Bullet,block and spring

  • Thread starter huskydc
  • Start date
  • #1
huskydc
78
0
A projectile of mass 20 g has an initial horizontal velocity of 100 m/s when it hits and stops in a wood block of mass 0.403 kg. The block is sitting on a horizontal frictionless surface and is attached to a massless spring, initially relaxed, with spring constant 148 N/m. What is the maximum compression of the spring?

since momentum is conserved, i used the equation
mv(initial) = mv (final), and found the final velocity of the block, as a result of the mass hitting it, is 4.963,

but I'm not sure where to go from here? ..or am i heading the right direction ?
 

Answers and Replies

  • #2
zwtipp05
107
0
You have the velocity of the block. Now you can use energy to solve for the compression of the spring. (KE = .5mv^2; Es=.5kx^2)
You're on the right track. Just try to think of the relationships between the various quantities (often there is more than one, i.e. velocity is related to momentum, but it is also related to energy)
 
  • #3
huskydc
78
0
ok, I've set up the equations as follow:

.5kx^2 = KE (in this case, also KE final)
thus,

.5kx^2 = .5m(4.963^2)

and solve for x.... but still incorrect, something wrong with my equations? or are there other factors i haven't thought of?
 
  • #4
zwtipp05
107
0
huskydc said:
ok, I've set up the equations as follow:

.5kx^2 = KE (in this case, also KE final)
thus,

.5kx^2 = .5m(4.963^2)

and solve for x.... but still incorrect, something wrong with my equations? or are there other factors i haven't thought of?
Make sure you include the mass of the bullet since it embeds in the block.
 

Suggested for: Bullet,block and spring

Replies
2
Views
310
  • Last Post
Replies
29
Views
790
  • Last Post
Replies
7
Views
316
  • Last Post
Replies
6
Views
380
Replies
8
Views
284
  • Last Post
Replies
4
Views
282
  • Last Post
Replies
27
Views
3K
Replies
6
Views
1K
Replies
1
Views
340
Replies
6
Views
342
Top