Solving Impulse & Momentum Problems

In summary, the problem deals with Impulse and Momentum involving two balls of different masses and initial speeds. The first question asks for the speed of Ball A just before impact using the principle of conservation of mechanical energy. The second question asks for the velocities of both balls after an elastic collision. The equations used are .5 m(vf^2)+m(g)hf = .5m(v0^2)+m(g)h0 and vf=5.56. It is important to write out the equations and explain the reasoning behind them in order to show a good understanding of the concepts being tested.
  • #1
Jared944
10
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!
 

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  • #2
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?
 
  • #3
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!
 
  • #4
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.
 

Related to Solving Impulse & Momentum Problems

1. What is impulse and momentum?

Impulse and momentum are two concepts in physics that are closely related. Impulse refers to the change in momentum of an object, while momentum is a measure of an object's motion, determined by its mass and velocity.

2. How do you calculate impulse?

Impulse is calculated by multiplying the force applied to an object by the time interval over which that force is applied. The formula for impulse is J = F * Δt.

3. How do you solve impulse and momentum problems?

To solve impulse and momentum problems, you need to use the laws of conservation of momentum and conservation of energy. These laws state that the total momentum and total energy of a system must remain constant, even after a collision or other interaction between objects.

4. What are some real-world applications of impulse and momentum?

Impulse and momentum are important concepts in many fields, including sports, transportation, and engineering. For example, the force and time of impact in a car crash can be calculated using impulse and momentum, and understanding these concepts is crucial for designing effective safety features in vehicles.

5. How do you differentiate between elastic and inelastic collisions?

In elastic collisions, the total kinetic energy of the system is conserved, meaning that no energy is lost during the collision. In inelastic collisions, some of the kinetic energy is converted into other forms of energy, such as heat or sound. This results in a decrease in the total kinetic energy of the system.

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