 Phyllis Fleming Physics Physics 107
 Answers - Center of Mass, Momentum, and Collisions
 1 Impulse J = pf - pi = mvf - mvi. Taking the return direction as positive, J = 0.060 kg{(40 - (-50)}m/s = 5.4 kg-m/s.
 2 XCM = (m1x1 + m2x2 + m3x3 + m4x4)/( m1 + m2 + m3 + m4) = {(1.0 kg)(0) + (4.0 kg)(2.0 m) + (2.0 kg)(0) + (3.0 kg)(2.0 m)}/(10.0 kg) = 1.4 m. YCM = (m1y1 + m2y2 + m3y3 + m4y4)/( m1 + m2 + m3 + m4) = {(1.0 kg)(0) + (4.0 kg)(0) + (2.0 kg)(2.0 m) + (3.0 kg)(2.0 m)}/(10.0 kg) 1.0 m.
 3 Taking east in the +i direction and north in the +j direction, J = pf - pi = (1500 kg)(8.0 m/s)(j - i) = 12 x 103 kg-m/s(j - i). J = Fav t Fav = J/t = {12 x 103 kg-m/s(j - i)}/3.0 s = 4 x 103 N(j - i). By Newton's third law of motion, the average force of the car on the road = -4 x 103 N(j - i).
 4 Divide the system into three pairs of particles, consisting of 1 + 2, 3 + 4, and 5 + 6. The midway for 1 + 2 is at a and the midpoint for 5 + 6 is c. We can treat 1 + 2 as a point particle of mass of 2m at a and 5 + 6 as a point particle of mass 2m at c. We can combine these two point particles to be another point particle with mass 4m at b.  3 + 4 have their midpoint at b so we can replace them with a point particle of mass 2m at b. Finally the whole works acts like a point particle of mass 6m at b.
 5 pA = (1.0 kg)(4.0 m/s) = 4.0 kg-m/s. pB = (2.0 kg)(2.0 m/s) = 4.0 kg-m/s. KA = 1/2(1.0 kg)(4.0 m/s)2 = 8.0 J. KB = 1/2(2.0 kg)(2.0 m/s)2 = 4.0 J. While the momenta are equal, the kinetic energies are not.
 6 Take the distance to the edge of the rink as x. Then the velocity v1 after the collision for the skater of mass m1 equals +x/12 s and the velocity v2 of the other skater equals -x/18 s. From conservation of momentum, pi = pf 0 = (50 kg)(x/12 s) + (m2)(-x/18 s)  or (m2)(x/18 s) = (50 kg)(x/12 s) m2 = (50 kg)((18/12) = 75 kg
 7 Take to the right to be positive. From conservation of momentum,                                                                 pi = pf 0.050 kg(0.20 m/s) + (0.10 kg)((-0.04 m/s) = (0.05 + 0.10)kg vf                                 (0.010 - 0.004)kg-m/s = (0.15 kg)vf                                                                vf = 0.040 m/s to the right
 8 For Fig. 3a: (a) The area of the triangle is 1/2 (base)(altitude) = ba/2. (b) The mass per unit area σ = mass/area = M/(ba/2) = 2M/ba. (c) The area of the element dA = (y)dx. (d) Since σ = mass/area, the mass of any element equals σ times the area of the element.  dm = (σ)dA = (2M/ba)ydx. (e) dm x = (2M/ba)ydx (x) (f) In general, the equation of a line is y = mx + b, where m is the slope of the line and b is the intercept on the y-axis.  In Fig. 3, we see b = 0 and m = a/b, so y = (a/b)x. (g) From the equation of the line found in (f) we substitute y into dm x = (2M/ba) y dx x, or dm x = (2M/ba) (a/b)x dx x = 2M/b2 x2 dx, and (h) For Fig. 3b: (a) and (b) The solutions to both (a) and (b) are the same for both Fig. 3a and Fig. 3b. (c ) dA = (b - x)dy (d) dm = (2M/ba) (b - x)dy (e) and (f) The solutions to both (e) and (f) are the same for both Fig. 3a and Fig. 3b. (g) from the equation of the line y = (a/b) x,  x = by/a (h) y dm = y (2M/ba)(b - by/a)dy = (2M/ba) (b/a)(a - y) y dy = (2M/a2)(a - y) y dy YCM = ∫ dm y/M = ∫(2M/a2) (ay - y2) dy /M = 2/a2 |(ay2/2 - y3/3)| taking the integral from 0 to a, YCM = (2/a2) (a3/2 - a3/3) = (2/a2) (3a3/6 - 2a3/6) = a/3
 9 The center of mass of the four particles at the base of the pyramid is the center of the base. The center of mass of the five-particle system lies on the line perpendicular to the base and passes through the peak of the pyramid. This line passes through the center of mass of the four particles at the base and gives the x-coordinate of the center of mass at the center of the base. YCM = {(m)(h) + (4m)(0)}/5m = h/5.
 10 In the figure above, b represents the center of mass of the barge and Δb the distance the barge moves to the right as the man moves to the left. No external forces act on the barge-man system and the center of mass CM of this system remains at rest. Initially,  XCM = {(75 kg)(10 m) + (225 kg)(b)}/(75 + 225)kg      (Equation 1) Finally,  XCM = {(75 kg)x + (225 kg)(b + Δb)}/(75 + 225)kg      (Equation 2) Since the CM of the system remains at rest, equate Eq. 1 and Eq. 2 and cancel denominators, (75 kg)(10 m) + (225 kg)(b) = (75 kg)x + (225 kg)(b + Δb)      (Equation 3) Distance moved by man relative to shore = (10 m - x) = 2.0 m - Δb or   Δb = x - 8.0 m. Substituting this expression for Δb into Eq. 3: 750 N + (225 kg)(b) = (75 kg)x + (225 kg)(b + x - 8.0 m) 750 N + 1800 N = (300 kg)x  and  x = 8.5 m.
 11 Momentum is conserved:   pi = pf pin = p1f + p2f Since the final momentum of product 1,  p1f, = - p2f,  the negative of the final momentumof product 2,  the initial momentum pin of the nucleus must equal zero.
 12 (pi)x = (pf)x (3.0 kg)(1.0 m/s) = (5.0 kg)(vf)x     (vf)x = 0.60 m/s                     (pi)y = (pf)y (2.0 kg)(2.0 m/s) = (5.0 kg)(vf)y     (vf)y = 0.80 m/s vf = {(0.60)2 + (0.80)2}1/2 = 1.0 m/s.  tan Θ = 0.80/0.60.  Θ = 53o.

13. pi = pf mvo = mvo Ki = Kf 1/2 mvo2 = 1/2 mvo2 pi = pf mvo = mvo/2 + mvo/2 Ki ≠ Kf 1/2 mvo2 ≠ 1/2 m(vo/2)2 + 1/2 m(vo/2)2

Another conservation law is at work here. In an elastic collision, kinetic energy is also conserved. For an elastic collision, (b) cannot occur.

 14 For an elastic collision,        pi = pf m1v1i = m1v1f + m2v2f       (Equation 1) where I have assumed both final velocities to the right. In any calculation, if one of them is to the left, you will get a minus answer in your solution.                Ki = Kf 1/2 m1v1i2 = 1/2 m1v1f2 + 1/2 m2v2f2            (Equation 2) Solving Eq. 1 and Eq. 2 simultaneously for v1f  and  v2f  gives: v1f = {(m1 - m2)/(m1 + m2)} v1i  and v2f = {2m1/(m1 + m2)} v1i For v1i = 2.0 m/s,  m1 = 1.0 kg  and m2 = 1.0 kg,  v1f = 0  and  v2f = 2.0 m/s = v1f Example: Cue ball colliding with another billiard ball. m2 = 1000 kg,  v1f = - (999/1001)(2.0 m/s) ≈ (1)(2.0 m/s) = - v1i v2f = (2.0/1001)(2.0 m/s) ≈ 0.004 m/s, which is small compared to v1i Example: A basketball colliding with a bowling ball is an example of a relatively small mass colliding with a large mass. m2 = 0.001 kg,  v1f = (0.999)(2.0 m/s) ≈ (1)(2.0 m/s) = v1i v2f = (2.0/1.001)(1.0 m/s) ≈ 2 v1i Example: A bowling ball colliding with a basketball is an example of a relatively large mass colliding with a small mass.

15. Momentum is conserved, pi = pf (4.0 kg)(2.0 m/s) + (1.0 kg)(-3.0 m/s) = (4.0 kg)v1f + (1.0 kg)v2f 5.0 m/s = 4.0v1f + 1.0v2f or v2f = 5.0 m/s - 4.0v1f  (Equation 1) Kinetic energy is conserved, Ki = Kf 1/2(4.0kg)(4.0m2/s2)+1/2(1.0kg)(9.0m2/s2) = 1/2(4.0kg)v1f2+1/2(1.0kg)v2f2 1/2 (25)(m/s)2 = 1/2(4.0v1f2+ 1.0v2f2) or (25)(m/s)2 = 4.0v1f2 + 1.0v2f2  (Equation 2)

Substituting Eq. 1 into Eq. 2:
25 (m/s)2 = 4.0v1f2 + (5.0 m/s - 4.0v1f)2
25 (m/s)2 = 4.0v1f2 + 25 (m/s)2 - 40v1f m/s + 16v1f2
0 = 20v1f2 - 40v1f m/s  or
0 = v1f(v1f - 2 m/s)
Solutions are v1f = 0 and 2.0 m/s.  To decide which to choose, use Eq. 1.
For v1f = 0,  v2f = 5.0 m/s.  For v1f = 2.0 m/s,  v2f = - 3.0 m/s. The latter is not possible because they would collide. The correct solution is shown in Fig. for 15a.

 16 The velocity of the center of mass V = (m1v1i + m2v2i)/(m1 + m2) = {4.0(2.0)+1.0(-3.0)}kg-m/s/5.0 kg = 1.0 m/s. In the center of mass coordinate system moving with V = 1.0 m/s to the right, v1i’ = v1i - V = 2.0 m/s - 1.0 m/s = 1.0 m/s. v2i = v2i - V = -3.0 m/s - 1.0 m/s = -4.0 m/s. v1f’ = v1f - V = - 1.0 m/s = -1.0 m/s. v2f’ = v2f - V = 5.0 m/s - 1.0 m/s = 4.0 m/s. Prime velocities are shown in Fig. 16b above. In center of mass system, pi = (4.0 kg)(1.0 m/s) + (1.0 kg)(-4.0 m/s) = 0 = pf = (4.0 kg)(-1.0 m/s) + (1.0 kg)(4.0 m/s).  The center of mass system is a zero momentum system.
 17 If momentum, a vector quantity, is conserved, its components are conserved. (pi)x = (pf)x 10 kg(24 m/s) = (20 kg)(v2f)x (v2f)x = 12 m/s (pi)y = (pf)y 0=10 kg(10 m/s)+20 kg (v2f)y (v2f)y = -5.0 m/s v2f = {(v2f)x2 + (v2f)y2}1/2 v2f = (122 + -5.02)1/2 m/s = 13 m/s. tan Θ= -5/12.  Θ = -22.6o.

18.  (pi)x = (pf)x    cos 37o = 0.80  and  sin 37o = 0.60
2.0 kg(39 m/s) = 2.0 kg(v1f)x + 8.0 kg(10 m/s)(0.80)
(78 - 64)m/s = 2.0(v1f)x
(v1f)x = 7.0 m/s

(pi)y = (pf)y
0 = 2.0 kg(v1f)y + 8.0 kg(10 m/s)(0.6)
- 24 m/s = (v1f)y
v1f = {(7.0)2 + (-24)2}1/2 m/s = 25 m/s
tan Θ = - 24/7.
Θ = - 74o.

 19 Momentum is conserved (px)i = (px)f mv1i = m(v1f)x + m(v2f)x          (Equation 1) (py)i = (py)f 0 = mv1f sin Θ1 + mv2f sin Θ2.   Since v1f = v2f, Θ1 = - Θ2  and  (v1f)x = (v2f)x         (Equation 2) From Eq. 1, v1i = 2 (v1f)x = 2v1f cos Θ1     (Equation 3) From conservation of kinetic energy,  Ki = Kf 1/2 m v1i2 = 1/2 m v1f2 + 1/2 m v2f2 = mv1f2  or  v1f = v2f = v1i/(2)1/2 From Eq. (3),  v1i = 2 [v1i/(2)1/2] cos Θ1;  cos Θ1 = (2)1/2/2  and Θ1 =45o   and   Θ2 = - 45o
 20 Take the zero of gravitational potential energy at the bottom of the ramp, Uaf = 0.  At the top of the ramp, Uai = mgh, where h = 0.40 m.  On the ramp, from conservation of energy               Uai                 + Kai = Uaf +     Kaf mA(9.8 m/s2)(0.40 m)  +  0   =   0  + 1/2 mAvfa2              vfa = [19.6(0.4)]1/2m/s = 2.8 m/s When A collides with B, momentum is conserved.     pbi = pbf mAvib = (mA + mB)vfb vfb = [mA/(mA + mB)]vib = [2.0/(11.2)]2.8 m/s = 0.50 m/s Work done by friction = (Ucf + Kcf ) - (Uci + Kci) fd cos 180o = -(2.8 N)d = (0 + 0) - [0 - 1/2(11.2 kg)(0.50 m/s)2] -(2.8 N)d = -1.4 N-m  and  d = 0.50 m.
 21 For an elastic collision in one dimension, v1f = (m1 - m2)/(m1 + m2) v1i  and v2f = 2m1/(m1 + m2) v1i In this problem,  m1 = m2  and  v1i = vo. Thus the first block stops after the collision,  v1f = 0,  and the second block gains a velocity v2f = vo (Fig. 6a' above). Now the second block swings up to height h.  Energy is conserved. Taking the middle of the block as the zero of potential energy at the initial position i,  the potential energy at "final" position of height h, where the block comes momentarily to rest is mgh. Ui +       Ki       =  Uf   + Kf  0 + 1/2 mvo2 = mgh + 0                 and    h = vo2/2g For the inelastic collision, pi = pf mvo = 2mvb vb = vo/2 Uib +          Kib           =   Ufb   + Kfb   0  + 1/2(2m)(vo2/4) = 2mgh +  0     and   hb = vo2/8g
 22 pia          =         pfa 0.020 kg(vo1) = (0. 020 + 5.0)kg vaf                  vaf = (0.02/5.02)vo1 It takes 0.80 N to keep the 5.0 kg moving with constant velocity. The increase in the normal force due to the weight of the bullet does not change the frictional force to two significant figures. Take the gravitational potential at the surface = 0 = Ufb = Uib. Work by friction = (Ufb + Kfb) - (Uib + Kib) -(0.80 N)(1.5 m) =   (0 + 0)     - {0   + 1/2(0.502 kg)[0.02vo1/5.02]2}   -1.2 kg(m/s)2  = -3.98 x 10-6 vo12 kg vo1 = 549 m/s
 23 Take the potential energy at the bottom of swing to be zero. The kinetic energy at the top of the swing is zero since the velocity of A there vbi = 0. See the figure for #23 above for parts (a), (b), (c) and (d). Ubf + Kbf = Ubi + Kbi 0 + Kbf = (2.0kg)(10 m/s2)(0.80 m) + 0 Kbf = 16 J = 1/2(2.0 kg)vbf2 vbf = 4.0 m/s pi = pf 2.0 kg(4.0 m/s) = (2.0 + 3.0)kg vcf vcf = 1.6 m/s Udf + Kdf = Udi + Kdi 5.0 kg(10 m/s2)h + 0 = 0 + 1/2(5.0 kg)(1.6 m/s)2 h = 0.13 m
 24 Parts (a) and (b) above have the same approach as #23 above. The figures for (a) and (b) are the same as for #23 above. Parts (a) and (b) above have the same approach as #23 above. The figures for (a) and (b) are the same as for #23 above. For the elastic collision, vBf = 2mA/(mA + mB)(4.0 m/s)= (4/5)(4.0 m/s) = 3.2 m/s. vAf = (mA - mB)/(mA + mB)(4.0 m/s) = (-1.0/5)(4.0 m/s) = -0.8 m/s. Udf + Kdf = Udi + Kdi For B, (3.0 kg)(10 m/s2)hB = 1/2(3.0 kg)(3.2 m/s)2.  hB = 0.51 m. For A, (2.0 kg)(10 m/s2)hA = 1/2(2.0 kg)(-0.8 m/s)2.  hA = 0.032 m.
 25 Do the problem in steps. First there is the collision between the bullet of mass m = 5.0 x 10-3 kg travelling with a speed of 400 m/s and the block of mass M = 1.0 kg.  After the collision, the bullet has a velocity v and the block a velocity vM.  From conservation of momentum,                                     pi = pf (5.0 x 10-3 kg)(400 m/s) = (1.0 kg)vM + (5.0 x 10-3 kg)v   (Equation 1) Just after the collision and just before the block compresses the spring, the initial potential energy of the block-spring system is 0 and the kinetic energy of the block is 1/2 (1.0 kg) vM2.  When the spring has been compressed to its maximum, the block momentarily comes to rest and its "final" kinetic energy is zero, but the potential energy of the block-spring system is 1/2 kx2 = 1/2(900 N/m)((5 x 10-2m)2.  From conservation of energy, Ui +           Ki            =                  Uf                         + Kf 0 + 1/2(1.0 kg) vM2 = 1/2(900 N/m)(5.0 x 10-2 m)2 + 0  or vM = 1.5 m/s Substituting this value of vM back into Eq. 1: (5.0 x 10-3 kg)(400m/s) = (1.0kg)(1.5m/s) + (5.0 x10-3 kg)v or  v = 100 m/s For the collision, Ki = 1/2 (5 x 10-3 kg)(400 m/s)2 = 400 J Kf = 1/2(5.0 x 10-3 kg)(100 m/s)2 + 1/2(1.0 kg)(1.5 m/s)2     = (25 + 1.12)J K lost = (400 - 26.12)J = 374 J
 26 This problem is a reminder to choose your system carefully. When you drop an object, its momentum does not stay constant because the gravitational attraction of the earth, an external force, acts on it. If you take your system to be the object and the earth, then momentum is conserved because the force of the earth on the object and the force of trhe object on the earth are internal forces. There is no net external force on the system of the object and the earth and momentum is conserved. Let the object be a textbook with mass mb = 2.0 kg and velocity vb. The earth has a mass Me = 5.98 x 1024 kg. Before you release the book, the momentum of the system is zero. pi = pf 0 = mbvb + Meve,  where ve = the speed acquired by the earth. ve = (mb/Me)vb = (2.0 kg/5.98 x 1024 kg)vb = 3.3 x 10-25 vb. So the earth does move up, but we do not detect the motion.

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 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming October 8, 2002 April 17, 2003