Simple question on Ratio of Kinetic to rest energy

AI Thread Summary
To find the ratio of kinetic energy to rest energy for a baseball thrown at 44 m/s, the kinetic energy can be calculated using the formula K=1/2mv^2 since the speed is much less than the speed of light. The rest energy is given by E=mc^2. The ratio is determined by dividing the kinetic energy by the rest energy. This exercise highlights the vast difference between kinetic energy at everyday speeds and the much larger rest energy. The discussion emphasizes the appropriateness of using classical mechanics for this scenario.
MillerGenuine
Messages
64
Reaction score
0

Homework Statement



A pitcher can throw a baseball at 44 m/s. what is the ratio of kinetic energy to the rest energy E=mc^2. (can you use k=1/2mv^2 ?)

Homework Equations



rest energy=mc^2
K=gamma*mc^2 - mc^2
k=1/2mv^2

The Attempt at a Solution


I am just not sure how to go about this problem. It seems i can use k=1/2mv^2 because the velocity is well below c. From there i would take me kinetic energy and divide it by the rest enregy to get the ratio..??
 
Physics news on Phys.org
MillerGenuine said:

Homework Statement



A pitcher can throw a baseball at 44 m/s. what is the ratio of kinetic energy to the rest energy E=mc^2. (can you use k=1/2mv^2 ?)

Homework Equations



rest energy=mc^2
K=gamma*mc^2 - mc^2
k=1/2mv^2

The Attempt at a Solution


I am just not sure how to go about this problem. It seems i can use k=1/2mv^2 because the velocity is well below c. From there i would take me kinetic energy and divide it by the rest enregy to get the ratio..??

Yep. They just want to show how huge the rest energy is...
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top