Ch 6 Momentum & Collisions

Updated 5/27/07

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6.1 Momentum & Impulse
Practice A, pg 199
     1] 2500 kg●m/s South    
     2a] 120 kg●m/s NW        [b] 94 kg●m/s NW   [c] 27 kg●m/s NW
     3] 46 m/s East

Ch 6 Review Q, pg 223:  11a] 8.35 x 10^-21 kg●m/s UP 
     [b] 4.88 kg●m/s right  [c] 750 kg●m/s SW 
     [d] 1.78x 10²⁹ kg●m/s forware

Practice B, pg 201
     1] 380 N left             [2] 1100 N up      [3] 16 kg●m/s South 
     4a] 9.0 m/s right         [b] 15 m/s left

Ch 6 Review Q, pg 223, 12] 160 N right     13] 18 N

Practice C, pg 203
     1a] 5.33 s,               [b]53.3 West 
     2a] 14 m/s North          [b] 42 m North     [c] 8.0 s
     3a] 12,200 N East         [b] 53.3 m West

Ch 6 Review Q, pg 223, 14] 0.010 s,  0.13 m

Section 6.1 Review, pg 204,
     1a] 31.0 m/s                [2] Bullet has greater KE    
     3a] 2.5 kg●m/s downfield    [b] 130 N downfield

6.2 Conservation of Momentum
Practice D, pg 209
     1] 1.90 m/s     [2] 1.66 m/s West     [3a] 12.0 m/s [b] 9.6 m/s 
     4] 38 kg
     Ch 6 Review Q, pg 224, 22a] 2.43 m/s forward  [b] 0.0797 m/s
     23] 0.037 m/s South

Section 6.2 Review, pg 211,
     1a] 7.0 m/s     c] 1.5 m/s right       3] 61 m/s

6.3 Elastic & Inelastic Collisions
Practice E, pg 214
     1] 3.8 m/s South        2] 1.8 m/s     3] 4.25 m/s North
     4] 4.2 m/s right       5a] 3.0 kg      b] 5.32 m/s
Ch 6 Review Q, pg 225,      28] 1 m/s      29] 3.00 m/s

Practice F, pg 216
     1a] 0.43 m/s west     [b] 17 J   2a] 6.2 m/s South     b] 3 J
     3] 4.6 m/s South      [b] 3900 J
Ch 6 Review Q, pg 225, 30a] 1.80 m/s  [b] 21600 J  
     31a] 0.81 m/s East    [b] 1400 J

Practice G, pg 219
     1a] 22.5 cm/s right       [b] KEi = 0.00062 J = KEf
     2a] V1f = 14.1 m/s right  [b] KEi = 3040 J = KEf    
     3a] V2f = 8.0 m/s right   [b] KEi = 130 J = KEf
     4a] V2i = 2.0 m/s right   [b] KEi = 382 J = KEf
     Ch 6 Review Q, pg 225, 32a] V2f = 12 cm/s  [b] 0.00011 J  
     33] 4.0 m/s     34] 17.2 cm/s right

Section 6.2 Review, pg 220,
     2a] Vf = 1.1 m/s South    [b] KE decreases by 1400 J
     3a] V1f = 3.5 m/s         [b] 0 J     [c] 0 J

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Mixed Review, pg 225, * = requires information from previous chapters.

35] 42.0 m/s         
36*] V = 10 m/s m 0 3 kg    
37a*] 0.0 kg●m/s       [b] 1.1 kg●m/s Up  
38] 6.00 kg 
39*] 23 m/s
40] 29 s
41] 400 N
42] 14.5 m/s North 
43*] 2.36 cm 
44] 254 s
45*] 0.413
46a] 0.83 m/s right    [b] 1.2 m/s left
47] 22 cm/s
48] 9.62 10^-24 m/s
49a*] 9.9 m/s down     [b] 1800 N up
50] 1.3 10⁷ m/s, 41^o below negative X axis

Ch 6 Standardized Test Prep, pg 228

     1] A  [2] J  [3] C  [4] G  [5] D  [6] G  [7] B  [8] J [9] C  [10] G 11] Yes      
[13] 340 m/s        [14] 1.5 J

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Solution Mixed Review Q 37, pg 225

37   A 0.1 kg ball of dough is thrown straight up into the air with an
initial speed of 15 m/s.
a] Find the momentum of the ball of dough at its maximum height.
   At its maximum height V = 0 ==> mV = 0
b] Given:  Vi = 15 m/s, Vf = 0
   ΔY = 1/2(Vf + Vi)t,
               2

   g = Vf - Vi ---> t = Vf - Vi  Substitute this for t above
          t                g

ΔY = 1/2(Vf² - Vi²)
            2g

ΔY = 1/2(-15²) = 5.73 m
      2(-9.81)

Use this to find Vf at 5.73 m
ΔY = (Vf² - Vi²)
         2g          

5.73 = (Vf² - 15²)
        2(9.81)g

Multiply both sides by 2(-9.81)

-112.5 = Vf² - 15² -112.5 = Vf² - 225
+225.0      + 225
112.5 = Vf²                      
Vf = √112.5 = 10.6 m/s

At 1/2 maximum height
P = mVf = 0.1*10.6 = 1.06 m/s--Round to 2 Sig. Fig--> 1.1 m/s

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