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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 8 / Problem 44E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: Combining Conservation Laws. A 15.0-kg block is

Chapter 8, Problem 44E

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QUESTION:

Combining Conservation Laws. A \(15.0-\mathrm{kg}\) block is attached to a very light horizontal spring of force constant \(500.0 \mathrm{~N} / \mathrm{m}\) and is resting on a frictionless horizontal table. (Fig. E8.44). Suddenly it is struck by a \(\text { 3.00-kg }\) stone traveling horizontally at \(8.00 \mathrm{~m} / \mathrm{s}\) to the right, whereupon the stone rebounds at \(2.00 \mathrm{~m} / \mathrm{s}\)horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

Equation Transcription:

$15.0-kg$

$500.0\ N/m$

$3.00-kg$

$8.00\ m/s$

$2.00\ m/s$

Text Transcription:

15.0-kg

500.0 N/m

3.00-kg

8.00 m/s

2.00 m/s

Questions & Answers

QUESTION:

Combining Conservation Laws. A \(15.0-\mathrm{kg}\) block is attached to a very light horizontal spring of force constant \(500.0 \mathrm{~N} / \mathrm{m}\) and is resting on a frictionless horizontal table. (Fig. E8.44). Suddenly it is struck by a \(\text { 3.00-kg }\) stone traveling horizontally at \(8.00 \mathrm{~m} / \mathrm{s}\) to the right, whereupon the stone rebounds at \(2.00 \mathrm{~m} / \mathrm{s}\)horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

Equation Transcription:

$15.0-kg$

$500.0\ N/m$

$3.00-kg$

$8.00\ m/s$

$2.00\ m/s$

Text Transcription:

15.0-kg

500.0 N/m

3.00-kg

8.00 m/s

2.00 m/s

ANSWER:

Step 1 of 5

Mass of stone = 3.0 kg

Mass of block = 15.0kg

Spring constant = 500.0 N/m

Initial velocity of stone = 8.00 m/s

Initial velocity of block = 0 m/s

After collision

Stone rebounds with velocity = -2.0m/s (opposite direction -x)

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