Help with a Simple Harmonic Motion problem

AI Thread Summary
A 25.0g bullet strikes a 0.600kg block attached to a spring, causing it to vibrate with an amplitude of 21.5cm. The spring constant is 6.70x10^3 N/m, and the combined mass of the block and bullet after impact is 0.625kg. The calculated velocity of the block and bullet together is 22.26 m/s, but the correct speed of the bullet before impact is 557 m/s. To find this, conservation of momentum must be applied, equating the bullet's momentum before impact to the combined momentum after impact. The discussion emphasizes the importance of using momentum conservation to solve for the bullet's speed accurately.
ceeforcynthia
Messages
11
Reaction score
0
1. A 25.0g bullet strikes a .600kg block attached to a fixed horizontal spring whose spring constant is 6.70x10^3 and sets it into vibration with an amplitude of 21.5cm. What is the speed of the bullet before impact if the two objects move together after impact?2. V=A\sqrt{k/m}
3. I used the equation above, adding the masses together (.625kg). I got the velocity to be 22.26m/s... but now i can't figure out how to get the speed of the bullet.


The answer in the book says that the speed of the bullet is 557m/s.
 
Last edited:
Physics news on Phys.org
ceeforcynthia said:
1. A 25.0g bullet strikes a .600kg block attached to a fixed horizontal spring whose spring constant is 6.70x10^3 and sets it into vibration with an amplitude of 21.5cm. What is the speed of the bullet before impact if the two objects move together after impact?


2. V=A\sqrt{k/m}



3. I used the equation above, adding the masses together (.625kg). I got the velocity to be 22.26m/s... but now i can't figure out how to get the speed of the bullet.


The answer in the book says that the speed of the bullet is 557m/s.

Use conservation of momentum. momentum of the bullet before the impact = momentum of the block plus bullet moving together at 22.26 m/s
 
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...

Similar threads

Back
Top