collision velocity problem?

Question

Two cars, car(1) with 1466 kg mass and a velocity of 10.56 m/s and car(2) with 2000 kg mass and unknown velocity approach an intersection at 90o with respect to each other and collide. Immediately after the collision the cars slide stuck together with a velocity of 16.34 m/s. What is the angle between the momentum of the car wreck and the initial direction of car (1).

The answer is 74.1364 but how do u get the answer?, please show formula used

um, if u have it, can u please show me how to do it, :-), ill give u the 10 pts

Answer 1

Ok, I have it now.

Look, you need to draw the first car, going from the left to the righ. And the second car, going from down to up, so they are gonna crash on the intersection at 90º.

Then, you must use the linear momentum. As you know, before and after the colission, the linear momentum is the same.

so :

1466*10.56 + 2000*V = 3466*16.34

V = unknown velocity

so after you solve that equation, V = 20,6 m/s

Now. you know that because of the colission, the cars, will be traveling in a different direction, and making an angle with the horizontal and vertical.

Now, let's say that the first car had a speed : Vx

The second car had a speed : Vy

After the crash, they together had a speed : Vx + Vy >> vector.

using the Vx, the X direction, in other words, using the horizontal speed from the first car before the colission and the Vx speed of the cars after the colission.

1466*10,56 = 3466*(16,34*cos(angle))

angle : angle between the momentum of the car wreck and the first car.

then, when you solve it :

cos(angle) = 0.2733

angle = 74.136º

Hope that might help you

Note : Try to make a graphic, that will help you, and as I said before, the Vx, is a vector, and is the speed of the first car. Vy is a vector too, and is the speed of the second car.

And after the colission they both had a speed : Vx + Vy, a vector too, but that speed makes an angle with the horizontal and the vertical

Answer 2

The preliminary 2d of each and every skater is shown under. East momentum of 1st skater = ninety * 5.6 = 504 E West momentum of 2d skater = 40 5 * 14 * cos 60 = 315 W complete East – West momentum = 504 – 315 = 189 E South momentum of 2d skater = 40 5 * 14 * sin 60 = 545.6 S on the grounds that momentum is conserved, the remarkable momentum of the two skaters, tangled jointly, is equivalent the excellent preliminary momentum. very final momentum = (189^2 + 545.6^2)^0.5 = 577.4 The tangent of the perspective south of east = East element ÷ South element Tangent = 545.6 ÷ 189 perspective = 70.9? south of east