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
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!
6 answers
what would I solve for when I do the conservation of momentum at the bottom of the block with the bullet?
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)
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)
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
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
Thank you so much!!!
You are welcome.