PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 7, Problem 79P
To determine

Whether or not any of the cars was speeding.

Expert Solution & Answer
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Answer to Problem 79P

The car with mass 1100kg crossed the speed limit.

Explanation of Solution

The mass of the car moving west is 1100kg. The mass of the car moving north is 1300kg. After the collision the car got stuck and move 17m in a direction 30° north of west. The coefficient of kinetic friction between the tire and road is 0.80. The speed limit is 70km/h.

The figure 1 below shows the initial and final velocities of the cars.

PHYSICS, Chapter 7, Problem 79P

The collision between the cars is perfectly inelastic. The final velocity of both the cars will be same. The final velocity of the collision will be the initial velocity of the sliding motion of the two cars stuck to each other. At the end of sliding, the system comes to rest.

Write the energy formula for the sliding motion of the two cars system when they are joined to each other.

12mvi2+fdcos180°=0 (I)

Here, m is the combined mass of the two cars, vi is the initial speed of sliding, f is the friction force, d is the distance travelled before coming to rest.

Write the formula for the friction force.

f=μkmg (II)

Here, μk is the coefficient of kinetic friction, g is the acceleration due to gravity.

Substitute equation (II) in equation (I).

12mvi2μkmgd=0

Re-write the above equation to get an expression for vi

12mvi2=μkmgdvi=2μkgd

Thus write the formula for the final speed of the collision.

vf=2μkgd (III)

Here, vf is the final speed of collision.

Refer figure 1 and write the equation for the conservation of linear momentum in x-direction.

m1vi1=(m1+m2)vfcos150°=(m1+m2)(2μkgd)cos150°

Here, m1 is the mass of the first car, m2 is the mass of the second car, vi1 is the initial velocity of the first car.

Re-write the above equation to get an expression for vi1.

vi1=(m1+m2m1)(2μkgd)cos150° (IV)

Refer figure 1 and write the equation for the conservation of linear momentum in y-direction.

m2vi2=(m1+m2)vfsin150°=(m1+m2)(2μkgd)sin150°

Here, m2 is the mass of the second car, vi2 is the initial velocity of the second car.

Re-write the above equation to get an expression for vi2.

vi2=(m1+m2m2)(2μkgd)sin150° (V)

Conclusion:

Substitute 1100kg for m1, 1300kg for m2, 17m for d, 0.80 for μk, 9.80m/s2 for g in equation (IV).

vi1=(1100kg+1300kg1100kg)(2(0.80)(9.80m/s2)(17m))cos150°=(1100kg+1300kg1100kg)(2(0.80)(9.80m/s2)(17m))(1km/h0.2778m/s)cos150°=110km/h

Substitute 1100kg for m1, 1300kg for m2, 17m for d, 0.80 for μk, 9.80m/s2 for g in equation (V).

vi2=(1100kg+1300kg1300kg)(2(0.80)(9.80m/s2)(17m))sin150°=(1100kg+1300kg1300kg)(2(0.80)(9.80m/s2)(17m))(1km/h0.2778m/s)sin150°=54km/h

The velocity of car with mass 1100kg is 110km/h. The velocity of the car with mass 1300kg is 54km/h. Thus the car with mass 1100kg crossed the speed limit.

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Chapter 7 Solutions

PHYSICS

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