An 8 g bullet travelling 315 m/s is shot into a 0.6 kg block of wood suspended f

Question

An 8 g bullet travelling 315 m/s is shot into a 0.6 kg block of wood suspended from the ceiling by a 2 m long light rope. The block is initially at rest before the bullet and the block have a perfectly inelastic collision.
(a) What is the velocity of the block of wood just after the bullet embeds itself?
m/s

(b) How high (above the starting position) does the block reach after the collision?
m

(c) Immediately after the collision, what is the centripatel acceleration (magnitude and direction) acting on the combination block+bullet.
m/s^2

Explanation / Answer

a) by conserving momentum ,
0.008 X 315 = (0.008 + 0.6) X v
v = 4.145 m/s  

b)K.E. will convert in potential energy ,

mv2 / 2 = mgh

h = 4.1452 / (2x 9.81) = 0.8756 m   or 87.56 cm

c) centripatel acceleration = v2 / r = 4.1452 / 2   = 8.59 m/s2   ......in downwards direction .

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