Conservation of momentum and energy

In summary, we used various equations such as the elastic collision formula, inelastic collision, impulse, work-energy theorem, and the change in kinetic energy formula to solve for the acceleration due to friction, frictional force, work done by friction and the change in kinetic energy for a cart at different times. We also used the elastic collision formula to calculate the final velocity of a cart after colliding with another cart of different masses, and found that the first cart's final velocity was dependent on the mass of the second cart. Finally, we saw that if the two carts stick together after the collision, the first cart's final velocity is dependent on the total mass of the two carts.
  • #1
JJBrian
20
0

Homework Statement


a) At t=1 second, a 0.5 kg cart is at x=0.5 m, traveling at 0.5 m/s. At t=3 seconds, the cart is
at 1.4 m, traveling at 0.4 m/s. What is the acceleration due to friction, the frictional force and
the work done by friction? What is the change in kinetic energy of the cart?
b) If a cart of mass 0.6 kg, traveling a 0.5 m/s collides elastically with a cart of mass 0.4 kg,
initially at rest, what is the final velocity of the first cart?
c) If the two carts above stick together after the collision, what is the final velocity of the first
cart?
d) What is the final velocity of the first cart in (b) if the second cart has mass 1.5 kg?
e) What is the final velocity of the first cart in (c) if the second cart has mass 1.5 kg?



Homework Equations


Elastic collison formula
Inelastic collision
Impulse
W=-FD
Change in kinetic energy formula

The Attempt at a Solution


ok here is my attempt. I need someone to check if I am using the right equation for each problem.

part a)
i used vf^2-vi^2 = 2ad
to get an acceleration of .05

To find the firctional force i used
F = -ma F= (.5)(.05)
Ff = -.025N

work done by friction
W = -Fd
W =-(-.025)(.05)
W = 0.0125J

change in ke
Ke =1/2mvf^2-1/2mvi^2
KE=1/2(.5)(0.5)^2-.5(0.5)(0.4)^2
Kef = .0225



part b)
v1f = vi*(m1-m2)/(m1+m2)
0.5*(0.6-0.4)/(0.6+0.4)
v1f=0.1m/s

part c)
v1f = vi*(m1)/(m1+m2)
(0.5)(.6)/(0.6+0.4)
v1f=0.3m/s


part d
v1f = vi*(m1-m2)/(m1+m2)
0.5*(0.6-1.5)/(0.6+1.5)
v1f=0.2143m/s


part e

v1f = vi*(m1)/(m1+m2)
(0.5)(.6)/(0.6+1.5)
v1f=0.143m/s
 
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  • #2


Your solutions for part a) seem to be correct. For part b), you are using the correct formula for elastic collisions, but I believe you made a mistake in your calculation. It should be v1f = vi*(m1-m2)/(m1+m2) = 0.5*(0.6-0.4)/(0.6+0.4) = 0.1 m/s. For part c), your solution is correct. For part d), you have the right formula, but I believe you made a mistake in your calculation. It should be v1f = vi*(m1-m2)/(m1+m2) = 0.5*(0.6-1.5)/(0.6+1.5) = -0.2143 m/s. The negative sign indicates that the first cart is moving in the opposite direction after the collision. For part e), your solution is correct. Good job!
 
  • #3


Your attempt is correct. You have used the appropriate equations for each part and your calculations are also correct. Well done!
 

Related to Conservation of momentum and energy

1. What is the conservation of momentum and energy?

The conservation of momentum and energy is a fundamental law of physics that states that the total momentum and energy of a closed system remains constant over time. This means that in any interaction or process, the total amount of momentum and energy before and after the interaction remains the same.

2. Why is the conservation of momentum and energy important?

The conservation of momentum and energy is important because it helps us understand and predict the behavior of objects and systems in motion. It allows us to accurately describe and analyze various physical phenomena, from the motion of planets to the collision of particles in a particle accelerator.

3. How is the conservation of momentum and energy related?

The conservation of momentum and energy are related in that they are both fundamental principles of physics that govern the behavior of objects and systems in motion. Momentum is defined as the product of an object's mass and velocity, while energy is the ability to do work or cause change. Both momentum and energy are conserved in any interaction or process.

4. Can the conservation of momentum and energy be violated?

No, the conservation of momentum and energy cannot be violated. This law has been extensively tested and has been found to hold true in all observed interactions and processes. If there appears to be a violation of this law, it is likely due to incomplete or inaccurate measurements rather than an actual violation of the law.

5. How does the conservation of momentum and energy apply to real-world situations?

The conservation of momentum and energy applies to a wide range of real-world situations. For example, it explains why a ball bounces off the ground after being dropped, why a rocket accelerates when its engines are fired, and even why the Earth continues to orbit the sun without losing energy. This law is essential in understanding the behavior of objects and systems in our everyday lives.

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