9.4 Types of collisions (Page 2/10)

University physics volume 1 Page 2 / 10

Many-to-many

The extreme case on the other end is if two or more objects approach each other, collide, and bounce off each other, moving away from each other at the same relative speed at which they approached each other. In this case, the total kinetic energy of the system is conserved. Such an interaction is called elastic .

In any interaction of a closed system of objects, the total momentum of the system is conserved $({\vec{p}}_{f} = {\vec{p}}_{i})$ but the kinetic energy may not be:

If $0 < K_{f} < K_{i}$ , the collision is inelastic.
If $K_{f} = 0$ , the collision is perfectly inelastic.
If $K_{f} = K_{i}$ , the collision is elastic.
If $K_{f} > K_{i}$ , the interaction is an explosion.

The point of all this is that, in analyzing a collision or explosion, you can use both momentum and kinetic energy.

Problem-solving strategy: collisions

A closed system always conserves momentum; it might also conserve kinetic energy, but very often it doesn’t. Energy-momentum problems confined to a plane (as ours are) usually have two unknowns. Generally, this approach works well:

Define a closed system.
Write down the expression for conservation of momentum.
If kinetic energy is conserved, write down the expression for conservation of kinetic energy; if not, write down the expression for the change of kinetic energy.
You now have two equations in two unknowns, which you solve by standard methods.

Formation of a deuteron

A proton (mass

1.67 \times 10^{−27} kg

) collides with a neutron (with essentially the same mass as the proton) to form a particle called a deuteron . What is the velocity of the deuteron if it is formed from a proton moving with velocity

7.0 \times 10^{6} m/s

to the left and a neutron moving with velocity

4.0 \times 10^{6} m/s

to the right?

Strategy

Define the system to be the two particles. This is a collision, so we should first identify what kind. Since we are told the two particles form a single particle after the collision, this means that the collision is perfectly inelastic. Thus, kinetic energy is not conserved, but momentum is. Thus, we use conservation of energy to determine the final velocity of the system.

Solution

Treat the two particles as having identical masses M . Use the subscripts p, n, and d for proton, neutron, and deuteron, respectively. This is a one-dimensional problem, so we have

M v_{p} - M v_{n} = 2 M v_{d} .

The masses divide out:

\begin{matrix} v_{p} - v_{n} & = & 2 v_{d} \\ 7.0 \times 10^{6} m/s - 4.0 \times 10^{6} m/s & = & 2 v_{d} \\ v_{d} & = & 1.5 \times 10^{6} m/s. \end{matrix}

The velocity is thus ${\vec{v}}_{d} = (1.5 \times 10^{6} m/s) \hat{i}$ .

Significance

This is essentially how particle colliders like the Large Hadron Collider work: They accelerate particles up to very high speeds (large momenta), but in opposite directions. This maximizes the creation of so-called “daughter particles.”

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Ice hockey 2

(This is a variation of an earlier example.)

Two ice hockey pucks of different masses are on a flat, horizontal hockey rink. The red puck has a mass of 15 grams, and is motionless; the blue puck has a mass of 12 grams, and is moving at 2.5 m/s to the left. It collides with the motionless red puck ( [link] ). If the collision is perfectly elastic, what are the final velocities of the two pucks?

Two hockey pucks are shown. The top diagram shows the puck on the left with 0 meters per second and the puck on the right moving to the left with 2.5 meters per second. The bottom diagram shows the puck on the left moving to the left at unknown v sub 1 f and the puck on the right moving with unknown v sub 2 f. — Two different hockey pucks colliding. The top diagram shows the pucks the instant before the collision, and the bottom diagram show the pucks the instant after the collision. The net external force is zero.

Strategy

We’re told that we have two colliding objects, and we’re told their masses and initial velocities, and one final velocity; we’re asked for both final velocities. Conservation of momentum seems like a good strategy; define the system to be the two pucks. There is no friction, so we have a closed system. We have two unknowns (the two final velocities), but only one equation. The comment about the collision being perfectly elastic is the clue; it suggests that kinetic energy is also conserved in this collision. That gives us our second equation.

The initial momentum and initial kinetic energy of the system resides entirely and only in the second puck (the blue one); the collision transfers some of this momentum and energy to the first puck.

Solution

Conservation of momentum, in this case, reads

\begin{matrix} p_{i} & = & p_{f} \\ m_{2} v_{2,i} & = & m_{1} v_{1,f} + m_{2} v_{2,f} . \end{matrix}

Conservation of kinetic energy reads

\begin{matrix} K_{i} & = & K_{f} \\ \frac{1}{2} m_{2} v_{2,i}^{2} & = & \frac{1}{2} m_{1} v_{1,f}^{2} + \frac{1}{2} m_{2} v_{2,f}^{2} . \end{matrix}

There are our two equations in two unknowns. The algebra is tedious but not terribly difficult; you definitely should work it through. The solution is

\begin{matrix} v_{1,f} & = & \frac{(m_{1} - m_{2}) v_{1,i} + 2 m_{2} v_{2,i}}{m_{1} + m_{2}} \\ v_{2 f} & = & \frac{(m_{2} - m_{1}) v_{2,i} + 2 m_{1} v_{1,i}}{m_{1} + m_{2}} . \end{matrix}

Substituting the given numbers, we obtain

\begin{matrix} v_{1,f} & = & 2.22 \frac{m}{s} \\ v_{2,f} & = & −0.28 \frac{m}{s} . \end{matrix}

Significance

Notice that after the collision, the blue puck is moving to the right; its direction of motion was reversed. The red puck is now moving to the left.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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