BASIC BIOMECHANICS
BASIC BIOMECHANICS
8th Edition
ISBN: 9781259913877
Author: Hall
Publisher: RENT MCG
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Chapter 12, Problem 6IP

Lineman A has a mass of 100 kg and is traveling with a velocity of 4 m/s when he collides head-on with lineman B, who has a mass of 90 kg and is traveling at 4.5 m/s. If both players remain on their feet, what will happen? (Answer: Lineman B will push lineman A backward with a velocity of 0.03 m/s.)

Expert Solution
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Summary Introduction

To determine: The kinetic energy of lineman A.

Answer to Problem 6IP

The kinetic energy of lineman A is 800J.

Explanation of Solution

Calculation:

Write an expression to find the kinetic energy of lineman A.

(KE)A=12mAvA2

Here, (KE)A is the kinetic energy of lineman A, mA is the mass of the lineman A, and vA is the velocity of the lineman A.

Substitute 100 kg for mA and 4m/s for vA to find the kinetic energy of lineman A.

(KE)A=12(100kg)(4m/s)2=800kgm2/s2(1Jkgm2/s2)=800J

Therefore, the kinetic energy of lineman A is 800J.

Conclusion

Therefore, the kinetic energy of lineman A is 800J.

Expert Solution
Check Mark
Summary Introduction

To determine: The kinetic energy of lineman B.

Answer to Problem 6IP

The kinetic energy of lineman B is 911J.

Explanation of Solution

Calculation:

Write an expression to find the kinetic energy of lineman B.

(KE)B=12mBvB2

Here, (KE)B is the kinetic energy of lineman B, mB is the mass of the lineman B, and vB is the velocity of the lineman B.

Substitute 90 kg for mB and 4.5m/s for vB to find the kinetic energy of lineman B.

(KE)B=12(90kg)(4.5m/s)2=911.25kgm2/s2(1Jkgm2/s2)911J

Therefore, the kinetic energy of lineman B is 911J.

Conclusion

Therefore, the kinetic energy of lineman B is 911J.

Expert Solution
Check Mark
Summary Introduction

To determine: The final velocity of combined mass.

Answer to Problem 6IP

Therefore, the kinetic energy of lineman B is higher than lineman A, thus lineman B will push lineman A backward with a magnitude of velocity 0.03m/s.

Explanation of Solution

Calculation:

Write an expression of conservation of linear momentum to find the final velocity of combined mass.

mAuAmBuB=(mA+mB)v

Here, uA is the velocity of lineman A, uB is the velocity of the lineman B, and v is the velocity of the combined mass.

Substitute 100 kg for mA, 4m/s for uA, 90 kg for mB, and 4.5m/s for uB to find the velocity of combined mass.

(100kg×4m/s)(90kg×4.5m/s)=(100kg+90kg)v

400kgm/s405kgm/s=(190kg)vv=5190m/s=0.026m/s0.03m/s

Therefore, the velocity of the combined mass is 0.03m/s.

Conclusion

Therefore, the kinetic energy of lineman B is higher than lineman A, thus, lineman B will push lineman A backward with a magnitude of velocity 0.03m/s.

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