Inelastic Collision and Finding Initial Velocity

AI Thread Summary
In an inelastic collision involving two gold balls of equal mass, one ball initially travels at 2.70 m/s. After the collision, their velocities are 2.49 m/s at 62.8 degrees N of W and 2.37 m/s at 69.2 degrees S of E. The conservation of momentum equation can be simplified by eliminating mass, focusing on the x and y components of the velocities. A vector diagram is essential for keeping track of the signs and ensuring the total momentum before the collision equals the total after. The expected initial velocity of the unknown ball is 3.00 m/s.
pitaaa
Messages
20
Reaction score
0

Homework Statement



Two rolling gold balls of the same mass collide. The velocity of one ball is initially 2.70 m/s [E]. After the collision, the velocities of the balls are 2.49 m/s [62.8 degrees N of W] and 2.37 m/s [69.2 degrees S of E]. What are the magnitude and direction of the unknown initial velocity?

Homework Equations



mv1 + mv2 = mv1' + mv2'

The Attempt at a Solution



First off, I drew the resulting velocities, and found their components.

The x and y components of the 2.49 m/s velocity are -2.49cos62.8 and 2.49sin62.8 respectively. The x and y components of the 2.37 velocity are 2.37 cos 69.2, and -2.37sin69.2 respectively. Knowing this, I figure I have to incorporate the components somewhere into the equation - I just don't know where! Given the above equation, since the masses are equal, I can eliminate the masses from the equation entirely - likewise, I can eliminate the 1/2, by multiplying the entire equation by two - thus, I have a simplified equation ... I just don't know where to go from here. Can anybody give me a hint as to what to do with the components? Any insight is greatly appreciated!

The answer is supposed to be 3.00 m/s, for the initial velocity of the unknown.
 
Physics news on Phys.org
Your inelastic collision will result in the conservation of momentum. Total before is total after as you have given.

Draw a careful vector diagram. Note that you can use vector addition by adding the x and y components separately and still insure that Total momentum before is the same as Total momentum after in each axis.

A careful diagram will help you keep the signs of the values straight.
 
Okay, I redrew the vector diagram.
But what do I do once I have my components? :/
I don't know what to do with them.
 
pitaaa said:
Okay, I redrew the vector diagram.
But what do I do once I have my components? :/
I don't know what to do with them.

Doesn't the sum of the x components of the velocity vectors (you've discarded mass as both are the same, but momentum vectors if you were still carrying the mass) before properly add to the sum of the after collision velocity (or momentum) components? The same for the y components?
 
Hmm I'll try it :/ Unless my diagram is totally wrong :P
 
i kinda have a proble like that but in mine i have a 5,000 kg railroad car moving at 2 m/s collides and connects to another identical car intially at rest.what is the finial velocity of the two connected train cars?
 
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 '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