A5.20g bullet moving at 672 m/s strikes a 700g wooden block atrest on a frictionless surface. the bullet emerges, travelingin the same direction with its speed reduced to 428 m/s.
a. what is the resulting speed of the block?
b. what is the speed of the bullet-block center of mass?

a) v=1.81 m/s; B) v=4.95 m/s

Explanation:

using momentum conservation

m1v1+m2v1=m1v2+m2v2

A)

5.2*672+700*0=5.2*428+700v2

The initial velocity of the block is 0, solving for v2

v2=1.81 m/s

B) Now both the bullet and the block travel together

m1v1+m2v1=(m1+m2)v2

5.2*672+700*0=(5.2+700)v

v=4.95 m/s

pe = weightxheight = 2000x1000=2,000,000joules

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