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nikklg [1K]
3 years ago
8

ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav

eling eastward. Car B weighs 490.0 kg and was traveling westward at 64.0 km/h. The cars locked bumpers and slid eastward with their wheels locked for 6.00 m before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750
Physics
1 answer:
Montano1993 [528]3 years ago
5 0

Answer:

    vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

             W = ΔK                             (1)

for this part the mass is

             M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

              W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

              N -W = 0

              N = W = Mg

friction forces have the expression

              fr =μ N

              fr = μ M g

we substitute in 1

               -μ M g x = 0 - ½ M v²

             v² = 2 μ g x

let's calculate

              v² = 2  0.750  9.8  6.00

              v = ra 88.5

              v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

         p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

         p_f = (mₐ + m_b) v

         p₀ = p_f

         + mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

           

         mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

          v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

         660 vₐ₀ = (660 +490) 9.39 + 490 17.778

         vₐ₀ = 19509.72 / 660

         vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

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\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}

\alpha = 36.469\,\frac{rad}{s^{2}}

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I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }

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M/s
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a. Final velocity, V = 2.179 m/s.  

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<u>Given the following data;</u>

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a. To find the velocity of the boat after it has traveled 4.75 m

Since it started from rest, initial velocity is equal to 0m/s.

Now, we would use the third equation of motion to find the final velocity.

V^{2} = U^{2} + 2aS

Where;

  • V represents the final velocity measured in meter per seconds.
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  • S represents the displacement measured in meters.

Substituting into the equation, we have;

V^{2} = 0^{2} + 2*0.500*4.75

V^{2} = 4.75

Taking the square root, we have;

V^{2} = \sqrt {4.75}

<em>Final velocity, V = 2.179 m/s.</em>

b. To find the velocity if the boat has traveled 50 m.

V^{2} = 0^{2} + 2*0.500*50

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Taking the square root, we have;

V^{2} = \sqrt {50}

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