Need help going over a momentum/spring problem

  • Thread starter Thread starter groundhog92
  • Start date Start date
AI Thread Summary
A pellet gun fires a bullet into a block of wood attached to a spring, resulting in simple harmonic motion with an amplitude of 11 cm. The bullet's initial speed is 650 m/s, and calculations show the spring constant is approximately 349.9 N/m. The total energy of the system after the collision is calculated to be 4.23 J, considering the maximum kinetic and potential energy. The maximum acceleration of the bullet/block system is determined to be 15.4 m/s². The velocity of the bullet/block reaches zero at the maximum displacement of ±11 cm.
groundhog92
Messages
7
Reaction score
0
1.Homework Statement
A pellet gun is fired into a block of wood that is attached to a spring on a frictionless surface. The bullet remains in the wood and the block of wood and the bullet enter into a simple harmonic motion with Amplitude=11cm. The 5g bullet was traveling at 650m/s before hitting the 2.5kg block of wood.
a)What is the spring constant?
b)What is the the total energy of the system after the collision?
c)What is the max acceleration of the bullet/block once it begins its oscillation?
d)Where will the bullet/block reach a velocity of 0?


Homework Equations


mv=mv
(1/2)mv^2=(1/2)kx^2
a=(V^2)/A
v=2pief(sqrt(A^2-x^2))



The Attempt at a Solution


I've already done the problem I just need someone to check my work because I'm not completely sure about the process.

Part a
(.005)(650)=(2.505)v
v=1.3m/s

(1/2)k(.11^2)=(1/2)(2.505)(1.3^2)
k=349.9N/m

Part b
E=(1/2)mv^2 + (1/2)kx^2
E=((2.505)(1.3^2)+(349.9)(.11^2))/2
E=4.23J

Part c
a=(v^2)/A
a=(1.3^2)/.11
a=15.4m/s^2

Part d
Where A=x
so where x=+-11cm

Thanks in advance for any help.
 
Physics news on Phys.org
In part b) you have added the max KE to the max PE.
When KE is max, PE is zero and vice versa.
 
So the total energy of the system would be equal to just the max potential or kinetic by themselves?
 
Yep!
 
alright thanks
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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