# Momentum & Energy

## Practice

### practice problem 1

The diagrams below represent generic objects before a collision followed by a set of outcomes to be considered. Comment on the outcomes, paying attention to the energy and momentum before and after each collision. Does the outcome describe a completely inelastic, partially inelastic, completely elastic, or impossible collision? Provide a brief explanation to accompany each answer.

#### solution

Here we are given the initial conditions of the two colliding objects.

∑ *p*=*m*_{1}*v*_{1}+ *m*_{2}*v*_{2}∑ *p*=(8 kg)(+6 m/s) + (4 kg)(−9 m/s) ∑ *p*=+12 kg m/s ∑ *K*=½ *m*_{1}*v*_{1}^{2}+ ½ *m*_{2}*v*_{2}^{2}∑ *K*=½(8 kg)(6 m/s) ^{2}+ ½(4 kg)(9 m/s) ^{2}∑ *K*=306 J Momentum is a vector quantity, so the total momentum is found by a vector sum. Since the momentums of the two objects are in opposite directions one of them is going to be negative. Since positive answers are preferred over negative ones, let's choose right as the positive direction. This gives us a total momentum of +12 kg m/s. Energy being a scalar is much easier to handle, especially here since the only energy that matters is kinetic (which is always positive). The total mechanical energy of this system in 306 J.

The first outcome we'll be examining has the two objects sticking together and moving off to the right.

∑ *p*′ =( *m*_{1}+*m*_{2})*v*′∑ *p*′ =(8 + 4 kg)(+1 m/s) ∑ *p*′ =+12 kg m/s ∑ *K*′ =½( *m*_{1}+*m*_{2})*v*′^{2}∑ *K*′ =½(8 + 4 kg)(1 m/s) ^{2}∑ *K*′ =6 J This is a sensible outcome since the initial total momentum is positive (to the right). Calculations show that the final total momentum is still +12 kg m/s, but the final total energy has dropped significantly to a relatively low value of 6 J. Momentum was conserved, but mechanical energy was lost. This a classic example of an inelastic collision. (Some would even call this a completely inelastic collision.) The lost energy has likely gone into plastic deformation of the two objects (given the distorted edges shown in the diagram).

Here the two objects have separated after collision and are moving in opposite directions. Each is moving more slowly than it was before the collision. This hints at a loss of mechanical energy.

∑ *p*′ =*m*_{1}*v*_{1}′+ *m*_{2}*v*_{2}′∑ *p*′ =(8 kg)(−1 m/s) + (4 kg)(+5 m/s) ∑ *p*′ =+12 kg m/s ∑ *K*′ =½ *m*_{1}*v*_{1}^{2}+ ½ *m*_{2}*v*_{2}^{2}∑ *K*′ =½(8 kg)(1 m/s) ^{2}+ ½(4 kg)(5 m/s) ^{2}∑ *K*′ =54 J Momentum was conserved as it should be, but mechanical energy was lost making this an inelastic collision. Since more energy was retained than in the previous outcome, some would call this a partially inelastic collision. Lost energy is not a big deal and does not violate the conservation of energy. The energy wasn't destroyed in this outcome. It just turned into a form that isn't easy to see with our eyes — internal energy. When kinetic energy transforms into internal energy, either the temperature of the system increases or it experience a phase change (melting, for example). Internal energy will be dealt with in more detail later in this book.

This outcome is similar to the previous one only now the objects are moving a bit more quickly. Still, their speeds after the collision are slower than before.

∑ *p*′ =*m*_{1}*v*_{1}′+ *m*_{2}*v*_{2}′∑ *p*′ =(8 kg)(−3 m/s) + (4 kg)(+9 m/s) ∑ *p*′ =+12 kg m/s ∑ *K*′ =½ *m*_{1}*v*_{1}^{2}+ ½ *m*_{2}*v*_{2}^{2}∑ *K*′ =½(8 kg)(3 m/s) ^{2}+ ½(4 kg)(9 m/s) ^{2}∑ *K*′ =198 J Momentum was conserved and energy was lost, but to a lesser extent than in the previous outcome. Of all the outcomes so far, this inelastic collision is the least inelastic. Whether one would call it partially inelastic or partially elastic doesn't really matter.

Here we see an elastic collision. (Some would even call this a perfectly elastic collision.)

∑ *p*′ =*m*_{1}*v*_{1}′+ *m*_{2}*v*_{2}′∑ *p*′ =(8 kg)(−4 m/s) + (4 kg)(+11 m/s) ∑ *p*′ =+12 kg m/s ∑ *K*′ =½ *m*_{1}*v*_{1}^{2}+ ½ *m*_{2}*v*_{2}^{2}∑ *K*′ =½(8 kg)(4 m/s) ^{2}+ ½(4 kg)(11 m/s) ^{2}∑ *K*′ =306 J Momentum and total mechanical energy of the system were both conserved. The total kinetic energy of the two objects after the collision is the same as it was before. In the macroscopic world such an outcome would never happen. The results above show the limit of what could happen. That is, macroscopic objects will always have less total mechanical energy after a collision than before. Never equal to or greater than.

This outcome is difficult to explain.

∑ *p*′ =*m*_{1}*v*_{1}′+ *m*_{2}*v*_{2}′∑ *p*′ =(8 kg)(−6 m/s) + (4 kg)(+15 m/s) ∑ *p*′ =+12 kg m/s ∑ *K*′ =½ *m*_{1}*v*_{1}^{2}+ ½ *m*_{2}*v*_{2}^{2}∑ *K*′ =½(8 kg)(−6 m/s) ^{2}+ ½(4 kg)(15 m/s) ^{2}∑ *K*′ =594 J Momentum was conserved, but mechanical energy increased. How could this happen? Where did the extra energy come from? Since this is a problem about generic objects, we are free to contrive all sorts of explanations. Perhaps there was a compressed spring on one of the objects, or a chemical explosive, or the two objects were small mammals that kicked off of each other after they collided. I think "contrive" was an appropriate word choice on my part to describe what we're doing here. This outcome seems improbable as it violates the conservation of mechanical energy in a way different from the previous outcomes.

This outcome is also perplexing.

∑ *p*′ =*m*_{1}*v*_{1}′+ *m*_{2}*v*_{2}′∑ *p*′ =(8 kg)(−8 m/s) + (4 kg)(+5 m/s) ∑ *p*′ =−44 kg m/s ∑ *K*′ =½ *m*_{1}*v*_{1}^{2}+ ½ *m*_{2}*v*_{2}^{2}∑ *K*′ =½(8 kg)(8 m/s) ^{2}+ ½(4 kg)(5 m/s) ^{2}∑ *K*′ =306 J Mechanical energy was conserved in this outcome (which is rare, but not impossible), but momentum was not. This would require an impulse applied from outside of the system. Since the description of this problem makes no mention of the existence or even the possibility of a third object, we would have to conclude that this outcome is impossible as it violates the conservation of linear momentum.

### practice problem 2

- A 5 kg bowling ball moving at 8 m/s approaches a row of stationary balls lined up end to end in a ball return. Comment on the likelihood of the following outcomes.
- The incoming ball stops and one 5 kg ball leaves the row of stationary balls at a speed of 8 m/s.
- The incoming ball stops and two 5 kg balls leave the row of stationary balls at a speed of 4 m/s.

- Two 5 kg bowling balls moving at 8 m/s approach a row of stationary balls lined up end to end in a ball return. Comment on the likelihood of the following outcomes.
- The incoming balls stop and two 5 kg balls leave the row of stationary balls at a speed of 8 m/s.
- The incoming balls stop and one 5 kg ball leaves the row of stationary balls at a speed of 16 m/s.

#### solution

Compute the total linear momentum and mechanical energy of the bowling balls before and after each collision has occurred. Compare the values of these quantities to answer this question.

In the case of one bowling ball the initial momentum and energy are…

In the first hypothetical outcome shown below, both momentum and energy are completely conserved. Such a collision is said to be perfectly elastic. While totally fine from a theoretical perspective, such an outcome is practically impossible. In the macroscopic real world, momentum and energy are always dissipated. (In the microscopic world of atoms and molecules collisions are always elastic, but that is another story.)

This outcome, while highly improbable, is not theoretically impossible.

In this hypothetical outcome, momentum is conserved but mechanical energy is lost. Such a collision is said to be inelastic. While totally fine from a theoretical standpoint I think, from personal observations of bowling balls, that such an outcome is highly unlikely.

Of the two outcomes presented neither

*will*occur. The real world doesn't work out according to hypothetical ideals. The real questions should be, "Is the real outcome of this collision more like the first hypothetical outcome or the second?" Experimental observation confirms that the collisions between bowling balls are more like elastic collisions than inelastic collisions. If one bowling ball comes into a row of stationary balls, the most likely outcome is that one bowling ball will leave the other end.

In the case of two bowling balls the initial momentum and energy are…

This is a trivial solution to the problem. Obviously both momentum and energy are conserved. This is another example of a perfectly elastic collision.

This outcome is possible, but not probable.

This last possible outcome makes no sense. The momentum after collision is the same as before, but the mechanical energy has somehow increased. Miraculously, it doubled.

While energy can neither be created nor destroyed, it certainly can become "lost". This is why I have no problem with the second outcome of the first collision. However, this outcome is surely impossible. The kinetic energy of a system cannot increase without work being done by some outside agent. For a row of bowling balls sitting in a ball return I cannot see anyway for positive work to be done. Thus if two bowling balls approach a row of stationary balls the most likely outcome is that two will emerge from the other end.

In general, the collision between bowling balls is more like an elastic collision than an inelastic one. If one ball approaches a row of stationary balls, one ball will leave from the other side. If two balls approach, then two will leave. If three balls approach, three will leave. And so on. This rule is the basis for a popular desktop ornament — and when I say "desktop" I mean the top of a real desk, not the pattern on a computer monitor that one sees when no applications are running or documents are open. Often called "Newton's cradle" or (more hilariously) "Newton's balls" and identified by a whole host of trade names, it is standard issue for the executive desktop in movies and television. In fact, I once heard it called "the executive intelligence tester". The dialog below illustrates this application.

Human ResourcesWatch. One goes in. One comes out. Two go in. Two come out. Three go in. Three come out.ExecutiveWow.Human ResourcesKeep watching. Four go in. What happens next?ExecutiveUh… Four come out?Human ResourcesGood. Now five go in.ExecutiveOo, oo, I know. Five come out!Human ResourcesCongratulations, you're our new vice president.

### practice problem 3

- the kinetic energy of the bullet
- the recoil velocity of the pistol
- the kinetic energy of the pistol
- the fraction of the total energy delivered to the bullet

- the velocity of the man and bullet together
- the kinetic energy of the man and bullet together
- the fraction of the total kinetic energy lost in the collision

#### solution

Answer it.

### practice problem 4

*m*

_{1}and

*m*

_{2}) traveling in opposite directions (+

*v*

_{1}and −

*v*

_{2}) collide head on and stick together. Derive an expression for…

- the final velocity of the two objects stuck together (easy)
- the kinetic energy lost as a result of the collision (hard)

#### solution

The first part is the easy part.

∑ *p*= ∑ *p*′*m*_{1}*v*_{1}−*m*_{2}*v*_{2}= ( *m*_{1}+*m*_{2})*v*′*v*′= *m*_{1}*v*_{1}−*m*_{2}*v*_{2}( *m*_{1}+*m*_{2})The second part is the hard part. Start with the basics.

∆

*K*=*K*−_{f}*K*_{i}Both terms on the right side of this equation are huge vomit piles of symbols. Work on each term separately. Start with the final total kinetic energy.

*K*=_{f}1 ( *m*_{1}+*m*_{2})⎛

⎝*m*_{1}*v*_{1}−*m*_{2}*v*_{2}⎞ ^{2}

⎠2 *m*_{1}+*m*_{2}Simplify, if that's the right word.

*K*=_{f}*m*_{1}^{2}*v*_{1}^{2}− 2*m*_{1}*v*_{1}*m*_{2}*v*_{2}+*m*_{2}^{2}*v*_{2}^{2}2( *m*_{1}+*m*_{2})Go back to the initial total kinetic energy.

*K*=_{i}1 *m*_{1}*v*_{1}^{2}+ 1 *m*_{2}*v*_{2}^{2}2 2 We can't subtract until we have a common denominator with the final total energy.

*K*=_{i}*m*_{1}*v*_{1}^{2}(*m*_{1}+*m*_{2}) +*m*_{2}*v*_{2}^{2}(*m*_{1}+*m*_{2})2( *m*_{1}+*m*_{2})Simplify again.

*K*=_{i}*m*_{1}^{2}*v*_{1}^{2}+*m*_{1}*m*_{2}*v*_{1}^{2}+*m*_{1}*m*_{2}*v*_{2}^{2}+*m*_{2}^{2}*v*_{2}^{2}2( *m*_{1}+*m*_{2})We now have two monster fractions than need to be subtracted. Instead of writing the whole mess out, let's just work on the numerator.

(

*m*_{1}^{2}*v*_{1}^{2}− 2*m*_{1}*v*_{1}*m*_{2}*v*_{2}+*m*_{2}^{2}*v*_{2}^{2}) −( *m*_{1}^{2}*v*_{1}^{2}+*m*_{1}*m*_{2}*v*_{1}^{2}+*m*_{1}*m*_{2}*v*_{2}^{2}+*m*_{2}^{2}*v*_{2}^{2})A few terms cancel and the masses can be factored out along with the minus sign.

−

*m*_{1}*m*_{2}(*v*_{1}^{2}+ 2*v*_{1}*v*_{2}+*v*_{2}^{2})The term in parentheses is the square of the sum of the two speeds (their absolute values in this case)

−

*m*_{1}*m*_{2}(*v*_{1}+*v*_{2})^{2}Stack numerator on denominator and call it quits.

∆ *K*=− *m*_{1}*m*_{2}(*v*_{1}+*v*_{2})^{2}2( *m*_{1}+*m*_{2})Of what use is this? I forgot why I wrote this question.