Inelastic collision and explosive separation

Analyze example 3.5.2 from the point of view of the system’s kinetic energy. In particular, answer the following questions:

(a) What is the total kinetic energy of the system (i) before the players collide, (ii) right after the collision, when they are holding to one another, and (iii) after they separate. How much of this energy is translational (that is, center-of-mass kinetic energy), and how much is convertible? (b) Answer the same questions from the point of view of the player who is skating at a constant 1.5 m/s to the right (player 3)

(To avoid needless repetition, you may use already established results, such as conservation of momentum.)

Solution(a) Before the players collide, we have Ksys,i = 1 2m1v 2 1i+ 1 2m2v 2 2i= 1 2(80 kg) × & 3m s '2 + 1 2(90 kg) × & −2m_s '2 = 540 J (4.20) While they are still holding to each other, we know from the solution to example 3.5.2 that their joint velocity is 0.353 , and that this has to be also the velocity of their center of mass, which is unchanged by the collision. So, we have

Kcm= 1 2(m1+ m2)v 2 cm= 1 2(170 kg) & 0.353m s '2 = 10.6 J (4.21)

This is Kcmthroughout, as well as Ksysright after the collision, since the collision is totally inelastic

and that means that Kconv drops to zero. Also, subtracting this from (4.20) will give us the initial

value of the convertible energy, without the need for a separate calculation, so

Kconv,i= Ksys,i− Kcm= 540 J − 10.6 J = 529.4 J ≃ 529 J (4.22)

After the separation, the new total kinetic energy (for which I will use the subscript f ) is Ksys,i = 1 2m1v 2 1f + 1 2m2v 2 2f = 1 2(80 kg) × & −0.176m s '2 +1 2(90 kg) × & 0.824m s '2 = 31.8 J (4.23) where I have gotten the values for v1f and v2f from the solution to part (d) of Example 3.5.2.

Subtracting Kcm from this will give us the final value of the convertible energy:

Kconv,f = Ksys,f − Kcm= 31.8 J − 10.6 J = 21.2 J (4.24)

To summarize, then, we have:

• Before the collision:

Ksys,i = 540 J, Kcm = 10.6 J, Kconv,i= 529.4 J

• Right after the collision (players still holding to each other): Ksys = Kcm= 10.6 J, Kconv = 0

• After the (explosive) separation:

Ksys,f = 31.8 J, Kcm= 10.6 J, Kconv,i= 21.2 J

So, in the collision, approximately 529 J of kinetic energy “disappeared” from the system (or, we could say, were “converted into some form of internal energy”), whereas the players’ pushing on each other managed to put about 21 J of kinetic energy back into the system; we will explore these kinds of processes in more detail in the following chapter!

(b) We need to repeat all the above calculations with all the velocities shifted down by 1.5 m/s, to bring them to the reference frame of player 3. Instead of putting a subscript “3” on all the quantities, since we already have tons of subscripts to worry about, I’m going to follow an alternative convention and use a “prime” superscript (′

) to denote all the quantities in this frame of reference. In brief, we have K′ sys,i = 1 2m1(v ′ 1i)2+ 1 2m2(v ′ 2i)2 = 1 2(80 kg) × & 1.5m s '2 +1 2(90 kg) × & −3.5m_s '2= 641.3 J (4.25)

4.4. EXAMPLES 85 K′ cm = 1 2(m1+ m2)(v ′ cm)2 = 1 2(170 kg) & 0.353m s − 1.5 m s '2 = 111.8 J (4.26) K′ conv,i = K ′ sys,i− K ′ cm= 641.3 J − 111.8 J = 529.5 J ≃ 529 J (4.27)

This shows explicitly that the convertible energy, as I pointed out earlier in this chapter, is the same in every reference frame! (The equality is exact, if you keep enough decimals in the calculation.) Knowing this, we can simplify the calculation of the final kinetic energy, after the explosive sepa- ration: the convertible energy, K′

conv,f, will be the same as in the earth reference frame, that is to

say, 21.2 J, and the total kinetic energy will be K′

sys,f = K ′ cm+ K

′

conv,f = 111.8 J +21.2 J = 133 J.

So, in this frame of reference, we have (to three significant figures): K′

sys,i = 641 J, K ′

cm= 112 J, K ′

conv,i= 529 J (before the collision)

K′ sys = K

′

cm = 112 J, K ′

conv= 0 (right after the collision)

K′

sys,f = 133 J, K ′

cm = 112 J, K ′

conv,i= 21.2 J (after the separation)

So, even though the total kinetic energy is different in the two reference frames, all the (inertial) observers will agree as to the amount of kinetic energy “lost” in the collision, as well as the amount of kinetic energy put back into the system by the players’ pushing on each other.

4.5

Problems

Problem 1

A 71-kg man can throw a 1-kg ball with a maximum speed of 6 m/s relative to himself. Imagine that one day he decides to try to do that on roller skates. Starting from rest, he throws the ball as hard as he can, so it ends up moving at 6 m/s relative to him, but he himself is recoiling as a result of the throw.

(a) Assuming conservation of momentum, find the velocities of the man and the ball relative to the ground.

(b) What is the kinetic energy of the system right after the throw? (By the system here we mean the man and the ball throughout.) Where did this kinetic energy come from?

(c) Is the man’s reference frame inertial throughout this process? Why or why not? (d) Does the center of mass of the system move at all throughout this process? Problem 2

Analyze Problem 1 from Chapter 3 from the point of view of the system’s kinetic energy. In particular, answer the following questions:

(a) What is the total kinetic energy of the system before and after the collision? How much of this energy is translational (that is, center-of-mass kinetic energy), and how much is convertible? (b) What kind of collision is this? (Elastic, inelastic, etc.) What is the coefficient of restitution? Problem 3

Analyze Problem 2 from Chapter 3 from the point of view of the system’s kinetic energy. In particular, answer the following questions:

(a) What is the coefficient of restitution for the collision described in part (a) of the problem, and how much kinetic energy is “lost” in that collision?

(b) What is the coefficient of restitution for the collision described in part (b) of the problem, and how much kinetic energy is “lost” in that collision?

Problem 4

A 0.012-kg bullet, traveling at 850 m/s, hits a 2-kg block of wood that is initially at rest, and goes straight through it. Assume that the final velocity of the bullet relative to the block is 400 m/s, and that the system is isolated.

(a) What is the coefficient of restitution for this collision? (b) How much kinetic energy is “lost” in the collision? (c) What is the final velocity of the block?

Problem 5

A 2-kg object, moving at 1 m/s, collides with a 1-kg object that is initially at rest. Assume they form an isolated system.

4.5. PROBLEMS 87

(a) What is the initial kinetic energy of the system? How much of this is center of mass energy, and how much is convertible?

(b) What is the maximum amount of kinetic energy that could be “lost” (converted to other forms of energy) in this collision?

(c) If 60% of the amount you calculated in part (b) is in fact converted into other forms of energy in the collision, what are the final velocities of the two objects?

Chapter 5

Interactions and energy