- #1
Zaphia
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Homework Statement
A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm. (a) Find the velocity of the block right after impact (b) what was the initial speed of the bullet?
The book lists the answers to be (a) 2.60 m/s and (b) 325 m/s.
Homework Equations
Kenetic Energy = .5mv^2
Hooke's Law: F=kx
m1v1=m2v2
The Attempt at a Solution
Alright, I have tried it 2 different ways (and gotten 2 sets of completely different answers) but I think I'm missing something.
First train of thought:
1. Find the spring constant, k, using the calibration information, gives me a k=300
2. Find the impact of compression by solving F=kx with my compression distance (.15 meters) and 'k' value (F=45Newtons)
3. Find velocity for (a) using KE=(1/2)mv^2 which gives me the velocity=9.49 m/s.
4. Initial speed of bullet for (b) via m1v1=m2v2, Plug values in, giving me a v1=1186.25 m/s
Second train of thought:
1. Momentum of block with bullet=momentum of bullet
so .5V^2=.5(.008)v^2
2. Force constant of spring 'k' = (Force/Compression) = .75/.0025 = 300 Newton/meter (same as previous method)
3. Potential Energy in Spring: .5kx^2=3.375 Joules
4. Kinetic Energy of Block = Potential Energy of Spring, thus (from step 1), .004v^2=3.375 Joules ---> v=29.05 m/s (for part (b))
5. Velocity of Block w/ Bullet V=.008v = .8216 m/s (for part (a)).
Any suggestions would be most useful, thank you :)