Solving Impulse & Momentum Problems

  • Thread starter Thread starter Jared944
  • Start date Start date
  • Tags Tags
    Impulse Momentum
Click For Summary
The discussion focuses on solving impulse and momentum problems involving two balls, A and B, with specific masses and initial conditions. The first question requires finding the speed of Ball A just before it strikes Ball B using conservation of mechanical energy principles. The user initially struggled with the setup but later realized the correct approach involves applying the energy conservation equation. After clarification, they successfully calculated the final speed of Ball A as 5.56 m/s. The conversation emphasizes the importance of clearly writing out equations before substituting values to demonstrate understanding in academic settings.
Jared944
Messages
10
Reaction score
0
Ok, so this problem is kind of complicated, so ask me if you need clarification. This problem deals with Impulse and Momentum.

There are two balls hanging vertically from a horizontal plane. Ball A is pulled back and is set up to strike ball B.
Given for ball A -
mass = 1.5 kg.
Height = .3 m (measured from a horizontal plane from Ball B)
initial speed = 5.00 m/s (as it leaves your hand)
Given for Ball B -
mass = 4.60 kg
speed = 0 m/s (implied)

Question a) is -
Using the principle of conservation of mechanical energy, find the speed of Ball A just before impact.
Question b) is -
Assuming an elastic collision, find the velocities (magnitude and direction) of both balls just after the collision.

I can't quite set up part A, I know I have to integrate a kinematics equation and the principle of conservation of mechanical energy, but I can't quite get it right.
Thanks!
 

Attachments

  • balls.jpg
    balls.jpg
    2.5 KB · Views: 482
Physics news on Phys.org
By integrate do you mean calculus and take the integral? Because that is not needed here. This question is to test whether you understand what conservation of energy and conservation of momentum is. If you wouldn't mind, could you please briefly explain to me what these 2 concepts mean?
 
Ahh, I just figured it out. I was using the wrong set of equations, I am sorry! Part a) can be figured by using the equation -
.5 m(vf^2)+m(g)hf = .5m(v0^2)+m(g)h0

Or, in other words, .5(1.5)(vf^2)+(1.5)(9.8)(0)=.5(1.5)(5^2)+(1.5)(.3)
vf=5.56!

Thanks for the nudge in the right direction!
 
You should get into the habit of writing out the equation first, before putting in numbers. This way even if you accidently put in the wrong numbers or do the calculations wrong, at least this shows your teacher/professor that you understand what is going on. So he/she will still give you a good mark. And you could even take 1 step further and explain why you are equating the equations you are equating.
 
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
10
Views
1K
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
920
  • · Replies 4 ·
Replies
4
Views
899