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Introductory Calculus Based Physics, Help save Matt's life. He needs you.

  • #1
Hello all, this is my first of likely many posts here at the newly discovered Physics Forums. I'll do my best to help others out, and hopefully will receive the same in return.

I've gotten through about 3/4 of a large problem that I'm working through in my book (no solution manual FTL). Here's my question...




Homework Statement



A pole P of mass 300kg and length L of 12m is approaching Matt(80kg, no motion) at 8.28 m/s on a frictionless surface. Matt has a massless spring loaded launcher loaded with a mass N of .5kg.

Matt plans to save himself by shooting the mass N at the pole P approaching him, propelling himself away from the pole while simultaneously slowing the velocity of the pole when the masses hit the pole. Here I am assuming the mass N hits the pole at the poles center of mass and sticks.

The spring constant of the massless spring launcher is 3x102 J/m2


1.) Determine the speed of the pole when it reaches the frictionless surface.

-I've solved this already, it is 8.28m/s

2.)Determine Matt's velocity when he and the pole have the same velocity.

3.) Determine how fast the mass N must be going in order to slow down the pole appropriately.

4.)Determine how much Matt should compress the spring before releasing it.





Homework Equations



I'm assuming the use of conservation of energy equations will be useful in this problem. However some equations that I'm not seeing as relevant may be missing.

Kinetic energy:

E = mv2

Spring Potential energy:

E = kx2

-Where x is (Uncompressed - compressed) length.



The Attempt at a Solution



All the questions depend on one another, this becomes a problem when I can't solve the second of four questions in sequence.

Question 2: Determine Matt's velocity when he and the pole have the same velocity.

-I'm having trouble with this problem, it seems to me that we would need to know more information about the situation before being able to solve, however, this class is my first exposure to physics, and I am most likely wrong. Any help would be greatly appreciated.



Question 3: Determine how fast the mass N must be going in order to slow down the pole appropriately.

-Since I have no answer for number two, I will be assuming that both Matt and the pole P have the same velocity at 6.28m/s. Please check my logic.

LOGIC:

The velocity of Pole P needs to be changed by 2m/s. (8.28m/s initial velocity - 6.28m/s final velocity = 2m/s)

Initial energy of pole at 8.28m/s: .5 x 300kg x (8.28m/s)^2 = 10290 J

Energy of pole at needed velocity of 6.28m/s : .5 x 300kg x (6.28m/s)^2 = 5766 J

10290J - 5766J = 4524J

So the small mas N (.5kg) that Matt shoots from his launcher at the pole needs to transfer 4524J of energy to slow the pole by 2m/s.

4524J = .5 x .5kg x V2

V = 134.5 m/s


Question 4: Determine how much Matt should compress the spring before releasing it.


As determined in question three, we know that the mass N that is shot out of the spring launcher needs an energy of 4524J in order to slow the speed of the pole P by 2m/s. Here is where I run into a logical problem.

All of the problems I've solved in HW sets thus far related to spring energy had the spring attached to a static object. However, in this problem on one side of the spring is Matt (80kg), on the other is mass N(.5kg). I'm not sure how to model this using the potential spring energy equation because I know that when the spring is allowed to uncompress, some of the energy will be given to Matt, and some to the mass N. Any help getting me going here will also be greatly appreciated.



I understand the the length of my question is probably less than desirable. However, please know that any help given is greatly appreciated, and will be reciprocated if possible.(I'm more of a math and computer science guy)
 

Answers and Replies

  • #2
129
0
2. Ok so before Matt releases the mass the pole has a momentum of 300*8.28. After Matt releases the mass and the mass sticks to the center of the pole and Matt is flung backwards, the pole mass system has a momentum of 300.5*v and Matt has a momentum of 80*v.

Using conservation of momentum 300*8.28 = 380.5*v

v = 6.53 m/s

3. So initially the pole has a momentum of 300*8.28 and the mass has a momentum of -.5*v this is equal to 300.5*6.53 due to conservation of momentum.

v = 1043 m/s

4. The springs potential transfers into Matt's Kinetic energy plus the masses kinetic energy.

3x102x2 = 0.5*80*6.532 + 0.5*0.5*10432

x = 30.2 m

That's my crack at it.
 
  • #3
Thank you so much Atticus. Wow #2 was extremely simple, I was over thinking that problem. Does anyone have anything else to add to Atticus's solution or confirm it? Again, thank you, I really appreciate your time.
 
  • #4
129
0
Thank you so much Atticus. Wow #2 was extremely simple, I was over thinking that problem. Does anyone have anything else to add to Atticus's solution or confirm it? Again, thank you, I really appreciate your time.
No problem. Hopefully the solution is correct. To solve these types of problems I usually break the situation into states. For example in number 2 I would take State A as just the pole moving and State B as everyone moving. Then I would use conservation of momentum. For number 3 I would look just at the pole-mass system with State A being them both moving and State B being the two moving together. Finally for number 4 I would look at just the Matt-spring-mass system with State A being the string compressed (potential energy no kinetic) and State B being Matt and the mass moving (no potential all kinetic).
 

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