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Physics of Hearing

- 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.

We start with the elastic collision of two objects moving along the same line—a one-dimensional problem. An

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.

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)

or

*m*_{1}*v*_{1} + *m*_{2}*v*_{2} = *m*_{1}*v*′_{1} + *m*_{2}*v*′_{2} (*F*_{net} = 0),

$\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v′_1}^2+\frac{1}{2}m_2{v′_2}^2\left(\text{two-object elastic collision}\right)\\$

expresses the equation for conservation of internal kinetic energy in a one-dimensional collision.*p*_{1} = *p′*_{1} + *p′*_{2} or *m*_{1}*v*_{1}=*m*_{1}*v*′_{1}+*m*_{2}*v*′_{2}.

$\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v′_1}^2+\frac{1}{2}m_2{v′_2}^2\\$

Solving the first equation (momentum equation) for $v′_2=\frac{m_1}{m_2}\left(v_1-v′_1\right)\\$

.
Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable *v*′_{1} = 4 . 00 m/s and *v*′_{1}=−3.00 m/s.

$v′_2=\frac{m_1}{m_2}\left(v_1-v′_1\right)=\frac{0.500\text{ kg}}{3.50\text{ kg}}\left[4.00-\left(-3.00\right)\right]\text{m/s}\\$

or*v*′_{2}=1.00 m/s.

The equations for conservation of momentum and internal kinetic energy as written above can be used to describe any one-dimensional elastic collision of two objects. These equations can be extended to more objects if needed.

- An elastic collision is one that conserves internal kinetic energy.
- Conservation of kinetic energy and momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one dimensional two-body collisions.

- What is an elastic collision?

- Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.
**Professional Application.**Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00 × 10^{3}kg, and the second a mass of 7.50 × 10 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?- A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities be in this case?