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  • Describe an elastic collision of two objects in one dimension.
  • Define internal kinetic energy.
  • Derive an expression for conservation of internal kinetic energy in a one dimensional collision.
  • Determine the final velocities in an elastic collision given masses and initial velocities.

Let us consider various types of two-object collisions. These collisions are the easiest to analyze, and they illustrate many of the physical principles involved in collisions. The conservation of momentum principle is very useful here, and it can be used whenever the net external force on a system is zero.

We start with the elastic collision of two objects moving along the same line—a one-dimensional problem. An elastic collision    is one that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. [link] illustrates an elastic collision in which internal kinetic energy and momentum are conserved.

Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but not quite, elastic—some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. One macroscopic collision that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring bumpers on an air track. Icy surfaces and air tracks are nearly frictionless, more readily allowing nearly elastic collisions on them.

Elastic collision

An elastic collision    is one that conserves internal kinetic energy.

Internal kinetic energy

Internal kinetic energy is the sum of the kinetic energies of the objects in the system.

The system of interest contains a smaller mass m sub1 and a larger mass m sub2 moving on a frictionless surface. M sub 2 moves with velocity V sub 2 and momentum p sub 2 and m sub 1 moves behind m sub 2, with velocity V sub 1 and momentum p sub 1 toward the right direction. P 1 plus P 2 equals p total. The net force is zero. After collision m sub 1 moves toward the left with velocity V sub 1 while m sub 2 moves toward the right with velocity V sub 2 on the same frictionless surface. The momentum of m sub 1 becomes p 1 prime and m 2 becomes p 2 prime now. P 1 prime plus p 2 prime equals p total.
An elastic one-dimensional two-object collision. Momentum and internal kinetic energy are conserved.

Now, to solve problems involving one-dimensional elastic collisions between two objects we can use the equations for conservation of momentum and conservation of internal kinetic energy. First, the equation for conservation of momentum for two objects in a one-dimensional collision is

p 1 + p 2 = p 1 + p 2 F net = 0 size 12{ left (F rSub { size 8{"net"} } =0 right )} {}

or

m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 F net = 0 , size 12{ left (F rSub { size 8{"net"} } =0 right )} {}

where the primes (') indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic energies before the collision equals the sum after the collision. Thus,

1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 (two-object elastic collision)

expresses the equation for conservation of internal kinetic energy in a one-dimensional collision.

Calculating velocities following an elastic collision

Calculate the velocities of two objects following an elastic collision, given that

m 1 = 0 . 500 kg, m 2 = 3 . 50 kg, v 1 = 4 . 00 m/s, and v 2 = 0 . size 12{m rSub { size 8{1} } =0 "." "500"" kg, "m rSub { size 8{2} } =3 "." "50"" kg, "v rSub { size 8{1} } =4 "." "00"" m/s, and "v rSub { size 8{2} } =0 "." } {}

Strategy and Concept

First, visualize what the initial conditions mean—a small object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in [link] where both objects are initially moving. We are asked to find two unknowns (the final velocities v 1 and v 2 size 12{v rSub { size 8{2} } '} {} ). To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can be simplified by the fact that object 2 is initially at rest, and thus v 2 = 0 size 12{v rSub { size 8{2} } =0} {} . Once we simplify these equations, we combine them algebraically to solve for the unknowns.

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
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Someone
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Someone
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AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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