I with a collision problem

[Enoch]

The question is as follows:

A 4g particle moving at 30 m/s collides with a 0.99 g particle initially at rest. After the collision the two particles have velocities that are directed 27 degrees on either side of the original line of motion of the 4 g particle. What is the speed of the 0.99 g particle after the collision? Answer in units of m/s.

I hate to be a beggar for answers, but I have truly been stumped by this problem. I had relatively little problem with the other collision/momentum problems, but for some reason, I am not sure what angle to take to solve this problem.

To begin with, I converted the units into the standard mass units of Kg. I then drew the figure and broke down the motion of the second particle into components - Vfx(cos 27) and Vfy(sin 27). However, from here I do not know what to use. I looked at the momentum conservation theory/equations, but the ones I have require more variables than I know how to solve for.

Any help would be appreciated!

 

[Sirus]

Think of this vectorally. Draw the momentum vector for the motion of the first particle (before the collision). Because momentum is conserved, the momentum vectors of the two particles after the collision must add up vectorally to the initial vector of the first particle. Does this help?

 

[Enoch]

Think of this vectorally. Draw the momentum vector for the motion of the first particle (before the collision). Because momentum is conserved, the momentum vectors of the two particles after the collision must add up vectorally to the initial vector of the first particle. Does this help?

In a way, however, without the speed of either particle post-collision...I'm not sure how to set up an equation.

 

[aekanshchumber]

You said that you hate to be a begger, with respect to that i am not going to give you the complete answer but a hint.
Apply the law of conservation of kinetic energy, and conservation of linear as well as vertical momentum. Two variable and three equations, solve them for velocity of both particles. Take 27degrees from the orignal line of motion of the 4g particle.

 

[Enoch]

You said that you hate to be a begger, with respect to that i am not going to give you the complete answer but a hint.
Apply the law of conservation of kinetic energy, and conservation of linear as well as vertical momentum. Two variable and three equations, solve them for velocity of both particles. Take 27degrees from the orignal line of motion of the 4g particle.

I tried to apply those principles, but I wasn't able to put together the formulas correctly in time for the homwork's due date. Thanks for the help, though ;).

 

[ehild]

In a way, however, without the speed of either particle post-collision...I'm not sure how to set up an equation.

Do not set up one equation. Set up two. One for the x components of the momenta and one for the y components.

After the collision the two particles have velocities that are directed 27 degrees on either side of the original line of motion of the 4 g particle.

The velocity of both particles make +27 and -27 degrees with the original velocity of the 4 g particle. The direction of the original velocity is taken as x axis. If v1 is the speed of the 4 g particle after the collision and v2 is the speed of the other one, and both velocities make angles of equal magnitude but oposite sign with the x axis, and the sum of the y components of momenta should be zero after the collision, what is the relation between v1 and v2?


aekanshchumber :

You can not assume about the collision that it is elastic unless it was stated so. This one is not elastic, the kinetic energy is not conserved.

ehild