Conservation of Momentum, Energy

AI Thread Summary
A bullet with a mass of 20 grams and a velocity of 100 m/s collides with a stationary wooden block of 2 kg, which is connected to a spring with a spring constant of 800 N/m. The conservation of momentum is applied to find the final velocity after the collision, followed by using energy conservation to determine the maximum compression of the spring. The initial calculations led to confusion, with the incorrect answer of 0.49 meters being noted. After recalculating with more precision, the correct maximum compression of the spring was found to be approximately 0.0497 meters. This highlights the importance of accuracy in calculations when applying physics concepts.
SuperCass
Messages
60
Reaction score
0

Homework Statement



A bullet with mass 20 grams and velocity 100 m/s collides with a wooden block of mass 2 kg. The wooden block is initially at rest, and is connected to a spring with k = 800 N/m. The other end of the spring is attached to an immovable wall. What is the maximum compression of the spring?

Homework Equations



p=mv
K=.5mv^2
Spring potential energy = .5kx^2

The Attempt at a Solution



I tried doing M(bullet)V(initial) = [M(bullet) + M(block)][Vfinal], and then I used that final velocity in the energy equation .5mv^2 = .5kx^2 to find x. I keep getting the answer to be .49 meters, but this isn't correct.

What am I doing wrong?
Thanks!
 
Physics news on Phys.org
I think you have the concepts correct and are just doing the numbers wrong. With these inputs, 0.5mv^2~=1J = 0.5 kx^2, so x^2 ~= 1/400 m^2. How are you getting 0.49m?
 
Whoops, I mean I keep getting .049 meters.

But still, doing what you said isn't giving me the right answer!
 
Okay, I got it, never mind!

I was doing it correctly, I just didn't put in an exact enough answer!

It ended up being .0497!
 
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...
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...
Back
Top