Solve for Bullet Speed Before Hitting Block: A 47.00g Bullet

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A 47.00 g bullet embeds itself in a 5.10 kg wooden block, causing the block to swing through an arc of 23.50 degrees. The conservation of momentum is applied to find the bullet's speed before impact, with calculations indicating the block rises 0.26 meters after the collision. The initial approach involved using momentum equations, but confusion arose regarding the height the block reaches. By applying trigonometric principles, the correct height was determined, leading to a calculated bullet speed of approximately 506 m/s. The discussion emphasizes the importance of understanding both momentum conservation and the geometry of the pendulum motion.
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Homework Statement



A 47.00 g bullet hits and becomes embedded in a 5.10 kg wooden block which is hanging from a 2.50 m long string. This causes the block to swing through an arc of 23.50o. What was the speed of the bullet before it hit the block?
mb = 0.047kg
mw = 510kg
r = 2.50
theta = 23.5 degrees

Homework Equations



using the radius and theta i imagine to figure out that when the block is shot it swings back 1.09m
i did this by using Tan(23.5) = Opp / 2.50
however when i tried using mv + mv = ∑mv i couldn't get an answer since i don't know any speeds


The Attempt at a Solution



mv + mv = mv
(0.047)v + 5.10(0) = 5.147(v)
0.047v = 5.147v
i don't know the second v cause i don't know how fast the block moved when it was shot.
Help?
 
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View this problem as having two parts:
(1) The collision of bullet and block: Momentum is conserved.
(2) The rise of 'bullet and block' after the collision: What's conserved here?

Start with part 2 to figure out the speed of 'bullet and block' immediately after the collision.
 
Wait now, i think i might have something
Ef = 1/2mv22 + mgh2
Ei = 1/2mv12 + mgh1
1/2mv22 + mgh2 = 1/2mv12 + mgh1
1/2v22 + gh2 = 1/2v12 + gh1
v22 = v12 + gh1 - gh2
v22 = v12 + 2g(h2 - h1)
0 = v12 + 2(-9.81)(1.09 - 0)
-v12 = -19.6(1.09)
-v12 = -21.4
v1 = 4.63m/s

(0.047)(v) + 5.10(0) = 5.147(4.63)
0.047v + 0 = 23.8
0.047v = 23.8
v = 506m/s

does that make sense?
 
Right approach, but how high does the block rise after the bullet hits?
 
i'm not sure, i don't think that me finding 1.09 even makes sense anymore, it's a circle not a square, they aren't any right angles, so i don't know how far up it goes.
 
Hypnos_16 said:
i'm not sure, i don't think that me finding 1.09 even makes sense anymore, it's a circle not a square, they aren't any right angles, so i don't know how far up it goes.
Draw a diagram and make your own right triangle.
 
Okay so all I've gotten was that the block moves a distance of 23.5 meters around the circle, though, if you start at the bottom on the circle would 23.5 meters be the height?
 
Hypnos_16 said:
Okay so all I've gotten was that the block moves a distance of 23.5 meters around the circle, though, if you start at the bottom on the circle would 23.5 meters be the height?
The block swings through an angle of 23.5 degrees. (Think of the block as a pendulum bob hanging from a string.) That's not the height, but you can figure out the height with a little trig.

Draw a diagram.
 
okay, i got now that it swings up 0.26 meters.
 
  • #10
Hypnos_16 said:
okay, i got now that it swings up 0.26 meters.
How did you determine that?
 
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