Two cars, one of mass 1300 kg, and the second of mass 2400 kg, are moving at right angles to each other when they collide and stick together. The initial velocity of the first car is 12 m/s in the positive x direction and that of the second car is 18 m/s in the positive y direction.
What is the magnitude of the velocity of the wreckage of the two cars immediately after the collision?

Respuesta :

To solve this problem we will apply the momentum conservation theorem, that is, the initial momentum of the bodies must be the same final momentum of the bodies. The value that will be obtained will be a vector value of the final speed of which the magnitude will be found later. Our values are given as,

[tex]m_1 = 1300kg[/tex]

[tex]m_2 = 2400kg[/tex]

[tex]u_1 = 12m/s i[/tex]

[tex]u_2 = 18m/s j[/tex]

Using conservation of momentum,

[tex]m_1u_1+m_2u_2 = (m_1+m_2)v_f[/tex]

[tex]1300*12i-2400*18j = (1300+2400)v_f[/tex]

Solving for [tex]v_f[/tex]

[tex]v_f = 4.2162i-11.6756j[/tex]

Using the properties of vectors to find the magnitude we have,

[tex]|v| = \sqrt{(4.2162^2)+(-11.6756)^2}[/tex]

[tex]|v| = 12.4135m/s[/tex]

Therefore the magnitude of the velocity of the wreckage of the two cars immediately after the collision is 12.4135m/s

For the two cars the final velocity of the wreckage of the two cars immediately after the collision is 12.4135 m/s.

What is conservation of momentum?

Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.

The mass of the car one is 1300 kg and the mass of the second car is 2400  kg.

The initial velocity of the first car is 12 m/s in the positive x direction, and that of the second car is 18 m/s in the positive y direction.

Final velocity of the two cars after the collision is equal. Thus, the final velocity of the wreckage of the two cars immediately after the collision by the law of conservation can be given as,

[tex](1300)12\hat i-(2400)18\hat j=(1300+2400)v\\v=4.2162\hat i-11.6756\hat j[/tex]

Solve the above equation further using the property of vectors as,

[tex]v=\sqrt{(4.2162)^2+(-11.6756)^2}\\v=12.4135\rm m/s[/tex]

Hence, For the two cars the final velocity of the wreckage of the two cars immediately after the collision is 12.4135 m/s.

Learn more about the conservation of momentum here;

https://brainly.com/question/7538238