# Answers/Solutions: Giancoli Chapter 9

Q1: A basic point: Momentum is only conserved in an ISOLATED SYSTEM, where the net external force is zero. A moving object is an isolated system only if no forces such as friction etc., act, otherwise its momentum does change: p = F_{ext}t.

Q6: (d); the girl moves in the opposite direction at 2.0 m/s. Since momentum is conserved and is originally zero, it must remain zero. The boy has twice the mass so the girl must have twice the velocity, and in the opposite direction.

Q8: Yes. In a perfectly elastic collision KE is unchanged. The earth has an effective mass of infinity compared to the ball, so it does not speed up at all (certainly not enough to be measured). Thus the KE and thus the speed of the ball must not change during the collision. Its original PE, Mgh, is the source of the KE just before the collision and so this same amount of KE will lead to the same PE (and thus the same h) after the collision.

Q12: Absolutely. Impulse = p = Ft and so a small F can result in a larger impulse if it acts for a sufficiently longer time.

P6: Comment: This is probably not a good test problem due to the fact that the messy arithmetic takes too much time. But one like it might show up on the final exam, or I might give you a simplified, one-dimensional version - such as P22 below - on a test.

We use the fact that the change in momentum over time interval t_{1} = a to t_{2} = b is given by Newton's Second Law in the form

p =

and if we are interested in the *average* force over this time interval, we can use the simpler relation

**p** = **F _{AV}**t

In this case, the original momentum is in the +x-direction:

p_{1} = (0.145 kg)(32 m/s)**i** = 4.64**i** kg-m/s

The momentum immediately after the collision with the bat can be obtained from the conservation of energy. Since we know the ball rises to a maximum height of 36.5 m, we equate mgh with the KE just after the collision and solve for v. We obtain v = 26.7 m/s. Therefore the momentum right after the collision is

p_{2} = (0.145 kg)(26.7 m/s)**j** = 3.87**j** kg-m/s

So p = 3.87**j** - 4.64**i**

So the average force must be

F_{AV} = p/t = 1548**j** - 1856**i**

The magnitude of this force is sqrt[(1548^{2} + (-1856)^{2}] = 2416 Newtons at an angle of Arctan (1548/-1856) = - 39.8 + 180 = 140 degrees CCW from the +x axis. (We must add the 180 degrees: since the x-component of the vector is negative and the y-component positive, we know the angle is in the second quadrant.)

P13: The "system" here is the boat+child+package, and the initial momentum is zero since they are all at rest. *After* the package-toss, the package is moving rightward (call this the +x direction) at 10.0 m/s. The "final" x-momentum (i.e., right after the toss) must still be zero because we assume no *external* horizontal forces act on the system. So, after the toss

(M_{Boat} + M_{Child})V_{f} + M_{Package}(10.0) = 0

Solving, we obtain V_{f} = - 0.966 m/s

P19: For mass M_{A} = 2.0 kg, we have: (**V _{A}**)

_{initial}= 4

**i**+ 5

**j**- 2

**k**; and are given: (

**V**)

_{A}_{final}= - 2

**i**+ 0

**j**+ 3

**k**

Meanwhile, for mass B = 3.0 kg, (**V _{B}**)

_{initial}= 0.

We are to find the final velocity for mass B.

Momentum must be conserved in all three directions. The clearest way to approach the analysis, I think, is simply to treat each direction separately. Along *each axis*, we must have

M_{A}(**V _{A}**)

_{initial}+ M

_{B}(

**V**)

_{B}_{initial}= M

_{A}(

**V**)

_{A}_{final}+ M

_{B}(

**V**)

_{B}_{final}

So, along x:

(2.0)(4.0) + (3.0)(0) = (2.0)(-2.0) + (3.0)((V_{B})_{x})_{final}

(sorry about the triple subscript! It indicates the *final* velocity of *mass B* in the *x-direction*. The result is:

((V_{B})_{x})_{final} = 4.0 m/s

Likewise, along y:

(2.0)(5.0) + (3.0)(0) = (2.0)(0) + (3.0)((V_{B})_{y})_{final}

which yields: ((V_{B})_{y})_{final} = 3.33 m/s

and along z:

(2.0)(- 2.0) + (3.0)(0) = (2.0)(3.0) + (3.0)((V_{B})_{z})_{final}

which gives: ((V_{B})_{z})_{final} = - 3.33 m/s

In "condensed" notation: (**V _{B}**)

_{final}= 4

**i**+ 3.3

**j**- 3.3

**k**in units of m/s.

P22: Comment: This problem is a simpler version of P6! The reason it is simpler is that everything happens along one dimension (i.e., horizontally).

For M = 0.145 kg, V_{initial} = + 35.0 m/s and V_{final} = - 56 m/s. The change in momentum (i.e., the impulse given the baseball) is

p = MV_{final} - MV_{initial} = (0.145)(- 56 - 35) = - 13.2 kg-m/s

Newton's Second Law says that the average force (in this case, just "the force" since it is assumed to be constant) is

F_{AV} = p/t = - 13.2/0.005 = 2640 Newtons (= 593 pounds!)

P35: The conditions here are identical with those of Example 9-8, page 224. Be sure you understand this example. We will take advantage of the algebra done there and simply use the first two equations in the "Solution" to that example. With A to denote the projectile and B to denote the target, and choosing east as the positive direction:

(V_{B})_{final} = (V_{A})_{initial}[2M_{A}/(M_{A} + M_{B})]

(V_{A})_{final} = (V_{A})_{initial}[(M_{A} - M_{B})/(M_{A} + M_{B})]

with numbers:

(V_{B})_{final} = (4.80)[(2)(0.450)/(0.450 + 0.900)] = + 3.2 m/s (plus indicates east)

(V_{A})_{final} = (4.80)[(0.450 - 0.900)/(0.450 + 0.900)] = - 1.60 m/s (minus indicates west)

P41Comment: I love this problem. However, part (c) must be addressed before we can answer part (b). The answer to part (c) is that the collision is, indeed, elastic. This is true because there is no friction to remove mechanical energy from the system. So, although the spring briefly stores some of the original KE as PE, when it "springs back" to its relaxed position, that PE once again becomes KE. Thus the final KE will equal the original KE and this makes the collision elastic.

(a) The spring's maximum compression occurs at an instant when the two masses have the same velocity. Call this instant t_{2}. Then we demand that the *momentum* is conserved: p_{2} must equal the original momentum p_{o}:

(3.0 kg)(8.0 m/s) + (4.5 kg)(0) = (3.0 + 4.5)v_{2}

which yields v_{2} = 3.2 m/s

Now at this instant, the KE is (1/2)(3.0 + 4.5)(3.2)^{2} = 38.4 Joules, which is much less than the initial KE of (1/2)(3.0)(8.0)^{2} = 96 Joules. Where is the missing energy?

In the spring, of course: (PE)_{Spring} = (1/2)ks^{2} where s is the desired compression. So:

(0.5)(850)s^{2} = 96.0 - 38.4

which yields s = 0.368 m = 36.8 cm.

(b) After the collision, when the blocks are again separated, there is no PE in the spring, and so all the original KE is again KE, and the collision is simply a one-dimensional elastic collision, as in Example 9-8 or in P35 above. The answers are on page A-23 in the textbook.

P45 Comment: A lovely problem because it shows the "power" of the conservation laws. It seems, upon reading the problem, that there is not enough information to answer it. But the conservation laws demand a unique answer!

Let M_{1} be the lighter fragment, so M_{2} = 1.5M_{1}.

The object is at rest before the explosion. Momentum must be conserved. This is enough to ensure that the two fragments are moving away from each other along a straight line afterward (think about why this must be so). We will call that line the x-axis. And, demanding momentum conservation

M_{1}V_{1} = - M_{2}V_{2}

We are told that the total KE is 7500 Joules, so

(1/2)M_{1}(V_{1})^{2} + (1/2)M_{2}(V_{2})^{2} = 7500

Substituting M_{2} = 3M_{1}/2, the first equation gives V_{2} = -2V_{1}/3. Then, in the second equation"

(1/2)M_{1}(V_{1})^{2} + (1/2)(3M_{1}/2)(4/9)(V_{1})^{2} = 7500

or, simplifying,

(5/6)M_{1}(V_{1})^{2} = (5/3)(KE)_{1} = 7500

So (KE)_{1} = (3/5)(7500) = 4500 Joules, which leaves (KE)_{2} = 3000 Joules.

If we were given the masses of the fragments, we could solve for their speeds

P51: We imagine a horizontal spring with the block (mass M) attached free to slide on a table-top. The bullet (mass m) strikes and embeds in the block, which slides, compressing the spring. There is friction between the sliding block and the table-top. Let v_{o} be the bullet speed before the collision, V be the bullet+block speed immediately after the collision (before the block has moved any significant distance) and s be the distance the block slides (and therefore the amount of spring compression) before stopping momentarily. The spring constant is k.

Momentum is conserved during the collision:

mv_{o} = (M+m)V

After the collision, the KE of the bullet+block is divided. An amount F_{fr}s is dissipated (turned into heat) due to friction, and the rest is (at the moment of stoppage) stored in the compressed spring:

(1/2)(M + m)V^{2} = (1/2)ks^{2} + _{k}(M + m)gs

We are given m, M, k, s, and _{k}. The unknowns are v_{o} and V. Once we find these we can compute the KE of the bullet initially, the KE of the bullet+block immediately after the collision, and answer questions about how much of the original KE ends up where. The answers to the questions asked are (a) v_{o} = 890 m/s and (b) The KE of the bullet+block immediately after the collision is only 0.001 (i.e., 0.01%) of the bullet's original KE, so a fraction 0.999 (i.e., 99.9%) is lost to heat, mechanical work deforming the wood, etc).

P56: Look at the figure for this problem (Figure 9-43). There are three quantities we are asked to find, so you know we will need three equations. We are told that the collision is to be assumed elastic, so KE is conserved along with the x and y components of momentum. Comment: I will use the notation of Section 9-7: primes to denote "final" -i.e., after the collision - velocities. Our three equations are, therefore.

M_{A}V_{Ax} + M_{B}V_{Bx} = M_{A}V'_{Ax} + M_{B}V'_{Bx}

M_{A}V_{Ay} + M_{B}V_{By} = M_{A}V'_{Ay} + M_{B}V'_{By}

(1/2)M_{A}**V _{A}**

^{2}+ (1/2)M

_{B}

**V**

_{B}^{2}= (1/2)M

_{A}

**V'**

_{A}^{2}+ (1/2)M

_{B}

**V'**

_{B}^{2}

Comment: Remember that the square of a vector is just **A**^{2} = **A****A** = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}

These look pretty scary. But M_{A} = M_{B} so all the masses divide out. We also are told that V_{Ax} = V_{By} = V'_{Bx} = 0. This simplifies things a lot:

V_{Bx} = V'_{Ax}

V_{Ay} = V'_{Ay} + V'_{By}

V_{Ay}^{2} + V_{Bx}^{2} = V'_{Ax}^{2} + V'_{Ay}^{2} + V'_{By}^{2}

These are *still* pretty scary, but getting less so. We know that V_{Ay} = 2.0 m/s, that V_{Bx} = 3.7 m/s, and that V'_{Bx} = 0. Thus

3.7 = V'_{Ax}

2 = V'_{Ay} + V'_{By}

(2)^{2} + (3.7)^{2} = V'_{Ax}^{2} + V'_{Ay}^{2} + V'_{By}^{2}

Now the first equation gives us one of our three unknowns immediately, and its value can be substituted into the third equation. This leaves two equations in two unknowns (unfortunately one equation is a quadratic). Solve the second equation for one of its unknowns in terms of the other, getting (say):

V'_{By} = 2 - V'_{Ay}

Now this can be substituted into the third equation, and we can solve for V'_{Ay} (since the equation is quadratic, we get two solutions):

V'_{Ay} = 2 **OR** 0.

But if V'_{Ay} = 2 we obtain V'_{By} = 0 which we know is inconsistent with the given information. Thus we conclude that V'_{Ay} = 0. We then find that V'_{By} = 2.0 m/s.

So, after all that, the balls exchange velocities: the A ball ends up moving in the +x direction at 3.7 m/s and the B ball ends up moving in the +y direction at 2.0 m/s!

There is no chance you would get this one in its entirety on a test, or even on a final. It is much too long. But I might ask you to *set one up* that is somewhat like this one.

P62: Along the front-to-rear axis of symmetry of the car (call it the x-axis), the CM is defined by the equation:

M_{Total}X_{CM} = M_{i}X_{i}

In this case we have six masses (the car and five people). Let's use the front of the car as the origin. Then M_{Total} = 1250 + (5)(70) = 1600 kg, so if we let the origin be the front of the car:

1600X_{CM} = (1250)(2.5) + (2)(70)(2.80) + (3)(70)(3.9)

X_{CM} = 2.71 m

P65: Our numerical answer will certainly depend on our choice of an origin! Let's choose the SW corner of the raft as (0,0). The COM of the raft is in the center at (9,9), with all distances in meters. The cars will be at (18,18), (18,0) and (0,0) respectively. The total mass will be 6200 + 3(1350) = 10250 kg. Therefore:

X_{CM} = (1/10250)[(6200)(9) + (1350)(18) + (1350)(18) + (1350)(0)] = 10.2 m

Y_{CM} = (1/10250)[(6200)(9) + (1350)(18) + (1350)(0) + (1350)(0)] = 7.8 m

Comment: Why don't these answers agree with the ones in the back-of-the-book? The solution there takes the origin (0,0) to be the *center of the raft*! His answers (+1.2 m and -1.2 m) locate the CM 1.2 meters to the east and south of the raft's center. In MY coordinate system, the center of the raft is (9,9). Convince yourself that the two answers are the same, but expressed in different coordinate systems. The point is that you must be clear about where you take the origin before your answer can be interpreted!

P73: (a) Take the earth's center as the origin and let the x-axis be the line between the two centers (i.e., center of Earth and center of Moon). Then:

X_{CM} = (1/M_{Total})[M_{Earth}(0) + M_{Moon}D]

which yields X_{CM} = 4.66x10^{6} meters. The interesting point is that the Earth's radius is R_{E} = 6.38x10^{6} meters, so the CM of the Earth-Moon system is *inside the earth*, roughly 3/4 of the radius from its center!

(b) The Earth and the Moon both orbit this point once each month. The CM of the system is the point that orbits the Sun once each year.

P75: Take the woman's position as the origin. The total mass is 127 kg. Then

(a) X_{CM} = (1/127)[(55)(0) + (72)(10)] = 5.67 m

(b) There are no external forces, and the system is initially at rest, so the CM will not move. Using our same coordinate system, the man is now at x_{m} = 7.5 m. So our equation for X_{CM} becomes

5.67 = (1/127)[(55)x_{w} + (72)(7.5)]

which gives x_{w} = 3.27 m. So the distance between the man and the woman would be 7.5 - 3.27 = 4.23 m.

(c) The two will collide at the CM, so the man must have moved from x = 10 m to x = 5.67 m, a distance of 4.33 m.

P90: Before the explosion the puck is at rest, and during the explosion, only "internal" forces are relevant. The momentum must be zero, then, after the explosion: along each axis

M_{1}v_{1} + M_{2}v_{2}+ M_{3}v_{3} = 0

We are given that M_{1} = m, M_{2} = 2m, **v _{1}** = v along +x, and

**v**= 2v along +y.

_{2}So along the x-axis:

mv + (2m)(0) + m(v_{3})_{x} = 0

and along the y-axis:

m(0) + (2m)(2v) + m(v_{3})_{y} = 0

These are easily solved to get (v_{3})_{x} = -v and (v_{3})_{y} = -4v.

The answer can be written as v_{3} = -v**i** - 4v**j** or expressed as v_{3} = at an angle of Arctan (-4/-1) = 256 degrees CCW from +x axis.

P 102: After the completely inelastic collision the two objects move as one along **some** straight line, and we will call this the +x axis. So after the collision, there is only x-momentum, and its magnitude is (2M)V_{f} = (2M)(v_{o}/3). The y-momentum is zero.

Now, *before* the collision, the y-momentum must have been zero also. Since the two objects are of equal mass, their y-velocities must have been equal and opposite. Then if the objects have the same velocity magnitude (v) and the same y-component of velocity, the angles between their original velocities and the x-axis must have been equal. Call these angles . Then

(P_{x})_{initial} = (P_{x})_{final}

2Mv_{o}cos() = (2M)(v_{o}/3)

therefore

cos() = 1/3

and = 70.5 degrees. The angle between the two velocity vectors, then, is 141 degrees.

P105: Gravity is a conservative force, so no energy is "lost" in the "collision" between the spacecraft and the planet. The interaction is therefore elastic. We can use the equations at the bottom of page 224, with "A" referring to the spacecraft and "B" to Saturn. Here, M_{B} is much, much greater than M_{A}, so we can neglect M_{A} when it is added or subtracted from M_{B}, and also neglect the fraction when M_{A} is divided by M_{B}. So under these conditions, the equations reduce to:

v_{B}^{'} = v_{A}(0) + v_{B}(1)

v_{A}^{'} = v_{A}(- 1) + v_{B}(2)

Obviously v_{B} is unchanged, and

v_{A}^{'} = -10.4 + (- 9.6)(2) = - 29.6 m/s

So the speed of the spacecraft is almost tripled! Note that the mass of the spacecraft does not matter, so long as it is much, much less than that of Saturn.

Comment: NASA has used this effect to advantage on a number of missions, for example Mariner 10 (slingshot by Venus en route to Mercury), Voyager 1 (slingshot around Jupiter *and* Saturn on its way out of the solar system), and Galileo (slingshot around Venus and Earth *twice* en route to Jupiter). For more, see the Wikipedia article at http://en.wikipedia.org/wiki/Gravity_assist.