Calculating Force and Speed of a Snooker Ball

  • Thread starter Thread starter glasgowm
  • Start date Start date
  • Tags Tags
    Ball Force Speed
AI Thread Summary
A cue exerts a force of 7N on a stationary snooker ball with a mass of 200g over an impact duration of 45ms. The impulse-momentum principle indicates that the impulse (force multiplied by time) equals the change in momentum of the ball. The equation F·t = mv2 - mv1 can be used to find the final velocity (v2), with the initial velocity (v1) being zero. The calculations provided suggest that substituting the values into the equation should yield the speed of the ball after impact. The discussion emphasizes the importance of correctly applying the impulse-momentum formula to determine the ball's speed.
glasgowm
Messages
16
Reaction score
0
A cue exerts an average force of 7N on a stationary snooker ball of mass 200g if the impact lasts for 45ms, with what speed does the ball leave the que?

---

Ft = mv - mu
F = m(v-u/t)
F = 0.2(v-0/0.0045)
= 0.2v/0.0045 = 44.4444V

Can't figure out what the next step is from my notes.

Cheers
 
Physics news on Phys.org
glasgowm said:
A cue exerts an average force of 7N on a stationary snooker ball of mass 200g if the impact lasts for 45ms, with what speed does the ball leave the que?

---

Ft = mv - mu
F = m(v-u/t)
F = 0.2(v-0/0.0045)
= 0.2v/0.0045 = 44.4444V

Can't figure out what the next step is from my notes.

Cheers

Simply stated, the impulse equals the change of linear momentum, so, you have: F\cdot t = mv_{2}-mv_{1}, where m is the mass of the ball, v2 the final velocity and v1 the initial velocity (equals zero). From this equation you can easily retrieve v2.
 
radou said:
Simply stated, the impulse equals the change of linear momentum, so, you have: F\cdot t = mv_{2}-mv_{1}, where m is the mass of the ball, v2 the final velocity and v1 the initial velocity (equals zero). From this equation you can easily retrieve v2.
I already did that.
 
glasgowm said:
I already did that.

I saw you did that and I don't see where the problem is. Just plug in the force and solve to get the speed v.
 
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