How do you calculate the speed of a bullet using a ballistic spring system?

AI Thread Summary
To calculate the speed of a bullet using a ballistic spring system, the conservation of momentum and energy principles are applied. The maximum compression of the spring, denoted as d, along with the spring constant k, the mass of the block M, and the mass of the bullet m, are key variables in the equations. The initial speed of the bullet, vB, can be derived from the relationship involving the final velocity of the bullet-block system after the bullet is fired. The correct expression for the bullet's speed is vB = sqrt((k*d^2(m + M))/m^2). This formula incorporates the necessary parameters to accurately determine the bullet's velocity.
1831
Messages
13
Reaction score
0

Homework Statement



You have been asked to design a "ballistic spring system" to measure the speed of bullets. A spring whose spring constant is k is suspended from the ceiling. A block of mass M hangs from the spring. A bullet of mass m is fired vertically upward into the bottom of the block. The spring's maximum compression d is measured.

Find an expression for the bullet's speed.
Express your answer in terms of the variables m, M, k, d, and constant g.


Homework Equations



Ki + Ug + Usp = Kf + Ug + Usp

The Attempt at a Solution



I used conservation of momentum to find the final velocity of the bullet+block

(m+M)vf=mvi + Mvi

so...

vf=(m/m+M)*vB

where vB is the initial speed of the bullet.

Next, I used:
Ki + Ug + Usp = Kf + Ug + Usp

(1/2)mvi^2 + (1/2)kd^2 + Mgd = (1/2)mvf^2 + (1/2)kd^2 + Mgd
0 + (1/2)kd^2 + 0 = (1/2)(m+M)[(m+M)^2*vB^2] + 0 + Mgd

I found the answer to be:

((kd^2 - Mgd)(m+M))/m^2 = Vb

but this was not correct...please help me.
 
Physics news on Phys.org
I got it...nevermind
 
Could someone please explain this, I got the following, but it is not correct

vB = (k*d^2(m + M))/m^2

I figured it out, i was missing the square root. The answer is:

vB = sqrt((k*d^2(m + M))/m^2)
 
Last edited:
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