A head-on Collision (Example #7) page 205 Cutnell & Jhonson Physics 8e

  • Thread starter Thread starter gcombina
  • Start date Start date
  • Tags Tags
    Collision Physics
AI Thread Summary
In an elastic collision involving two balls, one with a mass of 0.250 kg moving at 5.00 m/s and another with a mass of 0.800 kg at rest, the conservation of kinetic energy is applied. The formula used, 1/2 m1vf1^2 + 1/2 m2Vf2^2 = 1/2 m1V01^2, represents the total kinetic energy before and after the collision. The factor of 1/2 is part of the kinetic energy formula, which includes the mass and the square of the velocity. The initial kinetic energy of the moving ball is calculated to be 3.125 J. Understanding this formula is crucial for determining the final velocities of both balls post-collision.
gcombina
Messages
157
Reaction score
3
One ball has a mass of m1 = 0.250 kg and an initial velocity of 5.00 m/s. The other has a mass of m2=0.800 kg and is initially at rest. No external forces act on the balls. What are the velocities of the balls after the collision? Elastic Collision.
On the book I have this formula.

STEP 1#

1/2 m1vf1^2 + 1/2 m2Vf2^2 = 1/2 m1V01^2 + 0

(I am confused for the 1/2 at the beginning and why is squares?)

WHAT?
 
Physics news on Phys.org
gcombina said:
1/2 m1vf1^2 + 1/2 m2Vf2^2 = 1/2 m1V01^2 + 0
(I am confused for the 1/2 at the beginning and why is squares?)
That is the formula of KE and states that the sum of the energy of the balls will be equal to 1/2 0.250*5^2 = 3.125J
 
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...

Similar threads

Replies
2
Views
2K
Replies
16
Views
3K
Replies
4
Views
5K
Replies
5
Views
2K
Replies
1
Views
1K
Replies
18
Views
3K
Replies
5
Views
3K
Back
Top