# physics help please. oscillations is the topic?

A 23-gram bullet traveling at 176 m/s is fired into a 0.438-kg wooden block anchored to a 100 N/m spring.

a) How far is the spring compressed to the nearest thousandth of a meter

b) If the speed of the bullet is not known but it is observed that the spring is compressed 43 cm, what was the speed of the bullet to the nearest m/s?

### 1 Answer

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- NCSLv 74 years agoFavorite Answer
a) First conserve momentum:

23g * 176m/s + 0 = (23 + 438)g * v

v = 8.78 m/s

giving the bullet/block

KE = ½mv² = ½ * 0.461kg * (8.78m/s)² = 17.8 J

which becomes spring potential energy

SPE = 17.8 J = ½kx² = ½ * 100N/m * x²

x = 0.596 m

b) work it in reverse!

SPE = ½ * 100N/m * (0.43m)² = 9.2 J = ½ * 0.461kg * v²

v = 6.33 m/s

23g * u = 461g * 6.33m/s

u = 127 m/s

Yes!

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