A10.0-g bullet is fired into, and embeds itself in, a 1.95-kg block attached to a spring with a force constant of 16.6 n/m and whose mass is negligible. how far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block and the block slides on a frictionless surface? note: you must use conservation of momentum in this problem because of the inelastic collision between the bullet and block.

Distance the spring is compressed, x = 0.52 m  

Explanation:

Given :

Mass of the bullet, m = 10 g = 0.01 kg

Mass of the block, M = 1.95 kg

spring force constant, k = 16.6 N/m

Distance the spring is compressed =x

Speed of the bullet, v = 300 m/s

Speed of the block = V

Therefore we know that according to the law of conservation of momentum,

m.v = ( m+M )V

or

              = 1.53 m/s

Now according to the law of conservation of momentum,

m

Thus, distance the spring is compressed, x = 0.52 m

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west