Cannon and coefficient of friction between ground

  • Thread starter fishfish
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  • #1
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Homework Statement



A 3000kg cannon fires a 14 kg shell horizontally with a muzzle velocity of 450m/s. The cannon recoils for a distance of 2.5m before coming to a stop. What is the coefficient of friction between the cannon and the ground that it rests upon?

mcannon= 3000kg
mball=14kg
v1ball=0
v2ball=450m/s
v1cannon = 0
v2cannon= ?

I dont really know where to begin with this question.. Is it a momentum/impulse question or something else? Please help..
 

Answers and Replies

  • #2
kuruman
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Use momentum conservation to find the initial velocity of the cannon.
 
  • #3
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Use momentum conservation to find the initial velocity of the cannon.

Wouldn't the initial velocity of the cannon be zero though since both the ball and the cannon start off at rest? If I were to find the v2 of the cannon though, how would I proceed with the rest of the solution?
 
  • #4
ideasrule
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Yes, initial velocity is 0. What kuruman meant was the thing you call v2, that is, the velocity of the cannon at the instant it fired the cannonball.

After finding v2, can you find the acceleration that friction exerts on the object?
 
  • #5
kuruman
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Yes, the initial velocity (and momentum) of the cannon and shell system is zero. The final momentum of the cannon and shell system will also be zero. However, before the shell is fired nothing is moving, but after the shell is fired, both cannon and shell are moving.
 
  • #6
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Yes, initial velocity is 0. What kuruman meant was the thing you call v2, that is, the velocity of the cannon at the instant it fired the cannonball.

After finding v2, can you find the acceleration that friction exerts on the object?

Would I use the formula v12 = v22 - 1/2ad?

Am I using the wrong formula..? Because I seem to get this massive number as the acceleration..
 
  • #7
kuruman
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What is the definition of momentum? Momentum conservation says

Momentum before = Momentum after.
 

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