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Bullet,block and spring

  • Thread starter huskydc
  • Start date
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 ?
 
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)
 
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?
 
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.
 

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