What is the Kinetic Energy of Bullets Fired at Different Speeds?

AI Thread Summary
The discussion focuses on calculating the kinetic energy of two bullets with masses of 3.5 grams, fired at speeds of 42.4 m/s and 69.6 m/s. The kinetic energy formula used is EK=(1/2)mv^2, but the initial calculation was incorrect due to not converting the mass from grams to kilograms. After correcting the mass to 0.0035 kg, the kinetic energy for the first bullet was calculated to be approximately 3.14 Joules. The correct approach emphasizes the importance of unit conversion in physics calculations. The thread highlights the significance of accuracy in applying formulas to achieve correct results.
PepeF.
Messages
16
Reaction score
0

Homework Statement



two 3.5g bullets are fired with speeds of 42.4 m/s and the 2nd one with 69.6 m/s

what is the kinetic energy of the first bullet?
the second one?


Homework Equations



EK=(1/2)mv^2

The Attempt at a Solution



EK=(1/2)3.6*42.4^2 = 3235.968

it is wrong.. why?
 
Physics news on Phys.org
units
 
you have not converted the mass of the bullet from grams to kgrams

0.50 x (3.5x10^-3) x 42.4^2 = 3.14J

this is the answer i got
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

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...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging 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...
View full post »

Similar threads

Conservation of Momentum Problem - Mechanical and Kinetic Energies
Replies
2
Views
898
Confused about a conservation of energy problem
Replies
4
Views
2K
What is the temperature change of a bullet upon impact?
Replies
28
Views
3K
A bullet collides perfectly elastically with one end of a rod
Replies
21
Views
2K
Change in the Kinetic Energy of a System
Replies
2
Views
2K
Potential and Kinetic energy equations including drag coefficient
Replies
1
Views
1K
Falling stick problem (no friction): What is the kinetic energy?
Replies
3
Views
1K
Using Orbital Energy to Calculate Velocity
Replies
3
Views
1K
Hooke's Law using Potential Energy
Replies
21
Views
1K
How Is Potential Energy Converted to Kinetic Energy in Physical Systems?
Replies
6
Views
1K

Hot Threads

Recent Insights

Back
Top