#### Quiz

**Q1)**

**Most collisions in the everyday world are**

**A. elastic collisions.**

B. inelastic collisions.

C. perfectly elastic collisions.

D. perfectly inelastic collisions.

**Answer) B.**

**Q2) A helium atom collides with another helium atom in an elastic collision. Which of the following is true?**

**A. Both momentum and kinetic energy are conserved.**

B. Momentum is conserved but kinetic energy is not conserved.

C. Kinetic energy is conserved but momentum is not conserved.

D. Neither momentum nor kinetic energy is conserved.

**Answer) A.**

#### Elastic Collisions in One Dimension Simulation(Virtual Experiment)

When the next simulation is not visible, please refer to the following link.

(https://helpx.adobe.com/flash-player/kb/enabling-flash-player-chrome.html)

**In elastic collisions, both kinetic energy and momentum are conserved.**

Energy and momentum are always conserved in a collision, no matter what happens. Momentum is easy to deal with because there is only “one form” of momentum, (p=mv), but you do have to remember that momentum is a vector. Energy is tricky because it has many forms, the most troublesome being heat, but also sound and light. If kinetic energy is conserved in a collision, it is called an elastic collision. In an elastic collision, the total kinetic energy is conserved because the objects in question “bounce perfectly” like an ideal elastic. An inelastic collision is one where some of the of the total kinetic energy is transformed into other forms of energy, such as sound and heat.

Any collision in which the shapes of the objects are permanently altered, some kinetic energy is always lost to this deformation, and the collision is not elastic. It is common to refer to a “completely inelastic” collision whenever the two objects remain stuck together, but this does not mean that all the kinetic energy is lost; if the objects are still moving, they will still have some kinetic energy.

*Figure 1. Body 1 moves along an x axis before having an elastic collision with body 2, which is initially at rest. Both bodies move along that axis after the collision.*

Figure 1 shows two bodies before and after they have a one-dimensional collision, like a head-on collision between pool balls. A projectile body of mass $\ m_{1}$ and initial velocity $\ v_{1i}$ moves toward a target body of mass $\ m_{2}$ that is initially at rest ($\ v_{1i}$ =0). Let’s assume that this two-body system is closed and isolated. Then the net linear momentum of the system is conserved, and from Eq. 9-51 we can write that conservation as

$\ m_{1} v_{1i} = m_{1} v_{1f} + m_{2} v_{2f} $ (linear momentum). --(1)

If the collision is also elastic, then the total kinetic energy is conserved and we can write that conservation as

$\frac{1}{2} m_{1}v_{1i}^2 = \frac{1}{2} m_{1}v_{1f}^2 + \frac{1}{2} m_{2}v_{2f}^2 $ (kinetic energy). --(2)

In each of these equations, the subscript i identifies the initial velocities and the subscript f the final velocities of the bodies. If we know the masses of the bodies and if we also know $\ v_{1i}$ , the initial velocity of body 1, the only unknown quantities are $\ v_{1f}$ and $\ v_{2f}$, the final velocities of the two bodies. With two equations at our disposal, we should be able to find these two unknowns.

To do so, we rewrite Eq. (1) as

$\ m_{1}(v_{1i} - v_{1f}) = m_{2}v_{2f} $ --(3)

and Eq. (2) as*

$\ m_{1}(v_{1i} - v_{1f}) (v_{1i} + v_{1f}) = m_{2}v_{2f}^2 $ --(4)

After dividing Eq. (4) by Eq. (3) and doing some more algebra, we obtain

$\ v_{1f} = \frac {m_{1} -m_{2}}{m_{1} + m_{2}} v_{1i} $ --(5)

$\ v_{2f} = \frac {2m_{1}}{m_{1} + m_{2}} v_{1i} $ --(6)

Note that $\ v_{2f}$ is always positive (the initially stationary target body with mass $\ m_{2}$ always moves forward). From Eq. (5) we see that $\ v_{1f}$ may be of either sign (the projectile body with mass $\ m_{1}$ moves forward if $\ m_{1} > m_{2} $ but rebounds if $\ m_{1} < m_{2} $).

Let us look at a few special situations(Elastic Collisions in One Dimension).

**1. Equal masses**

If $\ m_{1} = m_{2} $, Eqs. (5) and (6) reduce to

$\ v_{1f} = 0 $ and $\ v_{2f} = v_{1i} $

which we might call a pool player’s result. It predicts that after a head-on collision of bodies with equal masses, body 1 (initially moving) stops dead in its tracks and body 2 (initially at rest) takes off with the initial speed of body 1. In head-on collisions, bodies of equal mass simply exchange velocities. This is true even if body 2 is not initially at rest.

**2. A massive target**

In Fig. 9-18, a massive target means that $\ m_{2} \gg m_{1} $. For example, we might fire a golf ball at a stationary cannonball. Equations (5) and (6) then reduce to

$\ v_{1f} \approx v_{1i} $ and $\ v_{2f}\approx (\frac{2m_{1}}{m_{2}})v_{1i} $ --(7)

This tells us that body 1 (the golf ball) simply bounces back along its incoming path, its speed essentially unchanged. Initially stationary body 2 (the cannonball) moves forward at a low speed, because the quantity in parentheses in Eq. 9-69 is much less than unity.All this is what we should expect.

**3. A massive projectile**

This is the opposite case; that is, $\ m_{1} \gg m_{2} $. This time, we fire a cannonball at a stationary golf ball. Equations (5) and (6) reduce to

$\ v_{1f} \approx v_{1i} $ and $\ v_{2f} \approx 2 v_{1i} $ --(8)

Equation (8) tells us that body 1 (the cannonball) simply keeps on going, scarcely slowed by the collision. Body 2 (the golf ball) charges ahead at twice the speed of the cannonball. Why twice the speed? Recall the collision described by Eq. (7), in which the velocity of the incident light body (the golf ball) changed from $\ v$ to $\ - v$, a velocity change of $\ 2 v$. The same change in velocity (but now from zero to $\ 2 v$) occurs in this example also.