A 7 g bullet has a speed of 270 m/s and embeds itself into a 600 g block. The bl

Question

A 7 g bullet has a speed of 270 m/s and embeds itself into a 600 g block. The block is initially at rest sitting on a rough surface which has a coefficient of kinetic friction equal to 0.5
Part A) What is the speed of the block after the collison
B)What fraction of the initial kinetic energy of the bullet remains after the collision?
C)How far does the block slide after the collision before coming to rest?
D)If the bullet collides elastically with the block, how far would the block slide then?

Thank you!

Explanation / Answer

a) 0.007 x 270 = (0.600 + 0.007)v

v = 3.114 m/s

b) [(0.600 + 0.007) x 3.114^2 /2 ] / [0.007 x 270^2/2] = 0.0115

c) mg = ma

a = 0.5 x 9.81 = 4.91 m/s2

v2 - 0 = 2as

3.114^2 = 2 x 4.9 x s

s = 1 m

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