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Physics For Scientists & Engineers
Found in: Page 281
Physics For Scientists & Engineers

Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

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Short Answer

A 10.0 −gbullet is fired into a 200 −gblock of wood at rest on a horizontal surface. After impact, the block slides8.0 m before coming to rest. If the coefficient of friction between the block and the surface is,0.400 what is the speed of the bullet before impact?

(a) 106 m/s

(b)166 m/s

(c)226 m/s

(d)286 m/s

(e) none of those answers is correct

The speed of the bullet before impact is, vb=166 m/smakingchoice (b) the correct answer.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Concept

Let us assume that the masses and initial velocities of both particles are known. Then, we can solve the below equation to calculate the final velocities in terms of the initial velocities because there are two equations and two unknowns:

v1f=m1−m2m1+m2v1i+2m2m1+m2v2iv2f=2m1m1+m2v1i+m1−m2m1+m2v2i

Here,

m1=Mass of the first particle.

m2=Mass of the second particle

v1i= Initial velocity of the first particle

v2i=Initial velocity of the second particle

v1f=Final velocity of the first particle

v2f=Final velocity of the second particle

Step 2: Speed of bullet before impact

We find the velocity of mass-1 using the equation derived for an elastic collision.

Before the collision, the bullet with massm1=10.0 ghas speed, v1i=vband the block with mass m2=200 ghas speed v2i=0

After the collision, the objects have a common speed (velocity),v1f=v2f=v.

The collision of the bullet with the block is completely inelastic.

From the conservation of momentum theorem, we can write the momentum equation as follows:

m1v1i+m2v2=m1v1f+m2v2f

Put the values in the above equation.

m1vb=(m1+m2)v vb=vm1+m2m1 (1)

The kinetic friction can be written as follows:

fk=μkN

Here,is the coefficient of friction, and N is the normal/contact force.

It slows down the block with anacceleration of magnitude a=μkg.

The block slides to a stop through a distance. d=800 m

Using the law of motion equation vf2=vi2+2ad, we find the speed of the block after the collision.

v=2μkgd

Put the values in the above equation.

v=2(0.400)(9.8 m/s2)(8.00 m)=7.92 m/s

Put the values in the above equation.

vb=(7.92  m/s)10+20010.0=166  m/s

Thus, the speed of the bullet before impact is, vb=166 m/smaking choice (b) the correct answer.

Most popular questions for Physics Textbooks

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