Magnitude of an Impulse and Plotting it? (What am I doing?)

Click For Summary
The discussion focuses on calculating impulse and momentum in a physics problem. The initial calculation for impulse, J = m(2/3)v0 - mv0, raises uncertainty about its correctness. For part b, the user expresses confusion over the relationship between force, time, and momentum change, suggesting F * t = m(2/3)v0 - v0. The importance of considering direction in momentum is emphasized, particularly regarding the signs of initial and final velocities. The problem also involves treating the system as a point mass with a spring, despite this not being entirely accurate.
cjedow
Messages
1
Reaction score
0
Homework Statement
A small rubber ball of mass m is thrown at a brick
wall. Initially, the ball has a speed v0 and is traveling perpendicular to the wall. After
the ball hits the wall and rebounds, it is traveling perpendicular to the wall at speed
(2/3)v0. The ball is in contact with the wall for a time Δt. (Assume that Δt is so short
that gravity doesn’t appreciably change the ball’s momentum.)
a. What is the magnitude of the impulse from the force of the wall on the ball?
b. As quantitatively as possible, plot the instantaneous force
exerted by the ball on the wall versus time. On the plot,
indicate the average force exerted by the wall on the ball.
Relevant Equations
J= m * vf - m * vi
J=F * t
I wrote:
J = m(2/3)v0 - mv0 for part a but I'm not sure if that's correct or if I need something else?
I have no idea how to do part b. I wrote F * t = m(2/3)v0 - v0 but I honestly don't know what I'm doing.
 
Physics news on Phys.org
Momentum has direction. If it were to hit at speed v and bounce back at speed v there would be no change to the magnitude of the momentum, but the magnitude of the change in momentum would not be zero.
To put it another way, be careful with the signs of vi and vf.

For part b, assume it behaves like a point mass with a light uniform spring in front... though that is not strictly true.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
Replies
6
Views
787
  • · Replies 1 ·
Replies
1
Views
943
  • · Replies 4 ·
Replies
4
Views
919
  • · Replies 4 ·
Replies
4
Views
5K
Replies
7
Views
1K
Replies
1
Views
4K
Replies
38
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K