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Help with variable mass on a slope question

  • #1
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Homework Statement



An armoured car with a mass of 5 tonnes is located on a smooth plane which is inclined at an angle of tan-1 (5/12) to the horizontal as shown in Figure Q3. A missile of mass 15kg is fired horizontally from this armoured car at 650m/s. Determine the velocity with which the armoured car will begin to travel up the inclined plane.

upload_2016-5-12_11-12-12.png
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Homework Equations



I think that a momentum equation needs to be used, but im not sure how to apply that equation to a slope.

The Attempt at a Solution



The answer to the question is 1.8m/s. The closest I have got to it is 1.9m/s. This is my attempt.

0 = mdv - Mv
dv+v = 650m/s
mdv = Mv
mx(650-v) = Mv
15x(650-v) = 5000v
v = (15x650) / (15+5000)
v = 1.94 m/s

At which point do I taken into consideration the angle of the slope?
 

Answers and Replies

  • #2
Merlin3189
Homework Helper
Gold Member
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637
Whenever you choose.
When you draw up your equation, mdv = Mv, you are saying (I think?) that the momentum of the shell is equal to the momentum of the AC. Since v is a vector, you need to take the directions into account. But the shell is fired horizontally and the AC moves up the slope and you should take the angle into consideration here.
But you could consider momentum independently in horizontal and vertical directions. Then your equation is presumably looking only at the horizontal component of momentum and you have calculated the horizontal component of the AC velocity. So now you consider the slope.

I agree with your 1.95m/sec (I think you maybe got 1.96?) and I can see how someone might then get 1.8m/s by considering the slope, but I don't agree with that. I get 2.11m/sec for the AC velocity up the slope.
 
  • #3
9
0
Thank you for explaining that to me. I have now taken into consideration the angle of the slope at the very end of my equation.
My new equation is now

0 = mdv - Mv
dv+v = 650m/s
mdv = Mv
mx(650-v) = Mv
15x(650-v) = 5000v
v = (15x650) / (15+5000)
v = 1.94 m/s x cos(tan^1(5/12))
v = 1.79 or 1.8 m/s

Solved.
Really appreciate your help.
 
  • #4
Merlin3189
Homework Helper
Gold Member
1,522
637
Yes. That's how I thought they got 1.8 m/s.
I hope a real physicist might take a look, because I don't believe that answer.
If 1.94m/s is the horizontal component of the AC velocity, then the velocity of the AC up the slope must be greater, in fact 1.95 / cos(tan-1(5/12)) or simply 1.95 x 13/12
If you were given the velocity of the AC up the slope, say 1m/s, then asked what were the horizontal and vertical components of its velocity, you'd get two smaller numbers (H=12/13 and v=5/13). So now you have its horizontal component, the actual velocity must be larger.
 

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