Final Velocity given only Initial Velocity and Angle

  • #1

Homework Statement

This is a homework that has me completely stumped. If anyone help me out here, I would greatly appreicate it.

A billard ball is east at 2.0 m/s. A second, idential ball is shot west at 1.0 m/s. The balls have a glancing collision (not head-on), deflecting the second ball by 90 degrees and sending it north at 1.41 m/s.

1) What is the speed of the first ball after the collision?
2) What is the direction of the first ball after the collision? (in degrees below the +x axis)

We're studying momentum and collisions in 1D and 2D, so I assumed that this should be that particular type of problem, but since no mass numbers are given for either ball, I'm lost as to where to start.

Homework Equations

The Attempt at a Solution


Answers and Replies

  • #2
The balls are said to be identical. What does that mean with regard it their masses?
  • #3
I assume it to mean that the masses are identical; however since no other information is given, I can't be for certain. How I have it worded above is the exact wording from the homework problem.
  • #4
Identical means everything is the same; that includes the mass.
  • #5
Ok. But I'm still unclear on what this problem expects.

I thought it might be an impulse problem; whereas the change in momentum of one ball would be equal in magnitude and opposite in direction to the momentum change in the other ball, which would make the angle on the first ball 90 degrees below the x axis (assuming I understand this concept correctly). However this is wrong according to the homework program.
  • #6
The total momentum does not change during a collision. Changes in individual momenta need not "mirror" each other.
  • #7
Here is the best thing I've come up with in regards to finding the post collision speed of the first ball.

Assuming the masses are identical, I call the masses for both balls simply "M". This is an isolated system so initial momentum = final system momentum.

initial p for ball one = M(2.0)
initial p for ball two = M(1.0)
total initial P for the system is M(3.0)

since we're looking for final velocity for ball one, I'll call final p for ball one "MV"

So I've come up with the equation:

M(3.0) = MV + M(1.41)

factor out the M

3.0 = v + 1.41
1.59 = v

Of course I'm being told that this is also incorrect. Any other ideas on how to tackle this?
  • #8
The problem is 2D; you have neglected directions in your analysis.
  • #9
Yeah. I noticed that after I posted it and went back and reworked it. Answers I'm coming up with are still wrong though. I used up all the allotted attempts at the problem that homework program allows and as such it displayed the correct answer for me. It was 1.73 m/s. Don't have any clue how they came up with that. Nothing I was coming up with was even close to that.

Anyway; thanks for trying to help. Apparantly this is something I need to go back and review some more. I thought I had a pretty good handle on it, but this problem really threw me a curve ball.

Oh well.
  • #10
The velocity of the first ball is (2, 0) - this means a vector whose x-component is 2 and the y-component is 0. The velocity of the second ball is (-1, 0). The total momentum is thus M(2, 0) + M(-1, 0) = M(1, 0). After the collision, the second ball's velocity is (0, 1.41). The first's ball velocity is unknown, let's denote it by (x, y). We must have M(x, y) + M(0, 1.41) = M(1, 0). Component wise, that means x + 0 = 1; y + 1.41 = 0.

Thus x = 1; y = -1.41 and the velocity of the first ball is (1, -1.41). The magnitude of this vector is the square root of 1*1 + 1.41*1.141 = square root of 3 = 1.73.
  • #11
Yep; really need to go back and review this concept then. It never even occurred to me to approach the problem in that matter.

Would you have any suggestions for online resources where I can review physics concepts in general? The textbook we're using in class isn't a very good explanation; of course, that could just me as well since I've been a better learner doing something rather than just reading about it.

Anyway thanks again for your help.
  • #12
Sorry, can't advise anything general.

Judging from this thread, I think you need to have some more practice with vectors.

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