07 - linear momentum conservation and rotational kinematics: In which we present problems to help with th the 7 homework assignment. Chapter 9, problem 53 (|| 9.52): In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviation MVC. Suppose a 1000 kg car slides into a stationary 500 kg moose on a very slippery road, with the moose being thrown through the wind-shield (a common MVC result). (a) What percent of the original kinetic energy is lost in the collision to other forms of energy? energy? (b) Suppose you are now 1 in Saudi Arabia . What percent of the original kinetic k inetic energy is lost if the car hits a 300 kg camel? (c) Generally, does the percent loss increase or decrease if the animal mass decreases?
With an initial speed of with a moose of mass
vi
, the initial kinetic energy of the car is K i mc vi2 / 2 . After a totally inelastic collision
mm ,
by momentum conservation, the speed of the combined system is
mc vi (mc mm )v f
v f
mc vi mc mm
,
with final kinetic energy 2
mc vi 1 mc2 K f (mc mm )v ( mc mm ) vi2 . 2 2 mc mm 2 mc mm 1
2 f
1
(a) The percentage loss of kinetic energy due to collision is
K K Ki K f 1 f 1 Ki
Ki
Ki
mc mc mm
mm mc mm
500 kg 1000 kg 500 kg
1 3
33.3%.
(b) If the collision were with a camel of mass mcamel 300 kg, then the percentage loss of kinetic k inetic energy would be
K K i
mcamel mc mcamel
300 kg 100 1000 kg 300 kg
3 13
23%.
(c) As the animal mass decreases, the percentage loss of kinetic energy also decreases. Chapter 9, problem 59 (|| 9.58): In Fig. 9-63, block 1 (mass 2.0 kg) is moving rightward at 10 m/s and block 2 (mass 5.0 kg) is moving rightward at 3.0 m/s. The surface is frictionless, and a spring with a spring constant of 1120 N/m is fixed to block 2. 2 . When the blocks collide, the compression of the spring is maximum at the instant
the blocks have the same velocity.
Find the maximum compression.
As hinted in the problem statement, the velocity v of the system as a whole, when the spring reaches the maximum compression x compression xm, satisfies m1v1i + m2v2i = (m (m1 + m2)v. The change in kinetic energy en ergy of the system is therefore 1
A similar danger occurs in Saudi Arabia because of camel – vehicle vehicle collisions (CVC).
1
1
1
K (m1 m2 )v m v m v 2
2
2
2 1 1i
2
2 2 2i
(m1v1i m2v2i )2 2( m1 m2 )
1
1
2
2
m1v12i m2 v22i
which yields K = – 35 J. (Although it is not necessary to do so, still it is worth noting that algebraic manipulation of the above expression leads to K
1 2
d iv m1m2 m1 m2
2 rel
where vrel = v1 – v2). Conservation of energy
then requires
1 2
kxm2 K xm
2 K k
2(35 J) 1120 N/m
= 0.25 m.
Chapter 9, problem 62 (|| 9.63): Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is 300 g, remains at rest. (a) What is the mass of the other sphere? (b) What is the speed of the two-sphere center of mass if the initial speed of each sphere is 2.00 m/s?
(a) Let m1 be the mass of one sphere, v1i be its velocity before the collision, and v1 f be its velocity after the collision. Let m2 be the mass of the other sphere, v2i be its velocity before the collision, and v2 f be its velocity after the collision. Then, according to Eq. 9-75, m m2 2m2 v1 f 1 v1i v2 i . m1 m2 m1 m2 Suppose sphere 1 is originally traveling in the p ositive 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 v1 f 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 c ompute the velocity of the center of mass:
vcom
m1v1i m2v2i m1 m2
300 g 2.00 m s 100 g 2.00 m s 300 g 100 g
1.00 m/s.
Chapter 9, problem 64 (|| 9.68): In Fig. 9-67, block 1 of mass m1 slides from rest along a frictionless ramp from height h 2.50 m and then collides with stationary block 2, which h as 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?
First, we find the speed v of the ball of mass m1 right before the collision (just as it reaches its lowest point of swing). Mechanical energy conservation (with h = 0.700 m) leads to 1 m1 gh m1v 2 v 2 gh 3.7 m s. 2 (a) We now treat the elastic collision using Eq. 9 -67: v1 f
m1 m2 m1 m2
v
0.5 kg 2.5 kg 0.5 kg 2.5 kg
(3.7 m/s) 2.47 m/s
which means the final speed of the ball is 2.47 m s. (b) Finally, we use Eq. 9-68 to find the final speed of the block: v2 f
2m1 m1 m2
v
2(0.5 kg) 0.5 kg 2.5 kg
(3.7 m/s) 1.23 m/s.
Chapter 9, problem 75 (|| 9.74): A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 60° from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?
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 m1v1 cos m2v2 cos 0 m1v1 sin m2v2 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 v 2 we obtain v2
v1 sin sin ( )
(500 m/s)sin(60 ) sin (90)
433 m/s.
We used the identity sin cos – cos sin = sin ( – ) in simplifying our final expression. (b) In a similar manner, we find v1
v1 sin sin ( )
(500 m/s)sin( 30) sin (90)
250 m/s .
Chapter 10, problem 6 (|| 10.4): The angular position of a point on the rim of a rotating wheel is given by,
(t ) 4.0t 3.0t 2 t 3 ; (a) What is the angular velocity at t = 2.0 s? (b)What is the angular velocity at t = 4.0 s. If we make the units explicit, the function is
g c
c
. rad / s3 t 3 4.0 rad / s t 3.0 rad / s2 t 2 10 but generally we will proceed as shown in the problem — letting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures. (a) Equation 10-6 leads to
c4t 3t dt d
2
h
t 3 4 6t 3t 2 .
Evaluating this at t = 2 s yields 2 = 4.0 rad/s. (b) Evaluating the expression in part (a) at t = 4 s gives 4 = 28 rad/s. (c) What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s? 2 avg 4 12 rad / s2 . 42 (d) What is the instantaneous angular acceleration at the beginning of this time interval? (e) What is the instantaneous angular acceleration at the end of this time interval? d d 4 6t 3t 2 6 6t . dt dt
c
h
2
Evaluating this at t = 2 s produces 2 = 6.0 rad/s .
2
Chapter 10, problem 15 (10.16): A wheel has a constant angular acceleration of 3.0 rad/s . During a certain 4.0 s interval, it turns through an angle of 120 rad. Assuming that the wheel started from rest, how long has it been in motion at the start of this 4.0 s interval?
We have a wheel rotating with constant angular acceleration. We can apply the equations given in Table 10-1 to analyze the motion. Since the wheel starts from rest, its angular displacement as a function of time is given by 12 t 2 . We take
t 1
to be the start time of the interval so that t2 t 1 4 . 0 s . The corresponding angular displacements at these times are 1 Given 2 1 , we can solve for
t 1 ,
1 2
t12 , 2
1 2
t 22
which tells us how long the wheel has been in motion up to the
beginning of the 4.0 s-interval. The above expressions can be combined to give 1 1 2 1 t22 t12 (t2 t1)(t2 t1 ) 2 2 With 120 rad , 3.0 rad/s 2 , and t2 t 1 4 . 0 s , we obtain t2 t 1
2( )
(t2 t 1 )
2(120 rad) (3.0 rad/s2 )(4.0 s )
20 s ,
12. 0s and t 1 8 . 0 s . So, the wheel started from rest 8.0 s before the which can be further solved to give t 2
start of the described 4.0 s interval. Note: We can readily verify the results by calculating 1 and 1 2
1 2 1 2
t 12 t 22
1 2 1 2
2 explicitly:
(3.0 rad/s 2 )(8.0 s) 2 96 rad (3.0 rad/s2 )(12.0 s)2 216 rad.
Indeed the difference is 2 1 120 rad . Chapter 10, problem 32 (|| 10.29): A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. The pulsar in the Crab neb-ula has a period of rotation of T = 5 0.033 s that is increasing at the rate of 1.26 10 s / yr
T T
. (a) What is the pulsar’s angular acceleration
?
(b) If is constant, how many years from now will the pulsar stop rotating? (c) The pulsar originated in a supernova explosion seen in the year 1054. Assuming constant , find the initial T. (a) A complete revolution is an angular displacement of = 2 rad, so the angular velocity in rad/s is given by = /T = 2/T . The angular acceleration is given by
d dt
2 dT T 2 dt
.
For the pulsar described in the problem, we have
dT dt
126 . 105 s / y
4.00 1013 .
316 . 10 s / y 7
Therefore,
2 G J(4.00 10 . s) K H(0033
13
2
) 2.3 10 9 rad/s2 .
The negative sign indicates that the angular acceleration is opposite the angular velocity and the pulsar is slowing down. (b) We solve = 0 + t for the time t when = 0: t
0
2
T
2 9
(2.3 10
2
rad/s )(0.033 s)
8.3 1010 s 2.6103 years 7
10
(c) The pulsar was born 1992 – 1054 = 938 years ago. This is equivalent to (938 y)(3.16 10 s/y) = 2.96 10 s. Its angular velocity at that time was 0 t
2 T
t
2 0.033 s
(2.3 109 rad/s 2 )( 2.96 1010 s) 258 rad/s.
Its period was T
2
2 258 rad / s
2.4 10 2 s.
Chapter 10, problem 31 (|| 10.34): A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round
through 800 rad, starting from rest, gaining angu-lar speed at the constant rate 1 through the first 400 rad and then losing angular speed at the constant rate 2 1 until it is again at rest. The magnitude of the 2
centripetal acceleration of any por-tion of the disk is not to exceed 400 m/s . (a) What is the least time required for the rotation? (b) What is the corresponding value of 1 ? (a) The upper limit for centripetal acceleration (same as the radial acceleration – see Eq. 10-23) places an upper limit of the rate of spin (the angular velocity ) by considering a point at the rim (r = 0.25 m). Thus, max = a/r = 40 rad/s. Now we apply Eq. 10-15 to first half of the motion (where o = 0):
o =
1 ( o + 2
)t 400 rad =
1 (0 2
+ 40 rad/s)t
which leads to t = 20 s. The second half of the motion takes the same amount of time (the process is essentially the reverse of the first); the total time is therefore 40 s. (b) Considering the first half of the motion again, Eq. 10-11 leads to = o + t
=
40 rad/s 20 s
2
= 2.0 rad/s .
Chapter 10, problem 35 (|| 10.39): Two uniform solid cylinders, each rotating abou t its cen-tral (longitudinal) axis at 235 rad/s, have the same mass of 1.25 kg but differ in radius. What is the rotational kinetic energy of (a) the smaller cylinder, of radius 0.25 m, and (b) the larger cylinder, of radius 0.75 m?
Since the rotational inertia of a cylinder is I
1 2
2
MR (Table 10-2(c)), its rotational kinetic energy is
K
1 2
I 2
1 4
MR2 2 .
(a) For the smaller cylinder, we have K 1
1 4
(1.25 kg)(0.25 m)2 (235 rad/s)2 1.08103 J 1.1103 J.
(b) For the larger cylinder, we obtain K 2
1 4
(1.25 kg)(0.75 m)2 (235rad/s)2 9.71 103 J 9.7 103 J.
