College Physics (Instructor's)
College Physics (Instructor's)
11th Edition
ISBN: 9781305965317
Author: SERWAY
Publisher: CENGAGE L
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Chapter 6, Problem 63AP

A 2.0-g particle moving at 8.0 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g. (c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy?

(a)

Expert Solution
Check Mark
To determine
The speed of each particle after collision.

Answer to Problem 63AP

The speeds of the particles are 2.67ms1 and 10.7ms1

Explanation of Solution

Explanation

Given Info:

Mass of the moving particle is 2.0g , speed of the moving particle is 8.0ms1 , mass of the particle which is at rest is 1.0g , the initial speed of particle at rest is zero.

Apply conservation of momentum for both particles before and after collision,

m1v1+m2v2=m1v1+m2v2

  • m1 is the mass of the moving particle
  • v1 is initial the speed of the particle
  • m2 is the mass of the particle which is at rest
  • v2 is the speed of particle at rest.
  • v1 is the speed of m1 after collision
  • v2 is the final speed of m2 after collision

Use 0m/s for v2 in the above equation and rewrite in terms of v2 .

v2=(m1m2)(v1v1) (I)

Apply conservation of mechanical energy for both particles,

12m1v12+12m2v22=12m1(v1)2+12m2(v2)2

Use 0m/s for (v2)2 in the above equation and rewrite in terms of v2 .

(v2)2=(m1m2)(v12(v1)2) (II)

Use (m1/m2)(v1v1) for v2 in the above equation and rewrite in terms of v1 .

((m1m2)(v1v1))2=(m1m2)(v12(v1)2)(m1m2)(v1v1)(v1v1)=(v1v1)(v1+v1)v1(m1m2m2)=v1(m1+m2m2)v1=v1(m1m2m1+m2) (III)

Substitute 8.0ms1 for v1 , 2.0g for m1 and 1.0g for m2 to find v1 .

v1=(8.0ms1)(2.0g1.0g2.0g+1.0g)=83ms1=2.67ms1

Substitute 2.67ms1 for v1 , 8.0ms1 for v1 , 2.0g for m1 and 1.0g for m2 in (1) to calculate v2 ,

v2=(2.0g1.0g)(8.0ms12.67ms1)=10.66ms110.7ms1

Conclusion:

Therefore the speeds of the particles are 2.67ms1 and 10.7ms1

(b)

Expert Solution
Check Mark
To determine
The speed of each particle after collision.

Answer to Problem 63AP

The speeds of the particles are 5.33ms1 and 2.67ms1

Explanation of Solution

Explanation

Given Info:

Mass of the moving particle is 2.0g , speed of the moving particle is 8.0ms1 , mass of the particle which is at rest is 10.0g , the initial speed of particle at rest is zero.

Formula to calculate speed of the moving particle after collision is,

v1=v1(m1m2m1+m2) (IV)

Substitute 8.0ms1 for v1 , 2.0g for m1 and 10.0g for m2 in (IV) to find v1 .

v1=(8.0ms1)(2.0g10.0g2.0g+10.0g)=6412ms1=5.33ms1

Formula to calculate speed of the particle of mass m2 at rest after collision is,

v2=(m1m2)(v1v1) (V)

Substitute 5.33ms1 for v1 , 8.0ms1 for v1 , 2.0g for m1 and 10.0g for m2 in the above equation to calculate v2 ,

v2=(2.0g10.0g)(8.0ms1+5.33ms1)=2.67ms1

Conclusion:

Therefore the speeds of the particles are 5.33ms1 and 2.67ms1

(c)

Expert Solution
Check Mark
To determine
The magnitude of kinetic energy of the moving particle for given velocities.

Answer to Problem 63AP

The kinetic energies of the moving particle are 7.13×103J and 28.41×103J (a)

Explanation of Solution

Explanation

Given Info:

The speeds of the moving particle are. 2.67ms1 and 5.33ms1 in part (a) and part (b) respectively,

Formula to calculate kinetic energy of the moving particle is,

KE=12m1(v1)2

Substitute 2.0g for m1 and 2.67ms1 for v1 from part (a) in the above equation to calculate KE.

KE=12(2.0g)(1g1000kg)(2.67ms1)2=7.13×103J

Substitute 2.0g for m1 and 2.67ms1 for v1 from part (b) in the above kinetic energy equation to calculate KE.

KE=12(2.0g)(1g1000kg)(5.33ms1)2=28.4×103J

Conclusion:

Therefore the kinetic energies of the moving particle are 7.13×103J and 28.4×103J

The incident particle loose more energy in part (a)

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Chapter 6 Solutions

College Physics (Instructor's)

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