HW06 - linear momentum: 9P4, 9P13, 9P22, 9P30, 9P46, 9P52, 9P58, 9P63, 9P68, 9P73. Chapter 9, Problem 4 SN: In the figure, three uniform thin rods, each of length L = 22 cm, form an inverted U. The vertical rods each have a mass
mv m 14 g 14 10 3 kg m = 14 g; the horizontal
rod has a mass mh 3m 3(14 g ) 42 g 3m = 3*(14g) = 42 g. What are (a) the x coordinate and (b) the y coordinate of the system's center of mass? Give your answer in terms of the variables given. We know that xCM
i ximi i mi
; in this formula, xi
xiCM ; e.g., it is as if all the individual rods rods are replaced by point-
masses whose center of mass is (obviously) located at the rod’s midpoint,
xCM
x1CM m1 x2CM m2 x3CM m 3 m1 m2 m3
yCM
0 m L2 3m L m m 3m m
y1CM m1 y2CM m2 y3CM m3 m1 m2 m3
L 2
0 L2 3 L 5
1
L
5
2
23 1 L
11cm ;
1 0 3 2L 1 L 22 cm; 1 3 1 5 5
(1.1)
(1.2)
Chapter 9, problem 13: A shell is shot with an initial velocity v0 of 20m/s, at an angle of 60 degrees with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. 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?
Need to find x and y coordinates at the top of the trajectory. The top of the trajectory occurs at time, v y 0 v0, y a y t v0, y gt t
v0, y g
v0 sin 0 g
;
(1.3)
Subsequently, the x and y coordinates at the top of the trajectory are, x v0, xt v0t cos 0
v0 2 g
sin 0 cos 0 17.7m; 2
v0, y v02 sin 2 0 2 1 1 2 g 15.3m; y v0, y t 2 ( g )t g g 2 g v0, y 2
(1.4)
(1.5)
Now, treat this position as the “new starting point” for new kinematics. Let M be the mass of the shell, and V0 be the velocity of the fragment, which has mass M/2. Then, this new velocity V 0 is,
p0, x px mi vi , x m f v f , x Mv0 cos 0 12 MV0 V0 2v0 cos 0 2(20 ms ) cos 60 20 ms ;
(1.6)
Now the problem is exactly the same as a projectile launched horizontally at time t = 0 with initial velocity 20 m/s as in (1.6), but with initial position r0 ( x0 , y0 ) (17.7,15.3) m from (1.4) and (1.5). Then,
y y0 12 gt 2 y (t ) y (tland ) 0 y0 12 gtland 2 tland 2y 0 / g ;
(1.7)
x-coordinate at this landing time is, x x0 V0tland x0 V0
2 y0 g
17.7 m 20 ms
2(15.3m) 9.81 sm2
53m ;
(1.8)
Cool Result: Notice that x 53m 3x0 3(17.7m) --that is, the total distance the particle travels is 3 times the st
distance the particle traveled on the 1 leg of the journey, before the boosting-explosion! Can you prove this? Chapter 9, problem 22 (|| 21): Figure 9-47 gives an overhead view of the path taken by a 0.165 kg cue ball as it bounces from a rail of a pool table.
The ball’s initial speed is 2.00 m/s, and the angle 1 is 30.0°. The bounce reverses the y-component of the ball’ s velocity but does not alter the xcomponent. What are (a) angle
2 and
(b) the change in the ball’s linear momentum in unit vector notation? (The fact that
the ball rolls is irrelevant to the problem.) (a) Since the force of impact on the ball is in the y direction, p x is conserved:
mvi sin 1 1 sin sin 1 1 ; (1.9) mvi
pi p f px ,i px , f mvi , x mv f , x mvi sin 1 mvi sin 2 2 sin 1
(b) The momentum change is,
p p f pi mvi cos 2 ( j) mvi cos 2 ( j) 2 0.165 kg 2.00 m/s cos 30 j (0.572 kg m/s)j ; (1.10) ˆ
ˆ
Chapter 9, problem 30 (|| 29) - Two average forces: A steady stream of 0.250 kg snowballs is shot perpendicularly into a wall at a speed of 4.00 m/s. Each ball sticks to the wall. Figure 9-49 gives the magnitude F of the force on the wall as a function of time t for two of the snowball impacts. Impacts occur with a repetition time interval tr = 50.0 ms, last a duration time interval td = 1 0 ms, and produce isosceles triangles on the graph, with each impact reaching a force maximum F_max =
ˆ
ˆ
200 N. During each impact, what are the magnitudes of (a) the impulse and (b) the a verage force on the wall? (c) During a time in-terval of many impacts, what is the magnitude of the average force on the wall?
(a) By Eq. 9-30, impulse can be determined from the “area” under the F (t ) curve. Keeping in mind that the area 1
of a triangle is 2 (base)(height), we find the impulse in this case is.
J 12 bh 12 td Fmax 12 (10 103 s )(200N ) 1.00 N s ;
(1.11)
(b) By definition (of the average of function, in the calculus sense) the average force must be the result of part (a) divided by the time (0.010 s), F
t d
1
td 0
F (t ) dt
0
1 1
td 2
bh
1
td
J
1 0.010s
1.00 N s 100 N ;
(1.12)
(c) Consider ten hits. Thinking of ten hits as 10 F (t ) triangles, our total time interval is 10(0.050 s) = 0.50 s, and the total area is 10(1.0 N s ). We thus obtain an average force of 10/0.50 = 20.0 N. One could consider 15 hits, 17 hits, and so on, and still arrive at this same answer. •• chapter 9, problem 46: A 4.0 kg mess kit sliding on a frictionless surface explodes into two 2.0 kg parts: 3.0 m/s, due north, and 5.0 m/s, 30° north of east. What is the original speed of the mess kit?
Original speed is given by vi vi , x 2 vi , y 2 , so we need to find these components of velocity. Momentum conservaton, after resolving v f ,1 v f ,1 j and v f ,2 v f ,2 (cos i sin j ) in which 30 , v f ,1 3.0 ms and ˆ
ˆ
ˆ
v f ,2 5.0 ms , tells us,
pi p f 2mvi m v f ,1 v f ,2 vi
1 2
v
f ,1
v f ,2
1
; sin
v f ,2 cos
2 v f ,1 v f ,2
(1.13)
In passing, we notice the components of the velocity we solved for are independent of thetotal mass of the Amazing Exploding Mess kit. Subsequently, we plug in numbers to compute a Pythagorean resultant of,
vi
1 2
v f ,2 cos
2
2
v f ,1 v f ,2 sin
1 2
2 2 5cos 30 3 5sin 30
m s
3.5
m s
;
(1.14)
Chapter 9, problem 52 (|| 9.53): 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?
We think of this as having two parts: the first is the collision itself – where the bullet passes through the block so quickly that the block has not had time to move through any distance yet – and then the subsequent “leap” of the block into the air (up to height h measured from its initial position).
The first part involves momentum conservation. The bullet ev idently “dumps” some momentum into the big block, which imparts to it a velocity of, m 0.01kg pi p f mbvb mbvb mBlv Bl v Bl b v b v b (1.15) 1000 ms 400 ms 1.2 ms ; m Bl 5.0kg
. m s . The second part involves either v y 2 v0, y 2 2a y y or K W , which have been which yields v 12
repeatedly shown to be equivalent. This yields, 1 2
mg y mv Bl 2
1 2
v
2 Bl
solve for y
g y y
vBl 2 2 g
Chapter 9, problem 58 (|| 9.59): In Fig. 9-62, block 2 (mass 1.0 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 200 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 2.0 kg), traveling at speed v1 = 4.0 m/s, collides with block
b vb
v mb m Bl
2 g
2
1.2m s
2 9.8m s
2
2
0.073m ; (1.16)
2, and the two blocks stick together. When the blocks momentarily stop, by what distance is th e spring compressed?
Knowing the compression of the spring is tantamount to knowing the initial kinetic energy of the system M=m1 1
+ m2 which, in turn, requires we know the initial velocity of M=m1+m2. Let the spring compress xS . Then, US
1 2
kxS xS 2
solve for xS
2 U S K U S k
2 k
K
K 12 Mv2
2 1
Mv
k 2
2
p Mv
M p k M
2 p p p m v i f 1 1
2
M m1v1 m1v1 ; k M kM
Plugging in numbers, xS
(2.0 1.0) kg (2.0kg )(4.0 sm ) 200 N m
2
64
(2.0 1.0) kg 600 m 0.327m 0.33m ;
(1.17)
Chapter 9, problem 63 (|| 9.62): Block 1 of mass m1 slides along a frictionless floor and into a onedimensional elastic collision with stationary block 2 of mass m2 = 3m1. Prior to the collision, the center of mass of the two-block system had a speed of 3.00 m/s. Afterward, what are the speeds of (a) the center of mass and (b) block 2? (a) Short answer: 3.00 m/s. Long (and much more informative) answer: the center of mass velocity 2 does not change in the absence of external forces, dp (1.18) F 0 pi p f ; dt
Notice the definition of the momentum of the center of mass allows us to say, using (1.18),
1
We think of this as having two parts: the first is (1) the collision itself, where the blocks “join” so quickly that the 1.0-kg block has not had time to move through any distance yet. During this part, momentum is conserved. (2) the subsequent motion of the 3.0 kg system as it compresses the spring to the maximum amount xm. During this part, we are interested in how much kinetic energy is dumped into the spring. 2 In this collision, only forces of one block on the other (both being part of the same system) are exerted, so the center of mass velocity is 3.00 m/s before and after the collision.
pCM mCM vCM mCM
d dt
xCM m1 m2
d m1 x1,i m2 x2,i m1 m2
dt
pi p f
m1v1,i m2v2,i p1,i p2,i ptot p1,f p2,f ; (1.19)
Subsequently, the center of mass velocity, in whic h we abbreviate M m1 m2 , is,
vCM ,i vCM , f vCM 3.00
m s
p f M
pi M
m1v1,i m1 3m1
1
M
4
m1
v1,i v1,i
vCM 4vCM 12
m s
;
(1.20)
(b) First, we use momentum conservation, and express the thing we’re looking for, v f ,2 , in terms of other quantities. Using (1.19) and subsequently (1.20), solve for v
2, f pi m1v1,i m2v2,i m1v1,i m1v1, f m 2v 2, f v 2, f
m1 m2
v
1,i
v 1, f
m1 m2
4v
CM
v 1, f ; (1.21)
The question becomes “what is v1, f ?”; this requires an additional piece of information. Indeed, since the collision is elastic, kinetic energy is conserved. The statement of this is, solve for v2, f 2 , then 2
2
use a b ( a b )( a b ) v 2, f 2 K i K f 12 m1v1,i 2 12 m1v1, f 2 12 m2v 2, f 2
m1 m2
(v1,i v1, f )(v1,i v 1, f ) v 2, f (v 1,i v 1, f ); (1.22)
Consequently, (1.22) tells us v2, f v1, i v1, f , which we use straight away to eliminate v1, f v2, f v1, i (in which v1,i 4vCM ) in (1.22), giving,
v2, f
m1 m2
4vCM v1, f
m1 m2
8vCM m12 m
v2, f 4vCM v2, f 4vCM solve for v2, f
1 m12 m
8vCM 13 1 13
2vCM ; (1.23)
Consistency check: Do the two velocities afterwards equal the center-of-mass momentum?
p f m1v1, f m2v2, f m(v2, f v1,i ) 3mv2, f m(2vCM 4vCM ) 3m 2vCM 2 6 mvCM 4mvCM ; (1.24) Indeed, they do, by definition (1.19) of the center of mass velocity. Chapter 9, problem 68 (|| 9.64): In Fig. 9-67, block 1 of mass m1slides from rest along a frictionless ramp from height h = 2.50 m and then collides with stationary block 2, which has mass m2 = 2.00m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction mu_k = 0.500 and comes to a stop in distance d within
that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic?
(a) If the collision is perfectly elastic, then Eq. 9-68 applies 2m1 2m1 2 v2 = v1i = 2 gh = 2 gh 3 m1+ m2 m1+ (2.00)m1 where we have used the fact (found most easily from energy conservation) that the speed of block 1 at the bottom of the frictionless ramp is 2 gh (where h = 2.50 m). Next, for block 2’s “rough slide” we use Eq. 8-37:
1 2 m v = E th = 2 2 2
f k d = k m2 g d
where k = 0.500. Solving for the sliding distance d , we find that m2 cancels out and we obtain d = 2.22 m. m1 (b) In a completely inelastic collision, we apply Eq. 9-53: v2 = v (where, as above, v1i = 2 gh ). Thus, m1+ m2 1i in this case we have v2 = 2 gh /3. Now, Eq. 8-37 (using the total mass since the blocks are now joined together) leads to a sliding distance of d 0.556 m (one-fourth of the part (a) answer). Chapter 9, problem 73: After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.
First of all, we notice something very important a bout the problem: the final velocity of the system is less than the initial velocity of the combined species, each of matching mass. The only wa y this can happen is if the angle st nd between them is greater than 90 , so we will note 90 . Let the 1 particle have a velocity, and the 2 particle have a velocity, each which respectively are,
; v1 v1i ; v2 v2 (cos i sin j ); v1 v2 v; 90 ˆ
ˆ
(1.25)
ˆ
Hence is the thing we are looking to compute. The final velocity of the two stuck together (i.e., necessary for a “completely” inelastic collision) is,
V V cos i sin j ; V 12 v1 12 v2 12 v; ˆ
ˆ
(1.26)
We need to eliminate somehow. Momentum conservation in two dimensions (containing 2 pieces of information, which is what we need for 2 un knowns , !), in which m1 m2 m and M m1 m2 2m , says, pi p f
mv mv cos MV cos
M 2 m V 12 v
1 cos cos
sin sin
mv sin MV sin
(1.27)
CAUTION: Remember that 90 , so 90 180 . Meanwhile, the arcsine-function is multivalued! Therefore, sin sin does not imply that equals , so let us write just to caution ourselves. We now substitute in 90 , and use trig-identities, and solve for which we subsequently add to 90 , 2nd equation implies
1 cos( 90) cos
sin( 90) sin
1 sin cos
cos 1 cos2 ,which in
turn implies 1-cos cos 1 sin cos 1 cos 2 cos sin (1.28) 2
1 1 cos2 1 cos 2 cos1 1
1 4
30;
1 Therefore, 30 120 . Notice, then, that cos (1 sin30) 60 , and so 60 30 , which
vindicates the caution we asserted before! Quiz question pool: 4, 13, 22, 30, 46, 52, 58.
