Inquiry into Physics
Inquiry into Physics
8th Edition
ISBN: 9781337515863
Author: Ostdiek
Publisher: Cengage
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Chapter 3, Problem 34Q
To determine

The rank of cars according to the stopping distance.

Expert Solution & Answer
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Answer to Problem 34Q

The rank of all cars,

(a)m=1000kgv=10m/s(b)m=1000kgv=20m/s(c)m=1000kgv=30m/s(d)m=3000kgv=10m/s(e)m=2000kgv=10m/s(f)m=2000kgv=20m/s(g)m=2000kgv=30m/s(h)m=3000kgv=20m/s p=1×104kg.m/sp=2×104kg.m/sp=3×104kg.m/sp=3×104kg.m/sp=2×104kg.m/sp=4×104kg.m/sp=6×104kg.m/sp=6×104kg.m/s Rank5Rank4Rank3Rank3Rank4Rank2Rank1Rank1.

Explanation of Solution

Given information:

The mass and velocity of different cars are given by,

(a)m=1000kgv=10m/s(b)m=1000kgv=20m/s(c)m=1000kgv=30m/s(d)m=3000kgv=10m/s(e)m=2000kgv=10m/s(f)m=2000kgv=20m/s(g)m=2000kgv=30m/s(h)m=3000kgv=20m/s

Concept Used:

Momentum, p=mv.

Calculation:

The stopping distance is directly proportional to the momentum.

The momentum of each car is,

(a)m=1000kgv=10m/s(b)m=1000kgv=20m/s(c)m=1000kgv=30m/s(d)m=3000kgv=10m/s(e)m=2000kgv=10m/s(f)m=2000kgv=20m/s(g)m=2000kgv=30m/s(h)m=3000kgv=20m/s p=1×104kg.m/sp=2×104kg.m/sp=3×104kg.m/sp=3×104kg.m/sp=2×104kg.m/sp=4×104kg.m/sp=6×104kg.m/sp=6×104kg.m/s

So, the rank of all cars according to stopping distance,

(a)m=1000kgv=10m/s(b)m=1000kgv=20m/s(c)m=1000kgv=30m/s(d)m=3000kgv=10m/s(e)m=2000kgv=10m/s(f)m=2000kgv=20m/s(g)m=2000kgv=30m/s(h)m=3000kgv=20m/s p=1×104kg.m/sp=2×104kg.m/sp=3×104kg.m/sp=3×104kg.m/sp=2×104kg.m/sp=4×104kg.m/sp=6×104kg.m/sp=6×104kg.m/s Rank5Rank4Rank3Rank3Rank4Rank2Rank1Rank1.

Conclusion:

The rank of all cars,

(a)m=1000kgv=10m/s(b)m=1000kgv=20m/s(c)m=1000kgv=30m/s(d)m=3000kgv=10m/s(e)m=2000kgv=10m/s(f)m=2000kgv=20m/s(g)m=2000kgv=30m/s(h)m=3000kgv=20m/s p=1×104kg.m/sp=2×104kg.m/sp=3×104kg.m/sp=3×104kg.m/sp=2×104kg.m/sp=4×104kg.m/sp=6×104kg.m/sp=6×104kg.m/s Rank5Rank4Rank3Rank3Rank4Rank2Rank1Rank1.

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Chapter 3 Solutions

Inquiry into Physics

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