Impulse from croquet mallete problem

  • Thread starter Thread starter in10sivkid
  • Start date Start date
  • Tags Tags
    Impulse
AI Thread Summary
The discussion revolves around calculating the speed of a croquet ball after being struck by a mallet that delivers an impulse of 8.83 N*s. The impulse is equated to the change in momentum, which is zero initially since the ball is at rest. Using the formula for momentum (p = mv), where mass is given as 0.44 kg, the equation simplifies to 8.83 N*s = (0.44 kg) * V. Solving for V reveals that the calculation is straightforward. The key takeaway is that impulse directly relates to momentum, allowing for the determination of velocity.
in10sivkid
Messages
36
Reaction score
0
A croquet mallete delievers an impulse of 8.83 N*s to a .44 kg croquet ball initally at rest. what is the speed of the ball immediately after being struck?

ok so I = Ft = change in momentum

and they give us just a mass...

so that would help with p = mv to find velocity...

but this problem seems there are two unknowns i would either need to know the force or the time...but they don't give us either...

help?
 
Physics news on Phys.org
all you need is the velocity, and you have I.
 
ok...how does having impulse help me with finding velocity
 
in10sivkid said:
ok...how does having impulse help me with finding velocity

The impulse is the change in momentum, and the initial momentum is zero. Momentum is mass times velocity. You know the mass.
 
ok so essentially then then impulse = momentum since the initial momentum = 0

so 8.83 N*s = (.44 kg) * V

then solve for V...??
if so wow that was kinda easy
 
in10sivkid said:
ok so essentially then then impulse = momentum since the initial momentum = 0

so 8.83 N*s = (.44 kg) * V

then solve for V...??
if so wow that was kinda easy

Looks good. Some of them are easy :smile:
 
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