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Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, while the second was traveling in a southeastern direction at an angle ϕ with respect to the east-west direction and at a speed v. After the collision, the two-car system travels in a northeastern direction at an angle θ with respect to the north-south direction and at a speed vfinal.

Find v_final, the speed of the joined cars after the collision.

Respuesta :

Answer:[tex]u=\frac{v}{2}\sqrt{5-4sin\phi }[/tex]

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

[tex]m(2v)+m(-vsin\phi )=2m(ucos\theta )[/tex]

[tex]2ucos\theta =v(2-sin\theta )[/tex]------1

Conserving momentum in x direction

[tex]mvcos\phi =2musin\theta [/tex]-----2

squaring and adding 1 &2

[tex](2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2[/tex]

[tex]4u^2=4v^2+v^2-4v^2sin\phi [/tex]

[tex]4u^2=5v^2-4v^2sin\phi [/tex]

[tex]u=\frac{v}{2}\sqrt{5-4sin\phi }[/tex]

Ver imagen nuuk

The final speed of the joined cars after the collision as stated in the question is;

v_f = (v/2)√(5 - 4sin ϕ)

From the law of conservation of linear momentum, we know that;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Where;

m₁ = mass of car 1

u₁ = Initial velocity of Car 1

m₂ = mass of car 2

u₂ = initial velocity of car 2

v₁ = final velocity of car 1

v₂ = final velocity of car 2

           We are told that before the collision, one car was travelling in the northern direction while the second car was travelling in the south eastern direction at an angle ϕ.

We are given initial speed of first car = 2v.

Thus;

Horizontal component of initial velocity of first car; u₁ₓ = 0

Vertical component of initial velocity of first car; u₁y = 2v j^

Similarly, for the second car we are given initial speed = v

Thus;

Horizontal component of initial velocity of 2nd car; u₂ₓ = v cos ϕ

Vertical component of initial velocity of first car; u₂y = v sin ϕ (-j^)

         Since both of their masses are m, after collision since they travel together, their combined masses will be 2m.

They now move together in a northeastern direction at an angle θ.

Thus, resolving the final velocities gives us;

vₓ = v_f sin θ (i^)

v_y = v_f cos θ (j^)

Thus, along the x-axis, we now have;

m₁u₁ₓ + m₂u₂ₓ = 2mvₓ

Plugging in the relevant values, we have;

m(0) + m(v cos ϕ) = 2m(v_f sin θ (i^))

v cos ϕ = 2(v_f sin θ (i^))  ---(eq 1)

Similarly, along the y-axis, we have;

m(2v j^) + m(v sin ϕ (-j^)) = 2m(v_f cos θ (j^))

⇒ 2v - v sin ϕ = 2v_f cos θ   ---(eq 2)

       We will square both eq 1 and eq 2 and add to each other to get;

v² cos² ϕ + (2v - v sin ϕ)² = 4(v_f)² (sin²θ  + cos²θ)

⇒ v² cos² ϕ + (2v - v sin ϕ)² = 4(v_f)²

⇒ v² cos² ϕ + 4v² - 4v² sin ϕ + v²sin²ϕ = 4(v_f)²

⇒ v²(cos² ϕ + sin²ϕ) + 4v² - 4v² sin ϕ = 4(v_f)²

cos² ϕ + sin²ϕ = 1. Thus;

⇒ v² + 4v² - 4v² sin ϕ = 4(v_f)²

⇒ 5v² - 4v² sin ϕ = 4(v_f)²

v_f = (v/2)√(5 - 4sin ϕ)

Read more at; https://brainly.com/question/16986657

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