Question

a 1200-kg pick-up truck traveling south at 15 m/s suddenly collides with a 750-kg car that is traveling east. the two vehicles stick together and slide along the road after colliding. a highway patrol officer investigating the accident determines that the final position of the wreckage after the collision is 25 m, at an angle of 50° south of east, from the point of impact. she also determines that the coefficient of kinetic friction between the tires and the road at that location was 0.40. what was the speed of the car just before the collision?

Answer 1

The conservation of momentum, Newton's second law and kinematics allows to find the result for the initial speed of the car is:

• The speed of the car is 40 m/s at the East direction.

Given parameters

• Mass of the truck M = 1200 kg.

• Truck speed v₀₁ = 15 m / s towards the South.

• Mass of the car m = 750 kg.

• Travel east

• The two vehicles unite.

• The braking distance d = 25m  at 50º SE

• The friction coefficient μ = 0.40

To find

• The initial speed of the car.

The momentum is defined by the product of the mass and the velocity of the body, so it is a vector quantity. In the case of an isolated system the momentum is conserved.

In the attachment we see a diagram of the vehicle crash, let's write the moment for each axis.

y-axis

Initial instant. Before the shock.

          $p_{oy}$  = M v₀₁

Final moment. After the crash

          $p_{fy}$ = (M + m) $v_{fy}$

The system is formed by the two vehicles for which during the crash it is isolated and the momentum is conserved.

          $p_{oy} = p_{fy} \\\\M v_{o1} = (M+m) v_{fy} \\v_{fy} = \frac{M}{M+m} \ v_{o1}$          

           

Calculate us

           $v_{fy} = \frac{1200}{1200+750} \ 15$  

           vfx = 9.23 m / s

x-axis

Initial instant. Before the crash.

           p₀ₓ = m v₀₂

Final moment. After the crash.

           $p_{fx}$ = ​​(M + m) $v_{fx}$  

The momentum is preserved.

          $p_{ox}= p_{fx} \\m v_{o2) = (M+m) \ v_{fx}$  

Let's find the velocity just after the collision, let's use Newton's second law,

y-axis

        N-W = 0

        N = W = (M + m) g

x-axis

       fr = (M + m) a

The friction force is the macroscopic manifestation of the interactions between the two bodies.

       fr = μ N

We substitute.

       μ  g = a

     

Now we can use kinematics.

        v² = v₀² - 2a d

When the vehicles stop their speed is zero.

        v₀² = 2 a x

        v₀ = $\sqrt{2\ \mu \ g \ d}$

Let's calculate.

       v₀ = $\sqrt{2 \ 0.40 \ 9.8 25 }$  

       v₀ = 14 m / s

This is the speed of the two vehicles just after the collision, that is, let's use trigonometry to find their components.

          cos 25 = $\frac{v_{ox}}{v_o}$  

          v₀ₓ = v₀ cos 25

          v₀ₓ = 14 cos 25

          v₀ₓ = 12.69 m / s

We substitute in the expression of the conservation of the momentum in the x-axis.

           m v₀₂ = (M + m) $v_{fx}$  

           $v_{o2} = \frac{M+m}{m} v_{fx}$  

           

Let's calculate.

           $v_{o2} = \frac{1200+750}{750} \ 12.69$  

           vfy = 32.99 m / s = 40 m / s

In conclusion using the conservation of momentum, Newton's second law and kinematics we can find the result for the initial speed of the car is:

• The speed of the car is 40 m/s at the East direction.

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