at Tech Rec. Two cue balls collide in a head-on collision. Both cue balls have equal mass of 0.165 kg. Before the collision, the first ball is travelling at 9.8 meters per second and the second ball is stationary, 0 meters per second. After the collision, the first ball travels backward at a velocity of 2 meters per second. What is the velocity of the second ball?

Solution

[math]\displaystyle{ \begin{align} m_1 = m_2 = m = 0.165 \,. \end{align} }[/math]

[math]\displaystyle{ \begin{align} v_1i = 9.8 \,. \end{align} }[/math]

[math]\displaystyle{ \begin{align} v_1f = -2 \,. \end{align} }[/math]

[math]\displaystyle{ \begin{align} V_2i = 0 \,. \end{align} }[/math]

[math]\displaystyle{ \begin{align} m * (v_1f - v_1i)= m (v_2f - v_2i) \,. \end{align} }[/math]

[math]\displaystyle{ \begin{align} v_2f = v_1f - v_1i + v_2i = 9.8 + 2 - 0 = 11.8 \,. \end{align} }[/math] The velocity of the second ball is 11.8 meters per second in the positive x direction.

Question

There is a perfectly inelastic collision between a big fish and a little fish. The big fish is initially going 2 m/s and the little fish is going 10 m/s. The big fish’s mass is 5 kg and the little fish is 0.5 kg. What is the velocity of the system after the collision?

Solution

Because there are no forces acting on the system, the system’s total momentum before the collision is equal to the momentum after the collision. [math]\displaystyle{ \begin{align} m_1 * v_1i + m_2 * v_2i = (m_1 + m_2) * v_f \,. \end{align} }[/math]

[math]\displaystyle{ \begin{align} 5 kg * 2 m/s + 0.5 kg * 10 m/s = (5 kg + 0.5 kg) * v_f \,. \end{align} }[/math] The final velocity of the big and little fish system is 0.909 m/s

Middling

Question

Figure 1

You are looking out the window of your dorm when you see a person on a Hoverboard collide with a person on a bike. The person on bike was going west with velocity 9 m/s and the person on the Hoverboard was traveling at 30 degrees north of west with velocity 5 m/s. After the collision, both the person on the Hoverboard and the person on the bike become entangled and travel together at some velocity [math]\displaystyle{ \begin{align} v_f \, \end{align} }[/math]. Find [math]\displaystyle{ \begin{align} v_f \, \end{align} }[/math].

Solution

[math]\displaystyle{ \begin{align} v_H=[9*cos(30), 9*sin(30), 0] \, \end{align} }[/math] [math]\displaystyle{ \begin{align} v_B=[-5,0,0] \, \end{align} }[/math]

[math]\displaystyle{ \begin{align} m_H = 60 kg \, \end{align} }[/math] [math]\displaystyle{ \begin{align} m_B = 50 kg \, \end{align} }[/math]

[math]\displaystyle{ \begin{align} m_H*v_H + m_bike*v_B = (m_H+m_B)*v_final \, \end{align} }[/math]

[math]\displaystyle{ \begin{align} v_f= (m_H*v_H + m_B*v_B)/ (m_H+m_B) \, \end{align} }[/math]

[math]\displaystyle{ \begin{align} v_f = [-1.5155, -4.8503, 0] \,. \end{align} }[/math] m/s

Question

An 80 kg running back runs in the -y direction at 1.5 m/s before getting tackled by a 95 kg linebacker traveling at 2.0 m/s in the +y direction. Both players bounce off each other after the collision. If the linebacker continues moving in the same direction at 0.5 m/s, what is the velocity and direction of the running back?

Solution

This is an elastic collision, and the total momentum is conserved.

[math]\displaystyle{ \begin{align} m_1 * v_1i + m_2 * v_2i = m_1 * v_1f + m_2 * v_2f \,. \end{align} }[/math] m/s

[math]\displaystyle{ \begin{align} 80 kg * [0,-1.5,0] m/s + 95 kg * [0,2,0] m/s = 80 kg * v_1f + 95 kg * [0,.5,0] \,. \end{align} }[/math] m/s

Then get v1f by itself and you find out that the linebacker is going [0,.282,0] m/s.

Difficult

Question

During a baseball game the 0.145 kg baseball is thrown straight upward with a velocity of 40 m/s. What is the recoil velocity of the earth? Why don’t we notice that the earth has gained velocity?

Solution

[math]\displaystyle{ \begin{align} m_b = 0.145 kg \, \end{align} }[/math] [math]\displaystyle{ \begin{align} m_E = 5.972 e 24 kg\, \end{align} }[/math] [math]\displaystyle{ \begin{align} v_b = 40 m/s \, \end{align} }[/math]

The system is the baseball plus the earth. So, the total momentum of the system must be conserved (i.e. the momentum before must be equal to the momentum after.)

[math]\displaystyle{ \begin{align} m_b*v_b=-m_E*v_E .\, \end{align} }[/math]

[math]\displaystyle{ \begin{align} v_e= - (m_b*v_b)/m_E .\, \end{align} }[/math]

[math]\displaystyle{ \begin{align} v_e=- 9.7120e-25 m/s .\, \end{align} }[/math]

Even though the magnitude of the momentum of the ball equals the magnitude of the momentum of the Earth, the Earth’s mass is so massive that the Earth recoils with a velocity so small we don’t feel it.

Question

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially upstretched and with a spring constant k = 2.0e4 N/M. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45 degrees above the horizontal. If the mass of the cannon and the carriage is 5,000 kg find the recoil speed of the cannon.

Solution

The initial velocity of the cannon and the projectile are both 0 m/s. [math]\displaystyle{ \begin{align} 0 = 200 kg * 125 m/s * cos(45) + 5000 kg * v_x .\, \end{align} }[/math]

v_x = -3.54 m/s

Application

Conservation of Momentum is particularly important in fluid flow and transport. In the field of biomedical engineering, engineers are often times working with fluids flowing through the body. So when a new design for a pacemaker is made, engineers need to find out how the new pacemaker diverts fluid flow in the body. To solve a problem like this, one would use Reynold's Transport Theorem, as well as the principle of conservation of momentum, since the basis for Reynold's Transport Theorem are the principles of conservation of mass and the conservation of momentum. No new device can be made for the human body without such an analysis. In addition, in certain disease states, such as atherosclerosis, when certain arteries become clogged; the fluid flow is diverted and there is a change in pressure. To model certain situations, engineers would start from basic principles such as Conservation of Mass and Conservation of Momentum.

Soyuz TMA-5 launch

Conservation of momentum is used to successfully propel a rocket through space. When a rocket is started it sends exhaust gases downward at a high velocity. The gases have mass and therefore have momentum. For momentum to be conserved, the rocket then moves upward at a velocity equivalent to the original momentum divided by the mass of the rocket. Nothing actually pushes the rocket while it's moving, it's just the momentum of the gases working with the momentum of the rocket itself.

In this way, the Conservation of Momentum (as well as the Conservation of Mass and the Conservation of Energy) plays important roles in engineering fields, such as Biomedical Engineering. These conservation laws are also important in other engineering fields. To name a few examples; chemical engineers are concerned with fluid flowing through a pipe, and mechanical engineers are concerned with fluid flowing through an engine.

History

Newton established Conservation of Momentum along with the other Conservation Laws (except Conservation of Energy). Newton published his theories in 1687 in Philosophiæ Naturalis Principia Mathematica (linked below in external links). When Newton was publishing his work, the challenge that he faced was describing his theories without using calculus. Newton did not publish his La Methode Dex Fluxions until 1736. His work on momentum was mostly focused on forces rather than energy and vectors, but Newton's Laws of Motion imply the conservation of momentum.

See also

Further reading

For extra concept quetions see: [2]

External links

Newton's Philosophiæ Naturalis Principia Mathematica: [3]

References

1. "Conservation of Momentum." Conservation of Momentum. NASA, 05 May 2015. Web. 05 Dec. 2015. <https://www.grc.nasa.gov/www/k-12/airplane/conmo.html>.

2. "Momentum Conservation Principle." Momentum Conservation Principle. The Physics Classroom, n.d. Web. 05 Dec. 2015. <http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle>.

3. Smith, George. "Newton's Philosophiae Naturalis Principia Mathematica." Stanford University. Stanford University, 20 Dec. 2007. Web. 05 Dec. 2015. <http://plato.stanford.edu/entries/newton-principia/>.

4. "Mathematical Treasure: Newton's Method of Fluxions." Mathematical Treasure: Newton's Method of Fluxions. Mathematical Association of America, n.d. Web. 05 Dec. 2015. <http://www.maa.org/press/periodicals/convergence/mathematical-treasure-newtons-method-of-fluxions>.

5. "Momentum Conservation Principle." Momentum Conservation Principle. N.p., n.d. Web. 26 Nov. 2016.

6. "Conservation Laws - Real-life Applications." Real-life Applications - Conservation Laws - Conservation of Linear Momentum, Firing a Rifle. N.p., n.d. Web. 26 Nov. 2016.