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Physics 140 HOMEWORK Chapter 9B Q3. Consider a box that explodes into two pieces while moving with a constant positive velocity along an x-axis. If one piece, with mass m 1 , ends up with positive velocity v 1 , then the second piece, with mass m 2 , could end up with (a) a positive velocity (Fig. 9-25a), (b) a negative velocity (Fig. 9-25b), or (c) zero velocity (Fig. 9-25c). Rank those three possible results for the second piece according to the corresponding magnitude of v 1 , greatest first. ——— b > c > a. If m 2 is stopped, then m 1 v 1 has to carry all of the original momentum. If v 2 > 0, then it makes a positive contribution. If v 2 < 0, then it makes a negative contribution, so v 1 must be greatest. Q12. Figure 9-34 shows four graphs of position versus time for two bodies and their center of mass.The two bodies form a closed, isolated system and undergo a completely inelastic, one-dimensional collision on an x axis. In graph 1, are (a) the two bodies and (b) the center of mass moving in the positive or negative direction of the x axis? (c) Which graphs correspond to a physically impossible situation? Explain. ——— (a) In graph 1, the two particles have positive slopes so are both moving in the positive direction. (b) The com slope of x vs t is positive, so its velocity is positive. (c) Graphs 2 and 3 are impossible. The com velocity has to be somewhere between the slopes of the particles. P6. Figure 9-39 shows a cubical box that has been constructed from uniform metal plate of negligible thickness. The box is open at the top and has edge length L = 40 cm. Find (a) the x coordinate, (b) the y coordinate, and (c) the z coordinate of the center of mass of the box. ——— First, let B refer to the bottom; and number the four sides. Let the side in the y-z plane be 1; the side in the x-z plane be 2; side 3 be opposite side 1; and 4 be opposite side 2. Let m 0 be the mass of any side, so that the total mass is 5m 0 . However, use the notation m B , m 1 , m 2 , etc to distinguish. Recall that any chunk (side, in this case) can be treated as concentrated at its com. (a) x com =[m B (L/2) + m 1 (0) + m 2 (L/2) + m 3 (L)+ m 4 (L/2)]/5m 0 = [(5/2)m 0 L]/5m 0 = L/2= 20 cm. This was clear from symmetry anyway. (b) y com = L/2= 20 cm, essentially the same as part (a) (c) z com =[m B (0)+m 1 (L/2)+m 2 (L/2)+m 3 (L/2)+m 4 (L/2)]/5m 0 = [(4/2)m 0 L]/5m 0 =0.4L = 16 cm. P13. A shell is shot with an initial velocity of 20 m/s, at an angle of 60 with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 9-42). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible? ——— Recognize the key concepts: projectile motion before the explosion, conservation of momentum through the explosion, and projectile motion again after the explosion. Remember that v x = v x0 and v y = v y0 - gt. v 0 = (20 m/s)(cos 60 ˆ ı + sin 60 ˆ ) = (10ˆ ı + 17.)m/s. At the explosion, v y =0 0= v 0y - gt 1 t 1 =1.767 s, where t 1 is the time of the explosion. In this time, the shell has moved downrange by an amount d 1 = v 0x t 1 = (10 m/s)(1.767 s) = 17.67 m. At the explosion, momentum is conserved, so that m 0 v 0x =(m 0 /2)(v 1 ) + 0. Here, v 1 is the speed of the moving fragment after the explosion. Solve for v 1 : v 1 =2v 0x =2 · 10 m/s = 20 m/s. The back fragment drops like the standard dropped rock, while the boosted fragment falls at exactly the same rate, so both hit the ground at the same time. We need its time in the air during this second

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Page 1: Physics 140 HOMEWORK Chapter 9B - University of …home.sandiego.edu/~rskelton/phyc140/hwch09B.pdf · Physics 140 HOMEWORK Chapter 9B ... explodes into two pieces while moving with

Physics 140

HOMEWORK Chapter 9B

Q3. Consider a box that explodes into two pieces while moving with a constant positive velocity along

an x-axis. If one piece, with mass m1, ends up with positive velocity ~v1 , then the second piece, with

mass m2, could end up with (a) a positive velocity (Fig. 9-25a), (b) a negative velocity (Fig. 9-25b),

or (c) zero velocity (Fig. 9-25c). Rank those three possible results for the second piece according to

the corresponding magnitude of ~v1, greatest first.

———

b > c > a. If m2 is stopped, then m1v1 has to carry all of the original momentum. If v2 > 0, then it

makes a positive contribution. If v2 < 0, then it makes a negative contribution, so v1 must be greatest.

Q12. Figure 9-34 shows four graphs of position versus time for two bodies and their center of mass.The

two bodies form a closed, isolated system and undergo a completely inelastic, one-dimensional collision

on an x axis. In graph 1, are (a) the two bodies and (b) the center of mass moving in the positive

or negative direction of the x axis? (c) Which graphs correspond to a physically impossible situation?

Explain.

———

(a) In graph 1, the two particles have positive slopes so are both moving in the positive direction.

(b) The com slope of x vs t is positive, so its velocity is positive.

(c) Graphs 2 and 3 are impossible. The com velocity has to be somewhere between the slopes of

the particles.

P6. Figure 9-39 shows a cubical box that has been constructed from uniform metal plate of negligible

thickness. The box is open at the top and has edge length L = 40 cm. Find (a) the x coordinate,

(b) the y coordinate, and (c) the z coordinate of the center of mass of the box.

———

First, let B refer to the bottom; and number the four sides. Let the side in the y-z plane be 1; the side

in the x-z plane be 2; side 3 be opposite side 1; and 4 be opposite side 2. Let m0 be the mass of any

side, so that the total mass is 5m0. However, use the notation mB, m1, m2, etc to distinguish.

Recall that any chunk (side, in this case) can be treated as concentrated at its com.

(a) xcom = [mB(L/2)+m1(0)+m2(L/2)+m3(L)+m4(L/2)]/5m0 = [(5/2)m0L]/5 m0 = L/2 = 20 cm.

This was clear from symmetry anyway.

(b) ycom = L/2 = 20 cm, essentially the same as part (a)

(c) zcom = [mB(0)+m1(L/2)+m2(L/2)+m3(L/2)+m4(L/2)]/5m0 = [(4/2)m0L]/5 m0 = 0.4L = 16 cm.

P13. A shell is shot with an initial velocity of 20 m/s, at an angle of 60◦ with the horizontal. At the

top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 9-42). One fragment,

whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the

other fragment land, assuming that the terrain is level and that air drag is negligible?

———

Recognize the key concepts: projectile motion before the explosion, conservation of momentum through

the explosion, and projectile motion again after the explosion.

Remember that vx = vx0 and vy = vy0 − gt.

~v0 = (20 m/s)(cos 60◦ ı̂ + sin 60◦ ̂) = (10 ı̂ + 17.3 ̂) m/s.

At the explosion, vy = 0 ⇒ 0 = v0y − gt1 ⇒ t1 = 1.767 s, where t1 is the time of the explosion.

In this time, the shell has moved downrange by an amount

d1 = v0xt1 = (10 m/s)(1.767 s) = 17.67 m.

At the explosion, momentum is conserved, so that m0v0x = (m0/2)(v1)+0. Here, v1 is the speed of the

moving fragment after the explosion. Solve for v1:

v1 = 2v0x = 2 · 10 m/s = 20 m/s.

The back fragment drops like the standard dropped rock, while the boosted fragment falls at exactly

the same rate, so both hit the ground at the same time. We need its time in the air during this second

Page 2: Physics 140 HOMEWORK Chapter 9B - University of …home.sandiego.edu/~rskelton/phyc140/hwch09B.pdf · Physics 140 HOMEWORK Chapter 9B ... explodes into two pieces while moving with

phase. We know its height h is

h = vy,avgt1 = [(17.32 m/s + 0)/2][1.767 s] = 15.31 m. The time to drop, released from rest from this

height, is t2 =√

2h/g = 1.767 s.

We might have recognized that t2 would equal t1 by symmetry.

So the horizontal distance from the explosion point d2 is

d2 = (20 m/s)(1.767 s) = 35.34 m. The total distance is:

d1 + d2 = 53.0m.

Now: the easy way, using com. Since the impact height is equal to the launch height, we know that

the com lands at the range given by Eq 4-26, R = v20 sin 2θ0/g = 35.35 m. Since the back fragment

lands at half this distance, the other fragment must land at 3/2 times this distance, or

d = (3/2)(35.35 m) = 53.0 m.

P15. Figure 9-44 shows an arrangement with an air track, in which a cart is connected by a cord

to a hanging block. The cart has mass m1 = 0.600 kg, and its center is initially at xy coordinates

(−0.500 m, 0 m); the block has mass m2 = 0.400 kg, and its center is initially at xy coordinates

(0,−0.100 m). The mass of the cord and pulley are negligible. The cart is released from rest, and

both cart and block move until the cart hits the pulley. The friction between the cart and the air track

and between the pulley and its axle is negligible. (a) In unit-vector notation, what is the acceleration

of the center of mass of the cart-block system? (b) What is the velocity of the com as a function of

time t? (c) Sketch the path taken by the com. (d) If the path is curved, determine whether it bulges

upward to the right or downward to the left, and if it is straight, find the angle between it and the x

axis.

———

First, we need to detemine the motion of the system.

FBD for m1: T to right; vertical forces balance; ~a to right. +x to right.

FBD for m2: T up; m2g down. ~a down. For the moment, call it +x downward.

x-eq for m1: T = m1a

x-eq for m2: m2g − T = m2a

Add the equations to get a = m2g/(m1 + m2) = 0.4g = 3.92 m/s2.

At this point, impose the given xy coordinate system on the block. That means that the a value is ax

for the cart, and −ay for the hanging block. Then:

xC = x0 + (1/2)at2 = −0.5 m + (1.96 m/s2)t2. yC = 0 at all times.

yB = y0 − (1/2)at2 = −0.1 m − (1.96 m/s2)t2. xB = 0 at all times.

The x- and y-coordinates of the com are now easily calculated:

xcom = [(.6 kg)(−0.5 m + 1.96 m/s2 t2) + (0.4 kg)(0)]/1 kg = (−0.3 + 1.176 t2) m.

ycom = [(.6 kg)(0) + (0.4 kg)(−0.1 m − 1.96 m/s2)t2)]/1 kg = (−0.04 − 0.784t2) m.

We can then write ~rcom in ı̂ ̂ notation as

~rcom = [(−0.3 + 1.176 t2) ı̂ + (−0.04 − 0.784 t2) ̂] m.

~vcom = d~r/dt = [2.35 ı̂ − 1.57 ̂] t m/s.

(a) ~acom = d~v/dt = [2.35 ı̂ − 1.57 ̂]m/s2.

We note that ~acom is a constant vector.

(b) ~vcom was calculated above: ~vcom = d~r/dt = [2.35 ı̂ − 1.57 ̂] t m/s.

(c) Line with slope (−1.57/2.35) = −0.667.

(d) Straight line, angle = Tan−1(−0.667) = −33.7◦.

Page 3: Physics 140 HOMEWORK Chapter 9B - University of …home.sandiego.edu/~rskelton/phyc140/hwch09B.pdf · Physics 140 HOMEWORK Chapter 9B ... explodes into two pieces while moving with

P52. In Fig. 9-59, a 10 g bullet moving directly upward at 1000 m/s strikes and passes through the

center of mass of a 5.0-kg block initially at rest. The bullet emerges from the block moving directly

upward at 400 m/s. To what maximum height does the block then rise above its initial position?

———

Recognize that vertical momentum is conserved during the collision: the impulse from gravity is negli-

gible, as we shall see. So, use conservation of momentum to get the initial upward velocity of the block;

then 1-D kinematics (or conservation of energy) to get its max height.

~pf = ~pi = (0.01 kg)(1000 m/s) = 10 kg m/s.

~pf = (0.01 kg)(400 m/s) + (5 kg)(vB) = 10 kg m/s ⇒ vB = 1.2 m/s.

v2 − v20 = 2a∆y ⇒ ∆y = (0 − 1.22)/(2 · (−9.8 m/s2)) = 0.0735m.

How about the influence of gravity during the collision?

Notice that taking the bullet’s average speed of 700 m/s and a rasonable block thickness of 0.1 m gives

a collision duration of (0.1 m)/(700 m/s) = 1.4× 10−4 s. The impulse on the bullet due to gravity then

is

Jgrav = (0.01 kg)(10 m/s2)(1.4 × 10−4 s) = 1.4 × 10−5 N · s (or kg m/s).

This is negligible compared to the pi = 10 kg m/s.

P71. In Fig. 9-21, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus.

The alpha particle is scattered at angle θ1 = 64.0◦ and the oxygen nucleus recoils with speed 1.20 ×105 m/s and at angle θ2 = 51.0◦. In atomic mass units, the mass of the alpha particle is 4.00 u and the

mass of the oxygen nucleus is 16.0 u. What are the (a) final and (b) initial speeds of the alpha particle?

———

Recognize a conservation of momentum problem, and in two dimensions. The x- and y-components are

conserved separately. Considering the alpha to be moving initially in the +x direction, we see that the

py equation will have only one unknown, viz the final speed of the alpha vyfα.

(a) The y-momentum equation is pyf,tot = pyi,tot = 0, or

0 = (16 u)(1.2 × 105 m/s)(sin 51◦) − (4 u)(vyfα)(sin 64◦) ⇒vyfα = [(16 u)(1.2 × 105 m/s)(0.7771)]/[(4 u)(0.8988)] = 4.150 × 105 m/s.

(b) The x-momentum eq is

(4 u)(v0) = (4 u)(4.150 × 105 m/s)(cos 64◦) + (16 u)(1.2 × 105 m/s)(cos 51◦) ⇒v0 = 4.84 × 105 m/s.

P86. Speed amplifier. In Fig. 9-75, block 1 of mass m1 slides along an x axis on a frictionless floor

with a speed of v1i = 4.00 m/s. Then it undergoes a one-dimensional elastic collision with stationary

block 2 of mass m2 = 0.500m1. Next, block 2 undergoes a one-dimensional elastic collision with

stationary block 3 of mass m3 = 0.500m2. (a) What then is the speed of block 3? Are (b) the speed,

(c) the kinetic energy, and (d) the momentum of block 3 greater than, less than, or the same as the

initial values for block 1?

———

We need to consider the collisions in order. When m1 strikes m2, eq. 9-68 applies:

v2f = v1i 2m1/(m1 + m2) = (4 m/s)(2/1.5) = 5.33 m/s

Similarly, v3f = (5.33 m/s)(4/3) = 7.11m/s.

(b) This speed is clearly greater than the initial speed of block 1.

(c) K1 = (1/2)m1(4 m/s)2 = 8m1 m2/s2, while K3 = (1/2)(0.25m1)(7.11 m/s)2 = 6.32m m2/s2. Some

kinetic energy was left in blocks 1 and 2. The m1 carries units of kilograms.

(d) p1 = m1(4 m/s) = 4m1 m/s, while p3 = (0.25m1)(7.11 m/s) = 1.78m m/s. Since blocks 1 and 2

are both moving forward after the collision, each has some momentum in the forward direction.

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P 69 Opt. A small ball of mass m is aligned above a larger ball of mass M = 0.63 kg (with a slight

separation, as with the baseball and basketball of Fig. 9-68a), and the two are dropped simultaneously

from a height of h = 1.8 m. (Assume the radius of each ball is negligible relative to h.) (a) If the larger

ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger

ball, what value of m results in the larger ball stopping when it collides with the small ball? (b) What

height does the small ball then reach (Fig. 9-68b)?

—————-

Immediately before the large ball hits the ground, va = −√

2gh = −v0 for each ball. The negative sign

means downward, and v0 is just an abbreviation for√

2gh. The basketball bounces elastically, so that

vb,M = +v0 where the “b” means immediately before the ball-ball collision.

The velocity of the baseball is still −v0, so apply eq 9-75 and 9-76. Take subscript 1 as corresponding

to the basketball (M) and 2 to the baseball (m).

(a) We want v1f = 0, where v1i = +v0 and v2i = −v0. Use Eq 9-75:

0 = [(M − m)/(M + m)] v0 + [2m/(M + m)](−v0) ⇒ m = M/3 = 0.21kg.

(b) We need v2f from Eq 9-76:

v2f = [2M/(M + m)]v0 + [(m − M)/(M + m)](−v0) = [(3M − m)/(M + m)]v0. This is positive,

meaning upward, unless m > 3M .

Substituting m = M/3, v2f = 2v0. This could have been gotten from the fact that, in an elastic

collision, speed of separation equals speed of approach.

The maximum height h2 is then given by

h2 = (2v0)2/2g = 4v2

0/2g = 4 · 2gh/2g = 4h = 7.2 m.

By the way, this is the basis of a supernova explosion.