Question

A 1500 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2500 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.750, and the cars slide to a halt at a point 5.48 m west and 6.37 in south of the impact point. How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?

Answer 1

Answer:

v = 19.33 m / s   South

Explanation:

To solve this exercise we must use the conservation of momentum, for which we must define a system formed by the two cars, therefore the forces during the collision are internal and therefore the moment is conserved.

Since it is a vector quantity, we are going to work on each axis, the x axis is in the East-West direction

initial instant. Before the crash

        p₀ = m 0 + M v₂ₓ

final instant. Right after the crash

        p_f = (m + M) vₓ

        p₀ = 0_pf

        M v₂ₓ = (m + M) vₓ

In this case m is the mass of the car and M the mass of the SUV

        vₓ = $\frac{M}{m+My}$  v₂ₓ            (1)

in the Y axis (North - South direction)

initial instant

       p₀ = m v_{1y} + M 0

final moment

       p_f = (m + M) v_y

       p₀ = p_f

       m v_{1y} + M 0 = (m + M) v_y

       v_y = $\frac{m}{m+M} \ v_{1y}$       (2)

With these speeds we can use the relationship between work and the variation of kinetic energy, in this part the two cars are already united.

         W = ΔK

friction force work is

         W = - fr d

the friction force is described by the equation

         fr = μ N

Newton's second law

         N-W = 0

         N = W

         

we substitute

         W = - μ (m + M) g d

as the car stops the final kinetic energy is zero and

the initial kinetic energy is

         K₀ = ½ (m + M) v²

we substitute

         - μ (m + M) g d = 0 - ½ (m + M) v²

            μ g d = ½ v²

            v² = 2 μ g d

the distance traveled can be found with the Pythagorean theorem

        d = $\sqrt{x^2+y^2}$

        d = $\sqrt{5.48^2 + 6.37^2}$

        d = 8.40 m

let's calculate the speed

         v² = 2 0.75 9.8 8.40

         v = √123.48

         v = 11.11 m / s

this velocity is in the direction of motion so we can use trigonometry to find the angles

          tan θ = y / x

          θ = tan⁻¹ y / x

          θ = tan⁻¹ (-5.48 / -6.37)

          θ = 40.7º

Since the two magnitudes are negative, this angle is in the third quadrant, measured from the positive side of the x-axis in a counterclockwise direction.

          θ'= 180 + 40.7

          θ’= 220.7º

In the exercise they indicate the the sedan moves in the y-axis, therefore

          sin θ'= v_y / v

          v_y = v sin 220.7

          v_y = 11.11 sin 220.7

          v_y = -7.25 m / s

the negative sign indicates that it is moving south

To find the speed we substitute in equation 2

          v_y = $\frac{m}{m+M} \ v_{1y}$

          v_{1y} = v_ y   $\frac{m+M}{m}$

           

let's calculate

         v_{1y} = -7.25    $\frac{1500+2500}{1500}$

         v_{1y} = - 19.33 m/s

therefore the speed of the sedan is v = 19.33 m / s with a direction towards the South