# Homework Help: Need help on a physics question [momentum in 2D]

1. Feb 28, 2004

### MtX

hi.. im having difficulty solving 2D momentum-collision questions.. i can do linear ones but somehow i dont know what to do for 2D ones.. can someone please help me?

1. A 3.2-kg hawk soaring at 20 m/s [N] collides with a 0.50-kg sparrow flying at 5.0 m/s [W]. If both the hawk and sparrow are on the same horizontal plane, find their velocity if the hawk hangs on to the sparrow after collision.

2. A 3000-kg car travelling at 20 m/s [N] collides with a 5000-kg truck moving east on an icy road. The bumpers of the two vehicles become entangled and the vehicles remain joined after the collision. Calculate the initial speed of the truck if both vehicles after collision go [E 30 N].

2. Feb 28, 2004

Just remember two things: momentum is conserved, and orthogonal directions (ones at right angles to each other, such as up-down and left-right or North-South and West-East) are independent.

So that means we can split up the directions and get:

$$p_\textrm{x,object1,before} + p_\textrm{x,object2,before} = p_\textrm{x,objects,after}$$
and
$$p_\textrm{y,object1,before} + p_\textrm{y,object2,before} = p_\textrm{y,objects,after}$$

By the way, shouldn't this be in the homework help forum?

3. Feb 29, 2004

### MtX

i used that formula to solve linear questions.. but i dont know what to do for 2D ones.. can someone show me a step to step solution please? i am trying to get a clear picture of things but its so confusing..

4. Feb 29, 2004

You're going to use the same equations. Except now there will be two of them, so you'll have to go through a bit more algebra to get to what you want.

5. Feb 29, 2004

### MtX

can you just show me how to do one?

6. Feb 29, 2004

Let an object of mass 100kg and velocity 10m/s travelling 30 degrees South of directly East collide with an object of mass 50kg and velocity 20m/s travelling 45 degrees North of directly East.

$$p_\textrm{x,object1,before} + p_\textrm{x,object2,before} = p_\textrm{x,objects,after}$$

$$m_1v_1\cos30 + m_2v_2\cos45 = (m_1 + m_2)v_\textrm{x,a}$$

$$(100\textrm{kg})(10\textrm{m/s})\cos30 + (50\textrm{kg})(20\textrm{m/s})\cos45 = (100\textrm{kg} + 50\textrm{kg})v_\textrm{x,a}$$

$$v_\textrm{x,a} = \frac{(100\textrm{kg})(10\textrm{m/s})\cos30 + (50\textrm{kg})(20\textrm{m/s})\cos45}{(100\textrm{kg} + 50\textrm{kg})}$$

Repeat for the y-component, keeping in mind that the first object's y component of velocity is negative.

Once you have both the x and y components, use the Pythagorean Theorem to get the total velocity.

$$v_\textrm{x,a}^2+v_\textrm{y,a}^2 = v_\textrm{a}^2$$

Last edited: Feb 29, 2004
7. Mar 1, 2004

### MtX

thanks.. i understand what you did, its just like linear ones but you add all the x-components and y-components... but for the examples i posted, the objects stick together after, so what will be the m1 and m2 (masses)? will there be two different masses, or will there just be one combined mass? for velocity, i know that since they stick together, the velocity will be the same, so i can use the same velocity notation..

8. Mar 1, 2004

### pedestrian

When two objects collide and stick together their masses combine to form one mass

9. Dec 7, 2010

m1v1+m2v2=m3v3

so 2,3*20+0,5*5=2,7*v3

232,5=2,7*v3

v3=86,1111

thats a) m1 is hawk, m2 is sparrow, m3 is sparrow+hawk, v1 is velocity of hawk ect

its simple momentum... the momentum before collision is the same as after collision