COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 7, Problem 91P

(a)

To determine

The change in momentum of the car due to the fly.

(a)

Expert Solution
Check Mark

Answer to Problem 91P

The change in momentum of the car due to the fly is 0.01kgkm/h opposite to the car’s motion.

Explanation of Solution

Write the expression for conservation of momentum.

  m1v1=m2v2        (I)

Here, m1 is the mass of the automobile, v1 is the speed of the automobile, m2 is the mass of the fly, and v2 is the speed of the fly.

Write the expression for car’s final momentum.

  pf=m1v2        (II)

Here, pf is the final momentum of the car.

Write the expression for car’s initial momentum.

  pi=m1v1        (III)

Here, pi is the initial momentum of the car.

Write the expression for change in momentum of the car due to the fly.

  Δp=pfpi        (IV)

Here, Δp is the change in momentum of the car due to the fly.

Conclusion:

Substitute 1000kg for m1, 100km/h for v1, and 0.1g+1000kg for m2 in the equation (1) to find v2.

  (1000kg)(100km/h)=[(0.1g)(0.001kg1g)+1000kg]v2100,000kgkm/h=(0.0001kg+1000kg)v2v2=100,000kgkm/h1000.0001kg=99.999990km/h

Substitute 1000kg for m1 and 99.999990km/h for v2 in the equation (1I) to find Pf.

  pf=(1000kg)(99.999990km/h)=99,999.990kgkm/h

Substitute 1000kg for m1 and 100km/h for v1 in the equation (1II) to find Pi.

  pi=(1000kg)(100km/h)=100,000kgkm/h

Substitute 99,999.990kgkm/h for Pf and 100,000kgkm/h for Pi in equation (V).

  Δp=99,999.990kgkm/h100,000kgkm/h0.01kgkm/h

Therefore, the change in momentum of the car due to the fly is 0.01kgkm/h opposite to the car’s motion.

(b)

To determine

The change in momentum of the fly due to the car.

(b)

Expert Solution
Check Mark

Answer to Problem 91P

The change in momentum of the fly due to the car is 0.01kgkm/h.

Explanation of Solution

Write the expression for change in momentum of the fly due to the car.

  Δp=mflyv1        (V)

Here, Δp is the change in momentum of the fly due to the car, mfly is the mass of the fly, and v1 is the speed of the automobile.

Conclusion:

Substitute 0.1g for mfly and 100km/h for v1 in the equation (V) to find ΔP.

  Δp=(0.1g)(0.001kg1g)(100km/h)=0.01kgkm/h

Therefore, the change in momentum of the fly due to the car is 0.01kgkm/h.

(c)

To determine

The number of flies takes to reduce the car’s speed into 1km/h approximately.

(c)

Expert Solution
Check Mark

Answer to Problem 91P

The number of flies takes to reduce the car’s speed into 1km/h approximately is 105flies.

Explanation of Solution

Write the expression for the number of flies takes to reduce the car’s speed.

  m1v1=[m1+nmfly]99        (VI)

Here, n is the number of flies takes to reduce the car’s speed.

Conclusion:

Substitute 0.1g for mfly, 1000kg for m1, and 100km/h for v1 in the equation (VI) to find n.

  (1000kg)(100km/h)=[1000kg+n(0.0001kg)]99km/h100,000kgkm/h=99000kgkm/h+n(0.0099kg)1000kgkm/h=n(0.0099kg)n=105flies

Therefore, the number of flies takes to reduce the car’s speed into 1km/h approximately is 105flies.

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Chapter 7 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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