Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN: 9780078807213
Author: Paul W. Zitzewitz
Publisher: Glencoe/McGraw-Hill
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Question 136 to 139
Transcribed Image Text:**Physics Problems on Momentum and Force**
**137.** What is the magnitude of the velocity of a 25-kilogram object that has a momentum of 100 kilogram meters per second?
Options:
(1) 0.25 m/s
(2) 2500 m/s
(3) 40. m/s
(4) 4.0 m/s
**138.** What is the momentum of a 1,200-kilogram car traveling at 15 meters per second due east?
Options:
(1) 80. kg·m/s due east
(2) 80. kg·m/s due west
(3) 1.8 x 10^4 kg·m/s due east
(4) 1.8 x 10^4 kg·m/s due west
**139.** A constant unbalanced force acts on an object for 3.0 seconds, producing an impulse of 6.0 newton-seconds east. Determine the magnitude and direction of the force.
**140.** A 10-newton force acts on an object for 0.010 second. What force, acting on the object for 0.050 second, would produce the same impulse?
Options:
(1) 1.0 N
(2) 2.0 N
(3) 5.0 N
(4) 10. N
**141.** What is the magnitude of the net force acting on a 2.0 x 10^3-kilogram car as it accelerates from rest to a speed of 15 meters per second in 5.0 seconds?
Options:
(1) 6.0 x 10^3 N
(2) 2.0 x 10^4 N
(3) 3.0 x 10^4 N
(4) 6.0 x 10^4 N
**142.** A 5.00-kilogram mass is traveling at 100. meters per second. Determine the speed of the mass after receiving an impulse of 50.0 N·s.
![The image contains a physics problem followed by a review question:
---
**Problem Solution:**
\[ v_{Bf} = \frac{(1.0 \, \text{kg})(0.0 \, \text{m/s}) + (0.20 \, \text{kg})(10.0 \, \text{m/s}) - (1.0 \, \text{kg})(3.0 \, \text{m/s})}{0.20 \, \text{kg}} \]
\[ v_{Bf} = -5.0 \, \text{m/s} \]
The velocity of cart B after the collision is 5.0 m/s to the left.
---
**Review Questions:**
136. As an object falls freely toward Earth, the object's momentum:
- (1) decreases
- (2) increases
- (3) remains the same](https://content.bartleby.com/qna-images/question/04561994-c58b-432c-94cc-dc98f116805a/8fdf9705-6449-45ba-b5b6-cb98e869b1c2/jfeegce.jpeg)
Transcribed Image Text:The image contains a physics problem followed by a review question:
---
**Problem Solution:**
\[ v_{Bf} = \frac{(1.0 \, \text{kg})(0.0 \, \text{m/s}) + (0.20 \, \text{kg})(10.0 \, \text{m/s}) - (1.0 \, \text{kg})(3.0 \, \text{m/s})}{0.20 \, \text{kg}} \]
\[ v_{Bf} = -5.0 \, \text{m/s} \]
The velocity of cart B after the collision is 5.0 m/s to the left.
---
**Review Questions:**
136. As an object falls freely toward Earth, the object's momentum:
- (1) decreases
- (2) increases
- (3) remains the same
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