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You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light when it was hit from behind by car A, of mass 1500 kg. The cars locked bumpers during the collision and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.30 m long, and inspection of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.65.
(a) What was the speed of car A just before the collision?
(b) If the speed limit was 35 mph, was car A speeding, and if so, by how many miles per hour was it exceeding the speed limit?

Sagot :

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x  2100 X 9.8  = 13377 N .

work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum  .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

v₁ = 17.676 m /s

= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.

snehashish65

(a) The speed of car A just before the collision is 51.58 mph.

(b) With the given speed limit of 35 miles per hour, car A was crossing the speed limit by 16.58 mph.

What is collision?

The event when two objects strike each other from either direction, then such event is known as a collision. During the collision, the speed of colliding objects may vary according to the direction of the approach.

Given data -

The mass of car A is,  mA = 1500 kg.

The mass of car B is, mB = 2100 kg.

The length of the skid mark is, d = 7.30 m.

The coefficient of kinetic friction between tires and road is, [tex]\mu = 0.65[/tex].

(a)

The combined kinetic energy of both cars is,

[tex]KE_{T}=\dfrac{1}{2} (mA+mB)v^{2}\\\\KE_{T}=\dfrac{1}{2} (1500+2100)v^{2}\\\\KE_{T}=1800v^{2}[/tex]

Applying the work-energy principle as,

Work done due to kinetic friction = Combined kinetic energy of cars

[tex]F \times d = KE_{T}\\\\(\mu \times (mA+mB)\times g) \times d = KE_{T}\\\\(0.65 \times (1500+2100)\times 9.8) \times 7.30 = 1800v^{2}\\\\v = 9.64 \;\rm m/s[/tex]

Converting into mph as,

[tex]v = 9.64 \times 2.23\\\\v = 21.49 \;\rm mph[/tex]

To know the speed of car A , we shall apply the law of conservation of momentum. Let the speed of car A before collision be v₁.

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 21.49

v₁ = 51.58 mph

Thus, we can conclude that the speed of car A just before the collision is 51.58 mph.

(b)

With the given speed limit of 35 mph, the obtained speed of car A before the collision is 51.58 mph. Clearly, car A is crossing the speed limit. And the difference is,

= 51.58 - 35 = 28.63 mph.

= 16.58 mph

Thus, we can conclude that car A was crossing the speed limit by 16.58 mph.

Learn more about the average speed here:

https://brainly.com/question/12322912

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