Solving an Elastic Collision: Billiard Ball Momentum

AI Thread Summary
In an elastic collision involving two billiard balls, one moving at 5.40 m/s strikes a stationary ball, resulting in the first ball moving at 4.48 m/s at a 34.0° angle. The conservation of momentum and kinetic energy principles are crucial for solving the problem. The user initially calculated the second ball's velocity to be approximately 4.709 m/s but struggles with applying direction to the momentum vectors. They seek clarification on setting up the problem and understanding how to resolve vectors into components. The discussion emphasizes the importance of determining the direction of net final momentum and the resultant of the vectors involved in the collision.
Juicy
Messages
2
Reaction score
0

Homework Statement



A billiard ball moving at 5.40 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.48 m/s at an angle of 34.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision.

Homework Equations



Conservation of momentum, conservation of KE

The Attempt at a Solution



Using conservation of momentum I got the final velocity of the second ball to be approximately 4.709

This is due in about 10 minutes, which is probably not possible for me to do, so at this point I give up on trying to figure it out on my own and need a bit of help understand these concepts as a whole for future homework/exams.
 
Physics news on Phys.org
Hello and Welcome to PF,

Could you please clarify as to where you are stuck in the problem?
 
Thank you,

and in all honesty I'm just having trouble setting things up. I understand the concepts of conservation of momentum and KE and where they do and don't apply, however it has always been a trouble for me applying direction to these sort of problems. I tried running through the notes my instructor provides both in-class and online help very little to not at all with this problem, as she barely mentioned changes in direction in collisions like this sort of problem.
 
I assume you are capable of resolving a vector into mutually perpendicular components. Ask yourself this: What would be the direction of the net final momentum of the system? How would you find the direction of the resultant of two vectors? If the resultant of the two and one of the vectors are known could you find the other?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top