How does the kinetic energy change for a varying mass?

Click For Summary
SUMMARY

The discussion centers on the change in kinetic energy for a particle with varying mass, specifically contrasting it with a particle of constant mass. The established equation for constant mass is derived as {dT/dt} = \vec F \cdot \vec v, where T represents kinetic energy, m is mass, and v is velocity. For varying mass, the equation changes to {d(mT)/dt} = \vec F \cdot \vec p, with p representing momentum. The initial proof for constant mass is confirmed, but further assistance is requested for proving the varying mass scenario.

PREREQUISITES
  • Understanding of classical mechanics principles, particularly Newton's laws of motion.
  • Familiarity with the concepts of kinetic energy and momentum.
  • Knowledge of calculus, specifically differentiation and its application in physics.
  • Experience with vector notation and operations in physics.
NEXT STEPS
  • Explore the derivation of the kinetic energy equation for varying mass systems.
  • Study the relationship between force, momentum, and mass in dynamic systems.
  • Investigate advanced topics in classical mechanics, such as Lagrangian mechanics.
  • Learn about the implications of variable mass in rocket propulsion and other applications.
USEFUL FOR

Students and professionals in physics, particularly those focused on mechanics, as well as engineers working with systems involving variable mass, such as rockets and fluid dynamics.

Reshma
Messages
749
Reaction score
6
Show that for a single particle with constant mass the equation of motion implies the following differential equation for the kinetic energy:
[tex]{dT\over dt} = \vec F \cdot \vec v[/tex]

while if the mass varies with time the corresponding equation is
[tex]{d(mT)\over dt} = \vec F \cdot \vec p[/tex]

Proof:

I was able to prove the first part:
[tex]T = {1\over 2}mv^2[/tex]
[tex]{dT\over dt} = {1\over 2} {d(mv^2)\over dt} = \vec v \cdot m{d\vec v \over dt} = \vec F \cdot \vec v[/tex]

But I am unable to prove the second part. Please help.
 
Physics news on Phys.org
what does mT equal in terms of m and v? in terms of p?
mv=p
F=dp/dt

try to play with the equation a little and substitute p in.
 

Similar threads

Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
2K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
7
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K