Question

A 0.15kg bullet is fired into a 1.85kg block that is hanging from a 45cm string. What minimum initial bullet speed is required if the bullet and block are to make one complete revolution?

Please show all work. I need this question answered as soon as possible!

Answer 1

at the top of the circle:
v^2/r = g
v^2 = g r = 9.81 * .45 = 4.4145
so at top v = 2.1 m/s

so Kinetic energy at the bottom must be enough to get up there at that speed

(1/2) m v^2 = m g h + (1/2) m (4.4145)
(1/2) v^2 = 9.81(0.90) + 2.207
v = 4.7 m/s at bottom after crash
NOW do your usual conservation of momentum thing at the bottom with the bullet in the block

Answer 2

what would I solve for when I do the conservation of momentum at the bottom of the block with the bullet?

Answer 3

LOL, figured you knew how to do all that

initial bullet speed = V
that is what we want
we know final speed is 4.7
conservation of momentum:

.15 V = (.15+1.85)(4.7)

Answer 4

the change in energy from the bottom of the circle to the top is... (m g h)
h = 90 cm

the centripetal force at the top is only gravitational... m g

velocity at top ... m v^2 / r = m g
___ v = √(r g) ... r is 45 cm

the KE at the bottom equals the KE
___ at the top, PLUS (m g h)

use the KE at the bottom to find the
___ block/bullet bottom velocity

momentum is conserved in the block/bullet collision

V bullet * M bullet equals
V block/bullet * M block/bullet

Answer 5

Thank you so much!!!

Answer 6

You are welcome.