 Phyllis Fleming Physics Physics 104
 Answers - Center of Mass, Momentum, 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, (a) J = pf - pi = (1500 kg)(8.0 m/s)(j - i) = 12 x 103 kg-m/s(j - i) (b) J = Fav t Fav = J/t = {12 x 103 kg-m/s(j - i)}/3.0 s =4 x 103 N(j - i) (c) 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 (a) 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 (b) 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 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.

9. In the figure above, b represents the center of mass of the barge and x 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) + (200 kg)(b)}/(75 + 200)kg        (Equation 1)

Finally,

XCM = {(75 kg)(8 m) + (200 kg)(b + x)}/(75 + 200)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) + (200 kg)(b) = (75 kg)(8 m) + (200kg)(b + x) (3) 75 kg(10 - 8)m = 200 kg x 150 m = 200 x x = 0.75 m

10.

Momentum is conserved:

 pi = pf pin = p1f + p2f

Since the final momentum of product 1, p1f, = -p2f,  the negative of the final momentum of product 2, the initial momentum pin of the nucleus must equal zero.

11. (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 = 0.8/0.6 Θ =530.

12. 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.

13.

Conservation of momentum:

 pi = pf (3 kg)(5 m/s) + (2 kg)(-10m/s) = (3 + 2)kg vcommon 15 kg-m/s - 20 kg-m/s = 5 kg vcommon vcommon = -1 m/s

After the collision, the blocks move to the left with a speed of 1 m/s.

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

(a) m2 = 1.0 kg, v1f = 0 and v2f = 2.0 m/s = v1f.
Example: cue ball colliding with another billiard ball.

(b) 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.
A basketball colliding with a bowling ball is an example of a relatively small mass colliding with a large mass.

(c) 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. A bowling ball colliding with a basketball is an example of a relatively large mass colliding with a small mass.

15. 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 = -22.60

16.  (pi)x = (pf)x  cos 370 = 0.80 and sin 370 = 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 = -820

17. 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)

So 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

From Eq. (3),

v1i = 2 (v1i/√2) cos Θ1
cos Θ1 = √2/2, and
Θ1 = 450
Θ2 = -450

18. (a) 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,

 Uaf + Kaf = Uai + Kai mA(9.8 m/s)(0.40 m) + 0 = 0 + 1/2 mAvfa2 vfa = [19.6(0.4)]1/2m/s = 2.8 m/s

(b) 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

(c) Work done by friction = (Ucf + Kcf ) - (Uci + Kci)
fd cos 1800 = (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.

19.

(a) 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 to the right) 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

(b) 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 20. pia = pfa 0.020 kg(vo1) = (0. 020 + 5.0)kg vfa vfa = (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

 21 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 #21, above, for parts (a), (b), (c) and (d). (a) Ubf + Kbf = Ubi + Kbi          0 + Kbf = (2.0kg)(10 m/s2)(0.80 m) + 0                Kbf = 16 J = 1/2(2.0 kg)vbf2. (b) vbf = 4.0 m/s (c) pi = pf 2.0 kg(4.0 m/s) = (2.0 + 3.0)kg vcf.  vcf = 1.6m/s. (d)                   Udf + Kdf = Udi + Kdi      5.0 kg(10 m/s2)h + 0 = 0 + 1/2(5.0 kg)(1.6m/s)2                                     h = 0.13 m

22. (a) 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.

(b) 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

 23 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. 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 September 25, 2002 September 29, 2002