Wood Block
What was the speed of the bullet? You can treat the wood block as a particle?
A 2.0 kg wood block hangs from the bottom of a 1.3 kg, 1.0-m-long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 11 g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 33 degrees angle.
What was the speed of the bullet? You can treat the wood block as a particle.
Can someone solve and explain this problem please? Thanks!!
Here's how we may conceptualize the solution:
First we have to find the CENTROID of the composite object, block, rod and bullet embedded in Wooden Block.
|←'------- 1.0M --------→|
|←'------- x------→|........|
|← .5M →|........|........|
======= ●===▄==='◙
...............↓.......↓......↓
...............Wr......R.......Wв+Wь
Where:
● <=== center of gravity of 1m-rod
▄<=== centroid of system of rod, block of wood and bullet
◙<=== center of gravity of block and bullet embedded
Wr = weight of 1m-rod
R = resultant (Wr + Wв+Wь)
Wв = weight of wooden block
Wь = weight of bullet
Finding the CENTROID of the system
Summation of moments of the weights = R(x) (Summation of moments of masses will be the same because after all "g" will cancel out, leaving the massses only. My choice is weights because masses are not vectors, so it seems ackward to represent them with arrows.)
∑M= Rx
Rx = Wr(.5) +( Wв+Wь )(1)
(Wr + Wв+Wь)x = Wr(.5) + (Wв+Wь) (1)
(1.3 + 2 + .011)gx =(1.3)g(.5) + (2 + .011)g(1) <==g will cancel out
(3.311)x = 0.65 + 2.011
x = .804 M (location of the CENTROID from the pivot. This is the point where the mass of the system could be considered as being concentrated.
.‾‾‾↑‾‾‾‾‾‾.......Ю
.....|.............||
.....|.............||..
....x.............||....
.....|.............||33º.
.....|.............||........
.....|.............||..........▄ ___
___↓__Vь→ ||▄→U __↕h_
From the figure h will be computed
cos 33º. = (x - h) / x
(x - h) = (cos 33º) x
...........= (.838)(.804)
...........= .673
h = .131 M
From considerations of energy transformation
KE ===> PE
½mV² = mgh
becomes
½(R/g)U² = (R/g)gh
½U² = gh
........._____..........___________
U =²√ 2gh...... = ²√ 2(9.81).131
U = 1.60 m/sec
..Vь→.......V = 0
●→.............▄ → U
mь..............(R/g)
mьVь + 0 = (R/g)U
(.011)Vь = (2 + 1.3 + .011)(1.60)
Vь = 481.6 M/sec
The real speed of the bullet will be (V)
______________
.....↑..............↑........../|
......|...............|......../..|
......|.....(.804M.)....../...|
.(1).|...............|....../....|
......|..............↓..../--→|Vь = 481.6 M/sec
......|................../.......|
.....↓................./-----→|V
(Take note that Vь is the theoretical bullet velocity as projected to hit the CENTROID .804M from the pivot. The real hitting point of the bullet is 1m from the pivot, specifically, on the block of wood at the end of the 1-M rod.)
By virtue of similar triangles
V.............Vь
----- = ----------
(1.0)......(.804)
V =( 481.6) / (.804)
V = 599.6M/sec
Did I get something wrong here? I was expecting a muzzle velocity of 1000M/sec or more.
-DP- sorry for the long solution, but it has to be, or else, you won't understand it. And we have to take into consideration all aspects possible. The diagrams are there because it would be very hard to describe everything in words alone.
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