Holt Physics Problem 6B - Cobb Learning

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Holt Physics

Problem 6B FORCE AND MOMENTUM PROBLEM

A student with a mass of 55 kg rides a bicycle with a mass of 11 kg. A net force of 125 N to the east accelerates the bicycle and student during a time interval of 3.0 s. What is the final velocity of the bicycle and student? Assume the student and bicycle are initially at rest. SOLUTION

Given:

ms = 55 kg mb = 11 kg F = 125 N to the east ∆t = 3.0 s vi = 0 m/s

Unknown:

vf = ?

Use the impulse-momentum theorem to solve for vf. F∆t = ∆p = mvf − mvi F∆t + mv vf = i m m = ms + mb = 55 kg + 11 kg = 66 kg (125 N)(3.0 s) − (66 kg)(0 m/s) vf =  66 kg

Copyright © by Holt, Rinehart and Winston. All rights reserved.

(125 N)(3.0 s) vf =  66 kg vf = 5.7 m/s to the east

ADDITIONAL PRACTICE 1. A net force of 10.0 N to the right pushes a 3.0 kg book across a table. If the force acts on the book for 5.0 s, what is the book’s final velocity? Assume the book to be initially at rest. 2. A 60.0 g egg dropped from a window is caught by a student. If the student exerts a net force of −1.5 N over a period of 0.25 s to bring the egg to a stop, what is the egg’s initial speed? 3. A child riding a sled is pulled down a snowy hill by a force of 75 N. If the child and sled have a combined mass of 55 kg, what is their speed after 7.5 s? Assume the child and sled are initially at rest. 4. A billiard ball with a mass of 0.195 kg and a velocity of 0.850 m/s to the right is deflected by the cushioned edge of the billiard table. The cushion exerts a force of 3.50 N to the left for 0.0750 s. What is the ball’s final velocity? Problem 6B

Ch. 6-3

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5. A 5.00 g projectile has a velocity of 255 m/s to the right. What force is required to stop this projectile in 1.45 s? 6. The Pacific walrus has an average mass of 1.1 × 103 kg and can swim with a speed of about 9.7 m/s. Suppose a walrus starting from rest takes 19 s to reach a velocity of 9.7 m/s to the east. What net force acts upon the walrus? 7. With a mass of 3.000 × 103 kg, the Russian-made Zil-41047 is the most massive automobile to have been manufactured on a regular basis. Suppose one of these cars accelerates from rest to a velocity of 8.9 m/s to the right in 5.5 s. Calculate the net force acting on the Zil-41047. 8. How much time would it take for a 0.17 kg ice hockey puck to decrease its speed by 9.0 m/s if the coefficient of kinetic friction between the ice and the puck is 0.050? 9. A girl pulls a 12.0 kg wagon along by exerting a force of 15.0 N on the wagon’s handle, which makes an angle of 20.0° with the horizontal. Friction provides a force of 11.0 N in the opposite direction. How long does it take for the wagon, which is initially at rest, to reach a speed of 4.50 m/s?

Copyright © by Holt, Rinehart and Winston. All rights reserved.

10. The compressed-air device at Sandia Laboratories in Albuquerque, New Mexico, accelerates small metal disks to a speed of 15.8 km/s. Suppose the compressed air exerts a force of 12.0 N on a 0.20 g disk that is initially at rest. How long will it take the disk to reach its maximum speed?

Ch. 6-4

Holt Physics Problem Bank

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Givens

Solutions

8. v = 50.0 km/h p = 0.278 kg•m/s

0.278 kg•m/s p m =  =  v (50.0 km/h)(103 m/km)(1 h/3600 s) m = 2.00 × 10−2 kg = 20.0 g

9. vavg = 96 km/h to the southeast pavg = 4.8 × 104 kg•m/s to the southeast 10. v = 255 km/s 36

p = 8.62 × 10 kg•m/s

4.8 × 104 kg•m/s p p vg =  m =  = a (96 km/h)(103 m/km)(1 h/3600 s) v vavg m = 1.8 × 103 kg

p 8.62 × 1036 kg•m/s m =  =  = 3.38 × 1031 kg v 255 × 103 m/s

Additional Practice 6B ∆p = mvf − mvi = F∆t

∆t = 5.0 s

(10.0 N)(5.05) + (3.0 kg)(0 m/s) F∆t + mv vf =  i =  = 17 m/s 3.0 kg m

vi = 0 m/s

vf = 17 m/s to the right

m = 3.0 kg

2. m = 60.0 g F = −1.5 N

∆p = mvf − mvi = F∆t

∆t = 0.25 s

mvf − F∆t (60.0 × 10−3 kg)(0 m/s) − (−1.5 N)(0.25 s) (1.5 N)(0.25 s) vi =   =  =  −3 m 60.0 × 10−3 kg 60.0 × 10 kg

vf = 0 m/s

vi = 6.2 m/s

3. F = 75 N m = 55 kg

∆p = mvf − mvi = F∆t

∆t = 7.5 s

(75 N)(7.5 s) + (55 kg)(0 m/s) F∆t + mv vf = i =  55 kg m

vi = 0 m/s

vf = 1.0 × 101 m/s

4. m = 0.195 kg vi = 0.850 m/s to the right = +0.850 m/s F = 3.50 N to the left = −3.50 N ∆t = 0.0750 s

5. m = 5.00 g vi = 255 m/s to the right

∆p = mvf − mvi = F∆t (−3.50 N)(0.0750 s) + (0.195 kg)(0.850 m/s) F∆t + mv vf =  i =  0.195 kg m −0.262 kg•m/s + 0.166 kg•m/s −0.096 kg•m/s vf =  =  = −0.49 m/s 0.195 kg 0.195 kg vf = 0.49 m/s to the left ∆p = mvf − mvi = F∆t

vf = 0 m/s

mvf − mvi (5.00 × 10−3 kg)(0 m/s) − (5.00 × 10−3 kg)(255 m/s)  =  = −0.879 N F= 1.45 s ∆t

∆t = 1.45 s

F = 0.879 N to the left

V

V Ch. 6–2

Holt Physics Solution Manual

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1. F = 10.0 N to the right

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Givens 6. m = 1.1 × 103 kg vf = 9.7 m/s to the east vi = 0 m/s ∆t = 19 s

Solutions ∆p mvf − mvi  F =  =  ∆t ∆t (1.1 × 103 kg)(9.7 m/s) − (1.1 × 103 kg)(0 m/s) F =  = 560 N 19 s F = 560 N to the east

7. m = 3.00 × 103 kg vi = 0 m/s vf = 8.9 m/s to the right ∆t = 5.5 s

8. m = 0.17 kg ∆v = −9.0 m/s g = 9.81 m/s2 mk = 0.050

∆p mvf − mvi  F =  =  ∆t ∆t (3.00 × 103 kg)(8.9 m/s) − (3.00 × 103 kg)(0 m/s) (3.00 × 103 kg)(8.9 m/s) F =  =  5.5 s 5.5 s F = 4.9 × 103 N to the right F∆t = ∆p = m∆v F = Fk = −mgmk m∆v ∆v −9.0 m/s ∆t =  =  =  −mgmk −gmk −(9.81 m/s2)(0.050) ∆t = 18 s

9. m = 12.0 kg Fapplied = 15.0 N q = 20.0° Ffriction = 11.0 N

mvf − mvi (12.0 kg)(4.50 m/s) − (12.0 kg)(0 m/s) ∆p ∆t =  =  =  Fapplied(cos q) − Ffriction (15.0 N)(cos 20.0°) − 11.0 N F

vi = 0 m/s

5.40 kg•m/s 5.40 kg•m/s − 0 kg•m/s ∆t =  =  3.1 N 14.1 N − 11.0 N

vf = 4.50 m/s

∆t = 1.7 s

10. vf = 15.8 km/s Copyright © by Holt, Rinehart and Winston. All rights reserved.

F = Fapplied(cos q) − Ffriction

vi = 0 km/s F = 12.0 N m = 0.20 g

∆p mvf − mvi  ∆t =  =  F F (0.20 × 10−3 kg)(15.8 × 103 m/s) − (0.20 × 10−3 kg)(0 m/s) ∆t =  12.0 N (0.20 × 10−3 kg)(15.8 × 103 m/s) ∆t =  12.0 N ∆t = 0.26 s

Additional Practice 6C 1. vi = 382 km/h to the right vf = 0 km/h mc = 705 kg md = 65 kg ∆t = 12.0 s

∆p (mc + md)vf − (mc + md)vi F =  =  ∆t ∆t [(705 kg + 65 kg)(0 km/h) − (705 kg + 65 kg)(382 km/h)](103 m//km)(1 h/3600 s) F =  12.0 s −(7.70 × 102 kg)(382 km/h)(103 m/km)(1 h/3600 s) F =  = − 6.81 × 103 N 12.0 s F = 6.81 × 103 N to the left 1

V

1

∆x = 2(vi + vf)∆t = 2(382 km/h + 0 km/h)(103 m/km)(1 h/3600 s)(12.0 s) ∆x = 637 m to the right

Section Five—Problem Bank

V Ch. 6–3