9 Chapter Homework Problem Solutions
1. We use Eq. 9-5 to solve for ( x3 , y3 ). (a) The x coordinates of the system's center of mass is: xcom = m1 x1 + m2 x2 + m3 x3 (2.00 kg)(-1.20 m) + ( 4.00 kg )( 0.600 m ) + ( 3.00 kg ) x3 = m1 + m2 + m3 2.00 kg + 4.00 kg + 3.00 kg
= -0.500 m. Solving the equation yields x3 = 1.50 m. (b) The y coordinates of the system's center of mass is:
ycom = m1 y1 + m2 y2 + m3 y3 (2.00 kg)(0.500 m) + ( 4.00 kg )( -0.750 m ) + ( 3.00 kg ) y3 = m1 + m2 + m3 2.00 kg + 4.00 kg + 3.00 kg
= -0.700 m.
Solving the equation yields y3 = 1.43 m.
Chapter 9 Homework Problem Solutions
2. Our notation is as follows: x1 = 0 and y1 = 0 are the coordinates of the m1 = 3.0 kg particle; x2 = 2.0 m and y2 = 1.0 m are the coordinates of the m2 = 4.0 kg particle; and, x3 = 1.0 m and y3 = 2.0 m are the coordinates of the m3 = 8.0 kg particle. (a) The x coordinate of the center of mass is xcom = m1 x1 + m2 x2 + m3 x3 0 + ( 4.0 kg )( 2.0 m ) + ( 8.0 kg )(1.0 m ) = = 1.1 m. m1 + m2 + m3 3.0 kg + 4.0 kg + 8.0 kg
(b) The y coordinate of the center of mass is ycom = m1 y1 + m2 y2 + m3 y3 0 + ( 4.0 kg )(1.0 m ) + ( 8.0 kg )( 2.0 m ) = = 1.3 m. m1 + m2 + m3 3.0 kg + 4.0 kg + 8.0 kg
(c) As the mass of m3, the topmost particle, is increased, the center of mass shifts toward that particle. As we approach the limit where m3 is infinitely more massive than the others, the center of mass becomes infinitesimally close to the position of m3.
Chapter 9 Homework Problem Solutions
15. We need to find the coordinates of the point where the shell explodes and the velocity of the fragment that does not fall straight down. The coordinate origin is at the firing point, the +x axis is rightward, and the +y direction is upward. The y component of the velocity is given by v = v0 y gt and this is zero at time t = v0 y/g = (v0/g) sin 0, where v0 is the initial speed and 0 is the firing angle. The coordinates of the highest point on the trajectory are
v2 ( 20 m/s ) sin 60 cos 60 = 17.7 m x = v0 x t = v0t cos 0 = 0 sin 0 cos 0 = g 9.8 m/s 2
2
and
2 1 2 1 v0 1 20 m / s y = v0 y t - gt = sin 2 0 = sin 2 60 = 15.3 m. 2 2 2 g 2 9.8 m / s
b
g
2
Since no horizontal forces act, the horizontal component of the momentum is conserved. Since one fragment has a velocity of zero after the explosion, the momentum of the other equals the momentum of the shell before the explosion. At the highest point the velocity of the shell is v0 cos0, in the positive x direction. Let M be the mass of the shell and let V0 be the velocity of the fragment. Then Mv0cos0 = MV0/2, since the mass of the fragment is M/2. This means
V0 = 2v0 cos 0 = 2 20 m / s cos 60 = 20 m / s. This information is used in the form of initial conditions for a projectile motion problem to determine where the fragment lands. Resetting our clock, we now analyze a projectile launched horizontally at time t = 0 with a speed of 20 m/s from a location having 1 coordinates x0 = 17.7 m, y0 = 15.3 m. Its y coordinate is given by y = y0 - 2 gt 2 , and when it lands this is zero. The time of landing is t = 2 y0 / g and the x coordinate of the landing point is
x = x0 + V0t = x0 + V0 2 y0 = 17.7 m + 20 m / s g
b
g
b
15. g 2b.8 m3/mg = 53 m. 9 s
2
Chapter 9 Homework Problem Solutions
16. We denote the mass of Ricardo as MR and that of Carmelita as MC. Let the center of mass of the two-person system (assumed to be closer to Ricardo) be a distance x from the middle of the canoe of length L and mass m. Then MR(L/2 x) = mx + MC(L/2 + x). Now, after they switch positions, the center of the canoe has moved a distance 2x from its initial position. Therefore, x = 40 cm/2 = 0.20 m, which we substitute into the above equation to solve for MC:
MC = 80 M R L / 2 - x - mx = L/2+ x
b
g
b gb
3.0 2
b
- 0.20 - 30 0.20 = 58 kg. 3.0 / 2 + 0.20
g b gb g g
Chapter 9 Homework Problem Solutions
27. We choose +y upward, which means vi = -25m s and v f = +10 m s. During the
collision, we make the reasonable approximation that the net force on the ball is equal to Favg the average force exerted by the floor up on the ball. (a) Using the impulse momentum theorem (Eq. 9-31) we find
J = mv f - mvi = 12 10 - 12 -25 = 42 kg m s. . .
b gb g b gb g
(b) From Eq. 9-35, we obtain Favg = J 42 = = 2.1 103 N. t 0.020
Chapter 9 Homework Problem Solutions
40. Our notation is as follows: the mass of the motor is M; the mass of the module is m; the initial speed of the system is v0; the relative speed between the motor and the module is vr; and, the speed of the module relative to the Earth is v after the separation. Conservation of linear momentum requires (M + m)v0 = mv + M(v vr). Therefore, v = v0 + 4m 82 km / h Mvr = 4300 km / h + = 4.4 103 km / h. M +m 4m + m
b gb
g
Chapter 9 Homework Problem Solutions
42. Our +x direction is east and +y direction is north. The linear momenta for the two m = 2.0 kg parts are then p1 = mv1 = mv1 j where v1 = 3.0 m/s, and p2 = mv2 = m v2 x i + v2 y j = mv2 cos i + sin j
e
j
e
j
where v2 = 5.0 m/s and = 30. The combined linear momentum of both parts is then
P = p1 + p2 = mv1 ^ + mv2 cos ^ + sin ^ = ( mv2 cos ) ^ + ( mv1 + mv2 sin ) ^ j i j i j = ( 2.0 kg )( 5.0 m/s )( cos 30 ) ^ + ( 2.0 kg ) ( 3.0 m/s + ( 5.0 m/s )( sin 30 ) ) ^ i j = 8.66 ^ + 11 ^ kg m/s. i j From conservation of linear momentum we know that this is also the linear momentum of the whole kit before it splits. Thus the speed of the 4.0-kg kit is
(
)
(
)
P v= = M
Px2 + Py2 M
=
( 8.66 kg m/s ) + (11 kg m/s )
2
2
4.0 kg
= 3.5 m/s.
Chapter 9 Homework Problem Solutions
50. (a) We choose +x along the initial direction of motion and apply momentum conservation: mbullet vi = mbullet v1 + mblock v2 (5.2 g) (672 m / s) = (5.2 g) (428 m / s) + (700 g)v2 which yields v2 = 1.81 m/s. (b) It is a consequence of momentum conservation that the velocity of the center of mass is unchanged by the collision. We choose to evaluate it before the collision:
vcom mbullet = vi (5.2 g) (672 m/s) = = 4.96 m/s. mbullet + mblock 5.2 g + 700 g
Chapter 9 Homework Problem Solutions
53. In solving this problem, our +x direction is to the right (so all velocities are positivevalued). (a) We apply momentum conservation to relate the situation just before the bullet strikes the second block to the situation where the bullet is embedded within the block. (0.0035 kg)v = (1.8035 kg)(1.4 m/s) v = 721 m/s. (b) We apply momentum conservation to relate the situation just before the bullet strikes the first block to the instant it has passed through it (having speed v found in part (a)). (0.0035 kg) v0 = (1.20 kg)(0.630 m/s) + (0.00350 kg)(721 m/s) which yields v0 = 937 m/s.
Chapter 9 Homework Problem Solutions
55. (a) Let v be the final velocity of the ball-gun system. Since the total momentum of the system is conserved mvi = (m + M)v. Therefore, v= mvi (60 g)(22 m/s) = = 4.4 m/s . 60 g + 240 g m+M
1 (b) The initial kinetic energy is Ki = 2 mvi2 and the final kinetic energy is
Kf =
1 2
bm + M gv
2
1 = 2 m2 vi2 m + M . The problem indicates E th = 0 , so the difference
b
g
Ki Kf must equal the energy Us stored in the spring: M m 1 1 2 1 m2 vi2 1 U s = mvi - = mvi2 1 - = mvi2 . m+ M m+ M 2 2 2 m+ M 2
b
g
FG H
IJ K
Consequently, the fraction of the initial kinetic energy that becomes stored in the spring is Us 240 M = = = 0.80 . K i m + M 60+240
Chapter 9 Homework Problem Solutions
59. As hinted in the problem statement, the velocity v of the system as a whole when the spring reaches the maximum compression xm satisfies m1v1i + m2v2i = (m1 + m2)v. The change in kinetic energy of the system is therefore (m v + m2 v2i ) 2 1 1 1 1 1 2 2 K = (m1 + m2 )v 2 - m1v12i - m2 v2i = 1 1i - m1v12i - m2 v2i 2 2 2 2(m1 + m2 ) 2 2 which yields K = 35 J. (Although it is not necessary to do so, still it is worth noting 2 1 m that algebraic manipulation of the above expression leads to K = 2 m11+m22 vrel where m
d i
vrel = v1 v2). Conservation of energy then requires
-2K -2(-35 J) 1 2 = 0.25 m. = kxm = -K xm = k 2 1120 N/m
Chapter 9 Homework Problem Solutions
60. (a) Let m1 be the mass of one sphere, v1i be its velocity before the collision, and v1f be its velocity after the collision. Let m2 be the mass of the other sphere, v2i be its velocity before the collision, and v2f be its velocity after the collision. Then, according to Eq. 9-75, m - m2 2m2 v2 i . v1i + v1 f = 1 m1 + m2 m1 + m2 Suppose sphere 1 is originally traveling in the positive direction and is at rest after the collision. Sphere 2 is originally traveling in the negative direction. Replace v1i with v, v2i with v, and v1f with zero to obtain 0 = m1 3m2. Thus, m2 = m1 / 3 = (300 g) / 3 = 100 g . (b) We use the velocities before the collision to compute the velocity of the center of mass: m v + m2 v2i ( 300 g ) ( 2.00 m s ) + (100 g ) ( - 2.00 m s ) vcom = 1 1i = = 1.00 m/s. m1 + m2 300 g + 100 g
Chapter 9 Homework Problem Solutions
71. We orient our +x axis along the initial direction of motion, and specify angles in the "standard" way -- so = +60 for the proton (1) which is assumed to scatter into the first quadrant and = 30 for the target proton (2) which scatters into the fourth quadrant (recall that the problem has told us that this is perpendicular to ). We apply the conservation of linear momentum to the x and y axes respectively.
m1v1 = m1v '1 cos + m2 v '2 cos 0 = m1v '1 sin + m2 v '2 sin
We are given v1 = 500 m/s, which provides us with two unknowns and two equations, which is sufficient for solving. Since m1 = m2 we can cancel the mass out of the equations entirely.
(a) Combining the above equations and solving for v2 we obtain
v2 =
v1 sin (500 m/s) sin(60) = = 433 m/s. sin ( - ) sin (90)
We used the identity sin cos cos sin = sin ( ) in simplifying our final expression. (b) In a similar manner, we find
v1 =
v1 sin (500 m/s) sin(-30) = = 250 m/s . sin ( - ) sin (-90)
Chapter 9 Homework Problem Solutions
92. One approach is to choose a moving coordinate system which travels the center of mass of the body, and another is to do a little extra algebra analyzing it in the original coordinate system (in which the speed of the m = 8.0 kg mass is v0 = 2 m/s, as given). Our solution is in terms of the latter approach since we are assuming that this is the approach most students would take. Conservation of linear momentum (along the direction of motion) requires mv0 = m1v1 + m2 v2 (8.0)(2.0) = (4.0)v1 + (4.0)v2
which leads to v2 = 4 - v1 in SI units (m/s). We require 1 1 2 1 2 K = m1v12 + m2 v2 - mv0 2 2 2 1 1 1 2 16 = (4.0)v12 + (4.0)v2 - (8.0) (2.0) 2 2 2 2
2 which simplifies to v2 = 16 - v12 in SI units. If we substitute for v2 from above, we find
(4 - v1 ) 2 = 16 - v12 which simplifies to 2v12 - 8v1 = 0 , and yields either v1 = 0 or v1 = 4 m/s. If v1 = 0 then v2 = 4 v1 = 4 m/s, and if v1 = 4 m/s then v2 = 0. (a) Since the forward part continues to move in the original direction of motion, the speed of the rear part must be zero. (b) The forward part has a velocity of 4.0 m/s along the original direction of motion.
Chapter 9 Homework Problem Solutions
120. (a) Since the center of mass of the man-balloon system does not move, the balloon will move downward with a certain speed u relative to the ground as the man climbs up the ladder. (b) The speed of the man relative to the ground is vg = v u. Thus, the speed of the center of mass of the system is vcom = This yields u= mv (80 kg)(2.5 m/s) = = 0.50 m/s. M + m 320 kg + 80 kg mv g - Mu M +m = m v - u - Mu = 0. M +m
b g
(c) Now that there is no relative motion within the system, the speed of both the balloon and the man is equal to vcom, which is zero. So the balloon will again be stationary.
Chapter 9 Homework Problem Solutions
129. Using Eq. 9-68 with m1 = 3.0 kg, v1i = 8.0 m/s and v2f = 6.0 m/s, then
v2 f = leads to m2 = M = 5.0 kg. 2v 2m1 v1i m2 = m1 1i - 1 v2 f m1 + m2