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

In summary, a bullet hit a 5.10 kg wooden block and caused it to swing back 1.09 m. The block then rises up 0.26 m after the bullet hits.
  • #1
Hypnos_16
153
1

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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  • #2
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.
 
  • #3
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?
 
  • #4
Right approach, but how high does the block rise after the bullet hits?
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
  • #8
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.
 
  • #9
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?
 

1. How can I calculate the bullet speed before hitting a block?

The bullet speed can be calculated using the formula v = √(2gh), where v is the speed in meters per second, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the block in meters.

2. What units should I use for the bullet's mass?

The bullet's mass should be measured in grams, as this is the unit used in the given information (47.00g).

3. Is the acceleration due to gravity constant?

Yes, the acceleration due to gravity is generally considered constant at 9.8 m/s^2, although it may vary slightly depending on location and altitude.

4. Can this formula be used for all types of bullets?

This formula can be used for most types of bullets, as long as the mass and height of the block are known. However, it may not be accurate for bullets with significantly different shapes or materials.

5. How precise will the calculated bullet speed be?

The calculated bullet speed will be an approximation, and the level of precision will depend on the accuracy of the given information and the precision of the calculations used. It is important to use accurate measurements and perform the calculations correctly to obtain a more precise result.

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