# Speed of a bullet in ballistic spring system

## Homework Statement

You have been asked to design a "ballistic spring system" to measure the speed of bullets. A bullet of mass 6.20 g is fired into a block of mass 2.10 kg. The block, with the embedded bullet, then slides across a frictionless table and collides with a horizontal spring whose spring constant is k=63.0 N/m. The opposite end of the spring is anchored to a wall. What was the speed of the bullet if the spring's maximum compression is 12.1 cm?

pinitial=pfinal
E=0.5*k*x2
E= 0.5*m*v2

## The Attempt at a Solution

m1= bullet and m2= block; v1= vbullet

i first solved m1v1+m2v2= mtotal*vfinal for Vfinal..so that looked like vfinal= m1*v1/mtotal

then i used the energy equations to get 0.5*k*x2= 0.5*m*v2
in v2 i substituted m1*v1/mtotal
and then tried to solve for v1 which is the speed of the bullet we are looking for ...and my hwk app. is telling me my answer is wrong...so i was wondering if i made a mistake in the workings somewhere..

## Answers and Replies

Doc Al
Mentor
Your method looks fine. Provide your final equation for v in terms of m1 and m2.

Also: Check that you haven't made an arithmetic error.

this is how my equation looks like
(k*x2)/mbullet= (m1*v1/mtotal)2

$$\sqrt{}(k*x^2)/mbullet$$ *Mtotal/mbullet = vbullet2

the square root only applies to the k*x2/mbullet

does that look right ??

Doc Al
Mentor
this is how my equation looks like
(k*x2)/mbullet= (m1*v1/mtotal)2
That should be:
(k*x2)/mtotal= (m1*v1/mtotal)2

i got the right answer but was wondering why it wouldnt be mass of bullet that would be divided by the energy of spring>>??

Doc Al
Mentor
i got the right answer but was wondering why it wouldnt be mass of bullet that would be divided by the energy of spring>>??
Because after the collision, the KE of the system is ½mtotalv2. The bullet and block are treated as one unit.