Physics, 09.08.2019 21:20 mistysallaberry

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.

Answers: 1

Show answers

Answers

Answer from: jacesmokesloud7254

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)

Another question on Physics

Physics, 21.06.2019 20:20

An asteroid is discovered in a nearly circular orbit around the sun, with an orbital radius that is 2.47 times earth's. what is the asteroid's orbital period, its "year," in terms of earth years?

Answers: 1

Answer

Physics, 21.06.2019 22:30

Matt is driving down 7th street. he drives 150 meters in 18 sec. assuming he doesn’t slow down or speed up, what is his speed?

Answers: 1

Answer

Physics, 22.06.2019 15:30

At 20∘c, the hole in an aluminum ring is 2.800 cm in diameter. you need to slip this ring over a steel shaft that has a room-temperature diameter of 2.804 cm . 1. to what common temperature should the ring and the shaft be heated so that the ring will just fit onto the shaft? coefficients of linear thermal expansion of steel and aluminum are 12×10−6 k−1 and23×10−6 k−1 respectively.

Answers: 1

Answer

Physics, 23.06.2019 02:30

Bonds within one water molecule are examples of

Answers: 2

Answer

You know the right answer?

Two cars, both of mass m, collide and stick together. prior to the collision, one car had been trave...

Questions