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Classical Mechanics Kleppner Problem

  • Thread starter ypudi
  • Start date
15
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1. Homework Statement

An Instrument carrying a projectile accidentally explodes at he top of its trajectory.The horizontal distance b/w the launch point and the point of explosion is L. The projectile breaks into 2 pieces which fly horizontally apart. The larger piece has three time the mass of the smaller one and returns to earth at the launching station. How far away does the larger piece land??

2. Homework Equations



3. The Attempt at a Solution
My answer is 4L/3. Just want to verify if this is correct...
 
15
0
1. Homework Statement

An Instrument carrying a projectile accidentally explodes at he top of its trajectory.The horizontal distance b/w the launch point and the point of explosion is L. The projectile breaks into 2 pieces which fly horizontally apart. The larger piece has three time the mass of the smaller one and returns to earth at the launching station. How far away does the larger piece land??

2. Homework Equations

Conservation of Momentum in the x direction at the top most point.
Multipying it by an unknown t (where t is the time of ascent)
v(small mass )x t = L

we need to find V(large mass)x t.


3. The Attempt at a Solution
My answer is 4L/3. Just want to verify if this is correct...
it comes out as L+ L/3.
 
15
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Can anyone solve this please and verify my answer!!!!!!!!!!!
 

Delphi51

Homework Helper
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Suppose there was no explosion. Then the whole mass would go another L, total of 2L before hitting the ground. The explosion increases the speed of the small piece, so it will go further than the 2L. So 4L/3 is too small.

You do have the right approach, doing a conservation of momentum at the turnover point where the explosion takes place. I agree with your initial speed of L/t. Maybe show me your momentum calc for the one mv before the collision and two after the collision, with 2 of the 3 speeds known. That must be where your difficulty is.
 

gneill

Mentor
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It may be profitable to think in terms of conservation of momentum for the center of mass of the system. Regardless of of what happens to the projectile, even if it explodes, the free-fall motion of its center of mass will remain the same! Where does that put the center of mass when the parts land? Where does that put the parts?
 
15
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An Instrument carrying a projectile accidentally explodes at he top of its trajectory.The horizontal distance b/w the launch point and the point of explosion is L. The projectile breaks into 2 pieces which fly horizontally apart. The larger piece has three time the mass of the smaller one and the smaller one returns to earth at the launching station. How far away does the larger piece land??
 

gneill

Mentor
20,497
2,616
An Instrument carrying a projectile accidentally explodes at he top of its trajectory.The horizontal distance b/w the launch point and the point of explosion is L. The projectile breaks into 2 pieces which fly horizontally apart. The larger piece has three time the mass of the smaller one and the smaller one returns to earth at the launching station. How far away does the larger piece land??
It lands just as far as the first time the question was asked :smile:
 
15
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My apologies. I had made an error in typing. It is the smaller piece that returns back to the launching station.
Thinking in terms of the conservation of momentum of center of mass. My answer seems correct. The centre of mass must be at the distance of L as it falls freely.

But shouldnt it be at a distance of 2L????
 

gneill

Mentor
20,497
2,616
The center of mass will fallow the same trajectory as the projectile would have followed if it had not exploded.
 
15
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Suppose there was no explosion. Then the whole mass would go another L, total of 2L before hitting the ground. The explosion increases the speed of the small piece, so it will go further than the 2L. So 4L/3 is too small.

You do have the right approach, doing a conservation of momentum at the turnover point where the explosion takes place. I agree with your initial speed of L/t. Maybe show me your momentum calc for the one mv before the collision and two after the collision, with 2 of the 3 speeds known. That must be where your difficulty is.
The smaller piece comes back top the launching station. Hence the
V(s)/V(l)= 3/1

IF V(s)xt= L then it implies that V(l)xt =L/3.
 
15
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wait let me show you the momentum conservation
that I had applied
 
15
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V - velocity of unbroken piece.
V(s)- velocity of smaller piece
V(l)- Velocity of larger peice.
t(f)- time of ascent or descent
all velocities in x axis.

before explosion momentum = MV
after explosion momentum = M/4 x[V-V(s)]+ 3M/4[V(l)+V]

equating and removing M and solving we get
3V(l)= V(s)

V(s)xt(f)= L

therfore V(l)xt = L/3
 

Delphi51

Homework Helper
3,407
10
I used m as the mass of the small piece, 3m for large piece, total 4m.
momentum before = momentum after
MV = MV + MV
4m*L/t = -m*L/t + 3m*v --> v = 5/3*L/t
Very tricky; better check to ensure that this results in the landed center of mass being 2L as Gneil saw so cleverly!
 
15
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Yes, it does. the answer comes as 8L/3 and this gives the centr of mass cordintes as (2L,0) The trick is if we take assume velocities (like I did) it is important to define wrt to ground. The relation that i derived for the velocities is correct but it is wrt to each other.
Hence

V(s)xt is not equal to L rather [V(s)-V]x t = L

the larger piece travels father than the smaller piece aided by the horizontal velocity making the velocity wrt ground of the larger piece as V + V(l).


An excellent problem sheerly because of its simplicity and I must thank you guys for this excellent discussion. I feel that I have learnt a lot more than this problem aimed to teach.
 

Delphi51

Homework Helper
3,407
10
My sentiments, too! I must try to remember the center of mass approach.
 

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