Velocity of 50g Putty+50g Mass After Collision

  • Thread starter Thread starter erinbrattin
  • Start date Start date
  • Tags Tags
    Mass Velocity
AI Thread Summary
In an inelastic collision involving a 50 g ball of putty moving at 3.0 m/s colliding with a stationary 50 g mass, the principle of conservation of linear momentum applies. The total momentum before the collision equals the total momentum after the collision. The combined mass after the collision is 100 g, and the final velocity can be calculated using the formula: (mass1 * velocity1 + mass2 * velocity2) / (mass1 + mass2). Substituting the values gives a final velocity of 1.5 m/s for the combined mass of putty and the stationary object. This demonstrates how momentum conservation is crucial in solving inelastic collision problems.
erinbrattin
Messages
5
Reaction score
0
a 50 g ball of putty moving with a velocity of 3.0 m/s has an inelastic collision head-on with a stationary mass of 50g and sticks to it. find the velocity of the mass and putty after the collision.
 
Physics news on Phys.org
Do you remember any equations to use for this type of problem?
 
just think about that inelastic collision means and use conservation of linear momentum.

Regards,

Nenad
 
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