Calculate Change in Momentum of Billiard Ball | CaptFormal

Click For Summary
To calculate the change in momentum of a billiard ball, the vertical and horizontal components of its initial and final velocities must be determined. The initial velocity components are calculated using the angle of incidence, while the final components use the angle of reflection. The user initially calculated the horizontal change incorrectly, not accounting for the direction of the components. The correct approach involves taking the negative of the initial horizontal component when calculating the change. The final step requires using the resultant velocity change and multiplying it by the mass to find the change in momentum.
CaptFormal
Messages
31
Reaction score
0

Homework Statement



A billiard ball of mass m = 0.15 kg strikes the cushion of a billiard table at θ1 = 48° and a speed v1 = 21 m/s. It bounces off at an angle of θ2 = 670 and a velocity of v2 = 16 m/s. What is the magnitude of its change in momentum (in kg·m/s)?

http://schubert.tmcc.edu/res/msu/mmp/kap6/picts/pool.gif


Homework Equations





The Attempt at a Solution


Not sure how to solve this one. Any help will be appreciated.


CaptFormal
 
Last edited by a moderator:
Physics news on Phys.org
Take the vertical and horizontal components of v1 and v2.
Find the difference in vertical and horizontal components. Take care of signs.Then take the resultant of these components.
 
Ok, so here is what I got so far.

Vx = 21cos(48) - 16cos(67) = 7.8

Vy = 21sin(48) - 16sin(67) = 0.8779

Now I am not sure what to do. I tried the following:

(7.8^2 + 0.8779^2)^(1/2)

and then took that answer and multiplied it by the mass but it was incorrect. What am I missing?
 
Vx = 21cos(48) - 16cos(67) = 7.8
This is wrong. vx components are in the opposite direction. So the change in vx is
Vx = - 21cos(48) - 16cos(67) = ?
 
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 1 ·
Replies
1
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
3K
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
10K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K