Bullet fired into a block attached to a spring

  • #1

Homework Statement



A bullet mass(b) .008 kg is fired (and embedded) into a block mass(B) 0.992 kg on a horizontal frictionless surface. The spring is compressed 0.15 m on impact. The spring's constant is 300 N/m.
Q: What is the block's velocity after impact
Q: What is the initial speed of the bullet?

Homework Equations



I want to say that this is a completely inelastic collision since the two masses have the same final velocity (2)? so in that case.... (k2/k1) = m(b)/[(m(b)+m(B)]
also,
m(b)v(b) + m(B)v(B) = [(m(b)+m(B)]v2
and U(el) ...potential energy of an elastic... is 1/2kx^2
W(el) ....work done by spring.... is -(change in potential energy).

The Attempt at a Solution



Im not sure what to do with so many unknowns. Also not sure if I need to use the work done by a spring or the potential energy of a spring. Any help to get me in the right direction is much appreciated.
 

Answers and Replies

  • #2
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
You know after impact, the block must have enough initial kinetic energy to depress the spring by 0.15 m. How much kinetic energy does that much elastic potential energy correspond to?
 
  • #3
so the elastic potential energy of the spring when it is compressed is 0.5(300)(0.15)^2 = 3.375 N*m. So are you saying that the initial kinetic energy of the block 0.5mv^2 = 3.375 ?
 
  • #4
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
Yeah, where "initial" in this context means, just after impact.
 

Related Threads on Bullet fired into a block attached to a spring

Replies
3
Views
3K
Replies
2
Views
12K
Replies
2
Views
6K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
7
Views
7K
Replies
2
Views
308
Top