Solving Elastic Collision Homework: Average Force on a 100g Ball

AI Thread Summary
To find the average force exerted by the wall on a 100 g ball bouncing elastically, the initial speed is 3 m/s, and the collision lasts 10 ms. The average force can be calculated using the formula F = m(vf - vi)/Δt, where vf is the final velocity. Since the collision is elastic, the final velocity will be -3 m/s (opposite direction), not the same as the initial velocity. This results in a non-zero average force, which is essential for solving the problem correctly. Understanding that momentum is a vector quantity and direction matters is crucial in this scenario.
slanderson113
Messages
2
Reaction score
0

Homework Statement


A 100 g ball bounces off a wall elastically. Its initial speed is 3 m/s perpendicular to the wall. If the collision lasts for 10 ms, what is the average force exerted by the wall on the ball?


Homework Equations


F=ma
Conservation of energy & momentum (?)
F=dP/dt


The Attempt at a Solution


I know to find average force the equation is m(vf-vi)/change in time. And in the equation were given the mass, time, and initial velocity, but to finish I need the final velocity but the conservation of energy and momentum, shows it should be the same as the initial making the force zero, however that's not one of the options. Please help!
 
Physics news on Phys.org
welcome to pf!

hi slanderson113! welcome to pf! :smile:
slanderson113 said:
… the conservation of energy and momentum, shows it should be the same as the initial …

no, momentum is a vector, and it'll be opposite :wink:
 
slanderson113 said:
And in the equation were given the mass, time, and initial velocity, but to finish I need the final velocity but the conservation of energy and momentum, shows it should be the same as the initial making the force zero, however that's not one of the options. Please help!
Realize that velocity is a vector and thus direction matters. The direction of the velocity is represented by its sign. (Do you still think that the initial and final velocities are the same?)
 
Oh my gosh thank you guys so much!
 
Last edited:
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