College Physics, Volume 1
College Physics, Volume 1
2nd Edition
ISBN: 9781133710271
Author: Giordano
Publisher: Cengage
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Chapter 3, Problem 95P

(a)

To determine

Acceleration of the bullet.

(a)

Expert Solution
Check Mark

Answer to Problem 95P

Acceleration is 5.6×105m/s2_.

Explanation of Solution

Then using the equation of motion

    v2=v02+2aS        (I)

Here v is the final velocity, v0 is the initial velocity, S is the distance covered and a is the acceleration

The final velocity is zero. Then rewrite (I) in terms of a

    a=v022S        (II)

Conclusion:

Substitute 300m/s for v0, 8.1cm for S in (II)

    a=(300m/s)22(8.1cm(102m1cm))=5.6×105m/s2

Acceleration is 5.6×105m/s2_

(b)

To determine

Time taken by the bullet to come to rest.

(b)

Expert Solution
Check Mark

Answer to Problem 95P

The time is 5.4×104s_.

Explanation of Solution

Write the equation of motion

    v=v0 +at        (III)

Substitute zero for final velocity and rearrange (III) in terms of t

    t=v0a        (IV)

Conclusion:

Substitute 300m/s for v0, 5.6×105m/s2 for a in (IV)

    t=(300m/s)(5.6×105m/s2)=5.4×104s

The time is 5.4×104s_.

(c)

To determine

Force exerted by the wood on the bullet

(c)

Expert Solution
Check Mark

Answer to Problem 95P

Force exerted by the wood on the bullet is 5.6×103N_.

Explanation of Solution

The only horizontal force acting on the bullet as it decelerated was a force of kinetic friction.

The net force is given by

    ΣF=Ffrictional=ma        (V)

Here m is the mass of the bullet

Conclusion:

Substitute 10g for m and 5.6×105m/s2 for a in (V)

    Ffrictional=(10g(103kg1g))(5.6×105m/s2)=5.6×103N

Force exerted by the wood on the bullet is 5.6×103N_.

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Chapter 3 Solutions

College Physics, Volume 1

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