# What is the total distance traveled by a bullet fragment after an explosion?

• Riman643
In summary: The 22 is meters per second...The 59 is degrees.In summary, The first thing I did was find the height of the apex. I found this by finding the x and y components of Vo: Vox = VoCos(θ) = 22cos(59) = 11.33 m/s Voy = VoSin(θ) = 22sin(59) = 18.86 m/s t = -Voy / ay = -18.86/-9.8 = 1.93 s Δy = Voyt + (1/2)ayt^2 = 18.86(1.93) + (1/2)(-9.8)(1.93^2) = 18.
Riman643
Homework Statement
A shell is shot with an initial velocity Vo of 22 m/s, at an angle of θ = 59° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?
Relevant Equations
Basic kinematics
p = mv
Okay so the first thing I did was was find the height of the apex. I found this by finding the x and y components of Vo:
Vox = VoCos(θ) = 22cos(59) = 11.33 m/s
Voy = VoSin(θ) = 22sin(59) = 18.86 m/s

Using the initial velocity of the y component I found the time it took to reach the apex:
t = -Voy / ay = -18.86/-9.8 = 1.93 s

Using the time I found the height of the apex:
Δy = Voyt + (1/2)ayt^2 = 18.86(1.93) + (1/2)(-9.8)(1.93^2) = 18.15 m

Using the height I found the time it took after the explosion for the bullet fragment to reach the ground. Because the bullet was at the apex, the initial velocity of the fragments is zero in the y direction
t = √(Δy/(1/2ay)) = √[(-18.15)/((1/2)(-9.8))] = 1.94s

Using the momentum equation I was able to find the initial velocity of the bullet fragment in the x direction:
Pi = Pf; 22m = (1/2)mVo; Vo = 44

Finally using the two times and two speeds of the fragment in the x direction I found (wrong according to the website) the total distance:
Δx = (Vox * t) + (Vox * t) = (11.33)(1.93) + (44)(1.94) = 107.23 m

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Riman643 said:
Problem Statement: A shell is shot with an initial velocity Vo of 22 m/s, at an angle of θ = 59° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?
Relevant Equations: Basic kinematics
p = mv

Using the momentum equation I was able to find the initial velocity of the bullet fragment in the x direction:
Pi = Pf; 22m = (1/2)mVo; Vo = 44
What is the initial velocity in the x-direction?

Before the collision the initial velocity is 11.33 m/s. Then treating the explosion as a separate event, I made the initial velocity in the x direction of the gun fragment 44 m/s.

Riman643 said:
Before the collision the initial velocity is 11.33 m/s. Then treating the explosion as a separate event, I made the initial velocity in the x direction of the gun fragment 44 m/s.
How did the horizontal velocity increase during the ascent to 22m/s?

haruspex said:
How did the horizontal velocity increase during the ascent to 22m/s?
It did not increase during the ascent. It increased when the bullet reached the apex because of the explosion.

Riman643 said:
It did not increase during the ascent. It increased when the bullet reached the apex because of the explosion.
How do you figure that it went from 11.33 m/s to 44 m/s? Were not the two fragments of the same mass?

gneill said:
How do you figure that it went from 11.33 m/s to 44 m/s? Were not the two fragments of the same mass?
I just realized I used Vo not Vox for the momentum in the x direction. I will try with that next. But I want to make sure my thinking on this is correct. The initial momentum must equal the final momentum (pi = pf) where
pi = 11.33m and pf = (1/2)mv. Setting them equal to each other it should be:

11.33m = (1/2)mV
V = 22.66

Riman643 said:
I just realized I used Vo not Vox for the momentum in the x direction. I will try with that next. But I want to make sure my thinking on this is correct. The initial momentum must equal the final momentum (pi = pf) where
pi = 11.33m and pf = (1/2)mv. Setting them equal to each other it should be:

11.33m = (1/2)mV
V = 22.66
Yes. Check your units for pi (see highlighted text in the quote). Momentum does not have units of only length.

Riman643 said:
I just realized I used Vo not Vox for the momentum in the x direction. I will try with that next. But I want to make sure my thinking on this is correct. The initial momentum must equal the final momentum (pi = pf) where
pi = 11.33m and pf = (1/2)mv. Setting them equal to each other it should be:

11.33m = (1/2)mV
V = 22.66
Yes.
But I strongly recommend you get into the habit of working entirely symbolically, only plugging in numbers at the end. It has many benefits.

You did not need to calculate the descent time. In the absence of drag, it must be the same as the ascent.

The problem in this thread can be solved very quickly if you know the range formula: http://www.softschools.com/formulas/physics/horizontal_range_formula/154/.
Since the explosion doubles the horizontal speed of the ongoing half, it doubles that half of the range, leading to 3/2 times the range calculated from the formula.

DaveE
gneill said:
Yes. Check your units for pi (see highlighted text in the quote). Momentum does not have units of only length.
The 11.33 is m/s. The m represents the mass in the momentum equation. Sorry for the confusion.

haruspex said:
Yes.
But I strongly recommend you get into the habit of working entirely symbolically, only plugging in numbers at the end. It has many benefits.

You did not need to calculate the descent time. In the absence of drag, it must be the same as the ascent.

The problem in this thread can be solved very quickly if you know the range formula: http://www.softschools.com/formulas/physics/horizontal_range_formula/154/.
Since the explosion doubles the horizontal speed of the ongoing half, it doubles that half of the range, leading to 3/2 times the range calculated from the formula.
Yeah, I've been given that advice before. I'm new to physics so I'm just trying to solve the problems the way I know how at the moment.

Riman643 said:
The 11.33 is m/s. The m represents the mass in the momentum equation. Sorry for the confusion.
Okay. I was confused because mixing numbers without units with variables that have implied units is somewhat confusing. I highly recommend @haruspex 's suggestion regarding doing the work symbolically until the very end.

haruspex said:
The problem in this thread can be solved very quickly if you know the range formula: http://www.softschools.com/formulas/physics/horizontal_range_formula/154/.
Since the explosion doubles the horizontal speed of the ongoing half, it doubles that half of the range, leading to 3/2 times the range calculated from the formula.
A small addendum to that: This works only (as is the case here) if the CM follows the projectile's original trajectory. The fragments must land simultaneously, which is the case if their vertical velocity is zero immediately after the explosion.
gneill said:
I highly recommend @haruspex 's suggestion regarding doing the work symbolically until the very end.
Amen to that.

haruspex said:
But I strongly recommend you get into the habit of working entirely symbolically, only plugging in numbers at the end. It has many benefits.
Me too.

## 1. What factors influence the distance traveled by a bullet fragment after an explosion?

The distance traveled by a bullet fragment after an explosion can be influenced by various factors such as the caliber and weight of the bullet, the type of explosive used, the distance between the bullet and the explosive, and any obstacles present in the path of the fragment.

## 2. Can the shape of the bullet affect its distance traveled after an explosion?

Yes, the shape of the bullet can play a role in determining its distance traveled after an explosion. A more aerodynamic shape can result in a longer distance traveled due to reduced air resistance.

## 3. Is the total distance traveled by a bullet fragment after an explosion predictable?

The total distance traveled by a bullet fragment after an explosion is not always predictable. It can be affected by various unpredictable factors such as wind, irregularities in the explosive material, and the angle of the explosion.

## 4. How does the type of explosive used affect the distance traveled by a bullet fragment?

The type of explosive used can have a significant impact on the distance traveled by a bullet fragment after an explosion. More powerful explosives can result in a longer distance traveled, while less powerful ones may not propel the fragment as far.

## 5. Can the distance traveled by a bullet fragment after an explosion be measured accurately?

The distance traveled by a bullet fragment after an explosion can be measured accurately using specialized instruments such as high-speed cameras and laser rangefinders. However, there may still be some margin of error due to the unpredictable nature of explosions.

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