3-5: Collisions
Introduction
We will continue our discussion of collisions and conservation of momentum by reviewing perfectly elastic and inelastic collisions.
After today’s lesson, you will be able to:
- Define a perfectly elastic collision, including the relationship of the relative velocities before and after the collision.
- Define an inelastic collision
- Define coefficient of restitution
- Apply conservation of momentum to perfectly elastic and inelastic collisions
Perfectly Elastic Collisions
A perfectly elastic collision occurs when no mechanical energy is lost during the collision. We can solve for two unknowns using Convervation of Momentum and Conservation of Energy.
$m_{1}v_{1} + m_{2}v_{2} = m_{1}v_{1}^{\prime} + m_{2}v_{2}^{\prime}$
$\frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2} = \frac{1}{2}m_{1}v_{1}^{\prime 2} + \frac{1}{2}m_{2}v_{2}^{\prime 2}$
With the help of a little algebra, we can obtain an easier equation to use.
$v_{1} - v_{2} = -(v_{1}^{\prime} - v_{2}^{\prime})$
This equation tells us that when objects experience a perfectly elastic collision, the relative velocities before and after the collision change signbut do not change magnitude.
Let’s look at a few examples of how to solve perfectly elastic collisions.
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Nuclear Reactor Example
In a uranium-fueled nuclear reactor it is important that neutrons be slowed in order to be most effective in causing the fission of U-235. Assuming a perfectly elastic collision, a neutron will more likely slow in speed if it bounces from a:
- heavy nucleus
- nucleus of moderate mass
- lightweight nucleus
Answer and Explanation
- bowling ball, akin to a heavy nucleus
- bocce ball, akin to a nucleus of moderate mass
- billiard ball, akin to a lightweight nucleus
For simplicity, assume the target balls are at rest or moving very slowly. For an elastic collision of a golf ball and bowling ball, the golf ball will bounce back with almost no loss in speed. The golf ball will slow a bit more in bouncing from a the lighter bocce ball. Bounce the golf ball from a billiard ball and it will slow even more, giving much of its energy to the billiard ball. So a neutron slows more when bouncing from a lightweight nucleus than a heavy one.
Application: That is why Enrico Fermi used blocks of graphite (composed of carbon atoms) to slow neutrons ejected in the fission of uranium. Carbon atoms behaved as billiard balls rather than bowling balls to slow energetic neutrons. Fermi first called the material to slow neutrons "slower downers." His colleague John Wheeler suggested "moderator."
Solving for Relative Velocities (only for algebra nerds)
Rewrite Conservation of Momentum equation.
$m_1 (v_{1} - v_{1}^{\prime}) = m_2(v_{2}^{\prime} - v_{2})$
Rewrite Conservation of Energy equation.
$m_{1}v_{1}^{2} - m_{1}v_{1}^{\prime 2} = m_{2}v_{2}^{\prime 2} - m_{2}v_{2}^{2}$
$m_{1}(v_{1} - v_{1}^{\prime})(v_{1} + v_{1}^{\prime}) = m_{2}(v_{2}^{\prime} - v_{2})(v_{2}^{\prime} + v_{2})$
Divide COE equation by COM equation:
$v_{1} + v_{1}^{\prime} = v_{2}^{\prime} + v_{2}$
or
$v_{1} - v_{2} = -(v_{1}^{\prime} - v_{2}^{\prime})$
Inelastic Collisions
We’ve talked about two types of collisions up to this point: perfectly inelastic and perfectly elastic collisions. Each represents one of two extremes. Real world collisions are typically inelastic collisions, or somewhere between these two extremes.
Coefficient of Restitution
A parameter that helps define an inelastic collision is called the coefficient of restitution. The coefficient of restitution is the negative of the ratio of the relative velocities before and after a collision. The equation for the coefficient of restitution, e is:
$e = \frac{-(v_{1}^{\prime} -v_{2}^{\prime})}{v_{1} - v_{2}}$
The value of the coefficient of restitution is as follows:
| Type of Collision | Coefficient of Restitution, e |
|---|---|
| Perfectly Inelastic | Zero |
| Inelastic (Real World) | 0 < e < 1 |
| Perfectly Elastic | 1 |
Calculating the Coffecient of Restitution
The cofficient of restitution between a ball and a surface can be calculated by bouncing the ball on the floor. For example, a regulation tennis ball must bounce between 53 and 56 inches when dropped from a height of 100 inches onto a concrete floor. The following video will show how the coefficient of restitution can be calculated from the bounce height. The result is:
$e=\sqrt{\frac{h_f}{h_i}}$
where hi is the initial drop height and hf is the maximum height of the bounce.
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Traffic Accident Reconstruction
We can use the information we have learned on conservation of energy and conservation of momentum to reconstruct what happened in an accident.
Two cars – same mass – collide head-on. After the collision the cars skid with their brakes locked as shown with
vA = 5 km/hr (1.39 m/s) right, and μk = 0.3.
Determine the speed of car B and the effective coefficient of restitution.
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Applications of Coefficient of Restitution
The following videos show how the coefficient of restitution is used in the design of golf clubs and bowling balls.
Review of Collisions
| Type of Collision | Final Relative Velocity | Solution Strategy |
|---|---|---|
| Perfectly Inelastic | Zero; Objects move together | COM: $m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2}){v}^\prime$ |
| Inelastic (Real World) | Ratio between 0 and 1; defined by coefficient of restitution | COM: $m_{1}v_{1} + m_{2}v_{2} = m_{1}{v}_{1}^\prime + m_{2}{v}_{2}^\prime$ Coefficient of Restitution: $e = \frac{-(v_{1}^\prime -v_{2}^\prime)}{v_{1} - v_{2}}$ |
| Perfectly Elastic | Same magnitude as original relative velocity; opposite sign | COM: $m_{1}v_{1} + m_{2}v_{2} = m_{1}{v}_{1}^\prime + m_{2}{v}_{2}^\prime$ Relative Velocities: $v_{1} - v_{2} = -(v_{1}^{\prime} - v_{2}^{\prime})$ |
Simulation
Spend a few minutes playing with this simulation.
Answer these concept questions on collisions.
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Summary
Conservation of Momentum
COM: $m_{1}v_{1} + m_{2}v_{2} = m_{1}{v}_{1}^\prime + m_{2}{v}_{2}^\prime$
Coefficient of Restitution
COR: $e = \frac{-(v_{1}^\prime -v_{2}^\prime)}{v_{1} - v_{2}}$
Perfectly Inelastic Collision
- Objects stick together
- $v_1’ = v_2’$
- e = 0
Inelastic (Real World) Collison
- 0 < e < 1
Perfectly Elastic Collision
- e = 1
- No mechanical energy loss
Hitting Baseball in Slow Motion