10.5 u = (10.5+104) v
F = m a = -.658 m g
so
a = -.658(g)
work done = F d
= m (.658)(g) (6.66)
work done = Ke after collision
so
(1/2) m v^2 = m(.658)(g)(6.66)
v^2 = 2(.658)(g)(6.66)
solve for v
then
u = (10.5+104) v/10.5
A 10.5 g bullet is fired into a 104 g wooden block initially at rest on a horizontal surface.The acceleration of gravity is 9.8 m/s2. After impact, the block slides 6.66 m before coming to rest.If the coefficient of friction between block and surface is 0.658 , what was the speed of the bullet immediately before impact?
Answer in units of m/s.
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