What is the magnitude and direction of the wreckage after the collision?

[kh_x2]

The car of mass 1500 kg is traveling east at 25 m/s and collides at an intersection with the truck of 2500 kg mass traveling north at 20 m/s if the wreckage moves as one mass after the collision (the collision is inelastic), calculate the magnitude and direction of the wreckage after the collision.

My problem is that I have used two methods of working the magnitude - both of which are in my notes and both of which give me a different answer and I don't know how to find the direction at all.


Before After
East > North ^ ?
1500 2500 4000
25 20 V

North was weird so I took it as positive ie east aswell

37500 + 50000 = 4000v
v= 87500/4000
v= 21. 875 m/s so it moves east direction


or it will move in west direction at 3.125 m/s if you take north as negative ( this answer is wrong i think)

 

[cryptoguy]

I'm not 100% sure about what you did. I would, for the moment not focus on positives and negatives because you are dealing with vectors that have direction. I believe you figured out the magnitude of the velocity correctly (21. 875 m/s). Now you have to figure out the direction, correct?

You can do this several ways, using trig. You know the x and y component of momentum, so you can figure out the resultant angle which will be the angle at which the object moves.

 

[Moetasim]

I would first clarify the given information and the problem statement. Are we assuming that the collision is happening in a two-dimensional plane? If so, then we can use vector addition to determine the magnitude and direction of the wreckage after the collision.


First, let's draw a diagram to visualize the situation. We have a car traveling east at 25 m/s and a truck traveling north at 20 m/s. After the collision, the wreckage moves as one mass in an unknown direction, which we will label as "V".


We can break down the velocities into their x and y components. The car's velocity is purely in the x-direction, so we can write it as (25 m/s, 0 m/s). The truck's velocity is purely in the y-direction, so we can write it as (0 m/s, 20 m/s).


To find the velocity of the wreckage after the collision, we can use the principle of conservation of momentum. This means that the total momentum before the collision must be equal to the total momentum after the collision.

Before the collision:
Total momentum = (1500 kg)(25 m/s) + (2500 kg)(20 m/s) = 37,500 kgm/s + 50,000 kgm/s = 87,500 kgm/s

After the collision:
Total momentum = (4000 kg)(V m/s)

Setting these equal to each other, we get:

87,500 kgm/s = 4000 kg(V m/s)

V = 87,500 kgm/s / 4000 kg = 21.875 m/s

So the magnitude of the wreckage's velocity after the collision is 21.875 m/s.

To find the direction, we can use trigonometry. We know that the x-component of the wreckage's velocity will be 21.875 m/s, since it is moving in the x-direction. To find the y-component, we can use the Pythagorean theorem:

V^2 = (x-component)^2 + (y-component)^2
(21.875 m/s)^2 = (25 m/s)^2 + (y-component)^2
y-component = √(21.875 m/s)^2 - (25 m/s)^2 = 13.125 m/s

So the y-component of the wreckage's velocity is 13