Conservation of Momentum I


by Kristin Ragsdale

Table of Contents

Quantities
Here's the quantities you can know when working with Conservation of Momentum:
These quantities are defined and explained on other pages.

 



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Previous Formulas
Here are formulas from previous lessons that we will use:

Linear Kinematics

    

  1. X = Vat


  2. V = Vo + at


  3. Va = (Vo + V)/2


  4. X = Vot + 1/2at2


  5. 2aX = V2 - Vo2


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Dynamics

    

  1. F = ma


  2. F = mg


  3. F = Mu(FN)


  4. F = -kX




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New Formulas

 
    

  1. p  = mV


  2. Ft = mV


  3. F  = p/t




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Mr. Murray's General Problem Solving Strategy:
  1. Read the problem.
  2. Go through the problem and figure out what is given or implied
    Make a list, and identify the quantities you know.
  3. Find any formula that will allow you to calculate
    anything that you don't know, and apply it.
  4. Add what you just found in the last step to your list of knowns.
  5. Check to see if you have found the answer. If not, repeat the
    previous two steps until you are done.

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Example Problem 1

A 67.2 g bullet leaves a gun at 328 m/s and tears through a 2.27 kg wooden block at rest on a (magical) frictionless surface. If the bullet is whistling along at a speed of 94 m/s after the collision what is the speed of the block?

Here's our initial table of knowns and unknowns: 
First let's find the momentum of the bullet before the collision 





p = mV





in numbers:





.0672 kg * 328 m/s = 22.0416 kg m/s





We can also calculate the momentum of the bullet after the collision





p= mV





.0672 kg * 94 m/s = 6.3168 kg m/s





So know we know:  




Since momentum is conserved the momentum of the block plus the momentum of the bullet after the


collision must equal the momentum of the bullet before the collision.





po(bullet) =	p(bullet) + p(block)





using the numbers we have found:

                   

22.0416 kg m/s = 6.3168 kg m/s + 15.7248 kg m/s 




Now we can use the first momentum formula to calculate the velocity of the block. 



p = mV



15.7248 kg m/s = 2.27 kg * 6.927 m/s 




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Example problem 2

Beth with a mass of 60 kg is standing on a 25 kg cart. Her 75 kg friend Seth is standing on a 30 kg cart 2 m to her right. Both carts run on a frictionless track. (Oh No! More sourcery) Seth picks up a 3 kg bowling ball and throws it at a speed of 1.7 m/s towards Beth, who catches it.
A) What is Beth's final velocity?
B) What is Seth's final velocity?
C) If Seth wants to travel at .15 m/s what velocity must Beth throw the ball back to him? (Within our frame of reference)

Here what the question gives us: 
The easiest thing to find first is the momentum of the bowling ball:


p = mV


p = 3.0 kg * -1.7 m/s = -5.1 kg m/s





So now we know:


We can also apply this information to Seth and Beth.  Beth will travel the same direction as the ball (thus having the 


same sign) while Seth will roll in the opposite direction.





So:


To find Beth's velocity we need to find her total mass: (Cart + Beth + Bowling Ball)


60 kg + 25 kg + 3 kg = 88 kg





Now using momentum we can find her velocity:


-5.1 kg m/s / 88 kg = -.05795 m/s





We answered the first question, so now we have:




Again we have to find the total mass that gets the momentum transfered to Seth:


30 kg + 75 kg = 105 kg



And using the first momentum formula:


p = mV


V(Seth) = 5.1 kg m/s / 105 kg = .04857 m/s





We've finished this senario, but we'll need to use some of the informaion for the next part:


To find the velocity at which the ball must be thrown back we need to find the desired momentum:


p = mV where m is seth plus his cart and the ball


p = 108 kg * .15 m/s = 16.2 kg m/s





Now that we know the momentum that we desire and the momentum we have, we need to find the momentum 


we need to gain:


16.2 kg m/s - 5.1 kg m/s = 11.1 kg m/s





Now that we know the momentum we need to gain we the speed needed to give the ball our desired momentum:
 

p = mV


V(ball) = 11.1 kg m/s / 3 kg = 3.7 m/s

 



After finding this go back and review the problem to make sure we've answered the right question!
 



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Sample Problems

The answers to each problem follow it in parentheses. They also link to a solution to the problem. Try the problem, check your answer, and go to the solution if you do not understand.

1.

 A 132.45 g bullet traveling at 386 m/s rips a hole through a 1.34 Kg block of wood at rest on some frictionless ice. The bullet is travelling 153 m/s following the collision, what is the speed of the block? (23.0m/s)

2.

 To what height would the a) bullet and b) block rise to in problem 1) if they were moving up? c) What if the bullet stuck in the block? (1.19 km, 27.1m, 61.5m)

3.

 Mrs Robinson is flying her 1256 Kg airplane along at 153.00 m/s. She has a cannon which shoots 5.348 Kg shells ahead at 497 m/s (With respect to the plane, of course). A) What is her velocity after firing the cannon? B) How many shells can she fire if the plane's stall speed is 134 m/s? C) What is the minimum speed she must have to fire one shell? (In WWII, this was a real consideration for airplanes with cannons) (150.87 m/s, about 9 times, 136.13m/s)

4.

 A bullet moving horizontally strikes a 800.0 g block of wood lying on a table. The bullet imbeds itself in the block, and the block slides along the table a distance of 6.345 m before coming to a halt. The coefficient of friction between the block and the table is .56, and the block and bullet together have a mass of 824.0 g. A. At what rate did the force of friction slow the block when it was sliding? (Calculate the deceleration of the block) B. What was the velocity of the block just after the bullet imbedded itself? C. What was the initial velocity of the bullet? (-5.5 m/s/s, 8.3 m/s, 290 m/s)

5.

 Moe and Joe each have a mass of 89.25 kg and each of them are initially at rest on a different 25.2 kg cart that moves on frictionless wheels. (Really!!) Initially they are at rest, and Joe is 1.2 m to the right of Moe. Joe, however, is holding a 5.00 kg shot put that he throws at a velocity of -1.25 m/s to Moe. (Negative is to the left). Moe catches the shot put. A) What is Joe's final velocity? B) What is Moe's final velocity? C) What velocity (in our frame of reference) must Moe give the shot put when he throws it back to Joe, so that when Joe catches it he is going .1760 m/s? (+.0546 m/s, -.0523 m/s, +2.95 m/s)

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Solutions to Sample Problems

1.

 A 132.45 g bullet traveling at 386 m/s rips a hole through a 1.34 Kg block of wood at rest on some frictionless ice. The bullet is travelling 153 m/s following the collision, what is the speed of the block?
(23.0m/s)

Here is what you start with: 
First we can calculate the initial and final momentum of the bullet.


p = mV


po(bullet) = .13245 kg * 386 m/s = 51.1257 kg m/s


p(bullet) = .13245 kg * 153 m/s = 20.2649 kg m/s





From this we can calculate the final momentum of the block.


po(bullet) - p(bullet) = p(block)


p(block) = 51.1257 kg m/s - 20.2649 kg m/s = 30.8609 kg m/s





Putting this back into the momentum formula we can find the block's final velocity


p = mV


V(block) = 30.8609 kg m/s /1.34 kg = 23.0 m/s





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2.

 To what height would the a) bullet and b) block rise to in problem 1) if they were moving up? c) What if the bullet stuck in the block?
(1.19 km, 27.1m, 61.5m)

Here is what you start with: 
Part A


To find the height of the bullet travels we need to use one of our Linear Kinematics formulas
 

2aX = V2 - Vo2


Because Vo = 0 and the only accelorating force is gravity it changes to: 

X = V2/2g


X(bullet) = (153 m/s)2/(2*9.8 m/s/s) = 1194.34 m





Part B


We follow the same formula to find the height of the block


X(block) = (23.0 m/s)2/(2*9.8 m/s/s) = 27.0614 m





Part C


To find the height the unit would travel we first need to find the total mass of the unit
 

m(block) + m(bullet) = m(unit)


1.34 kg + .13245 kg = 1.47245 kg





Using the first momentum formula we can find the velocity of the unit


p = mV 


Vo(unit) = 51.1257 kg m/s /1.47245 kg = 34.7215 m/s

 



Now the process is the same as in Part A and B


X = V2/2g


X(unit) = (34.7215 m/s)2/(2*9.8) = 61.5094 m



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3.

 Mrs Robinson is flying her 1256 Kg airplane along at 153.00 m/s. She has a cannon which shoots 5.348 Kg shells ahead at 497 m/s (With respect to the plane, of course).
A) What is her velocity after firing the cannon?
B) How many shells can she fire if the plane's stall speed is 134 m/s?
C) What is the minimum speed she must have to fire one shell?
(In WWII, this was a real consideration for airplanes with cannons)
(150.87 m/s, about 9 times, 136.13m/s)

Part A
Here is what you start with: 
We can start by finding the initial momentum of the plane:


p = mV


po(plane) = 1256 kg * 134 m/s = 192168 kg m/s





We can also find the final momentum of the shell:


p = mV


p(shell) = 5.348 kg * 497 m/s = 2657.96 kg m/s





Since moentum is conserved:


po(plane) = p(plane) + p(shell)


p(plane) = 192168 kg m/s - 2657.96 kg m/s = 189510 kg m/s





Now by substituting this back into the momentum formula we can find the final velocity for the plane:


p = mV


V(plane) = 189510 kg m/s / (1256) = 150.87



 

Part B


Here is what you start with:


Since we have both the final velocity and mass of the plane we can calculate it's final momentum:


p(plane) = mV


p(plane) = 1256 kg * 134 m/s = 168304 kg m/s





Now we can calculate the total momentum for all the shells:


po(plane) = p(plane) + p(shells)


192168 kg m/s - 168304 kg m/s = 23864 kg m/s





Since we know the momentum of a single shell we can divide that into our momentum for all the shells:


p(shells) / p(shell) = n
 

n = 23864 kg m/s / 2685.0 kg m/s = 8.978 or aproxiametly 9 shells





Part C


Here is what you start with:


To find the initial momentum of the plane we can add the final momentum of the plane and bullet:


po(plane) = p(plane) + p(shell)


po(plane) = 168304 kg m/s + 2658.0 kg m/s = 170962 kg m/s





Now all we need to do is substitute it back into our favorite formula:


p = mV


Vo(plane) = 170962 kg m/s / 1256 kg = 136.13 m/s





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4.

 A bullet moving horizontally strikes a 800.0 g block of wood lying on a table. The bullet imbeds itself in the block, and the block slides along the table a distance of 6.345 m before coming to a halt. The coefficient of friction between the block and the table is .56, and the block and bullet together have a mass of 824.0 g.
A. At what rate did the force of friction slow the block when it was sliding?
(Calculate the deceleration of the block)
B. What was the velocity of the block just after the bullet imbedded itself?
C. What was the initial velocity of the bullet?
(-5.5 m/s/s, 8.3 m/s, 290 m/s)

Here is what you start with: 
Part A


From the dynamics formulas we see that:


F = Mu(mg) = ma


Canceling the mass:


Mug = a


a = .65*-9.8m/s/s = -5.488 m/s/s





Part B


Using the acceloration we calculated in Part A and a linear kinematics formula:


2aX = V2 - Vo2


Vo2 = 6.345 m /(2*5.448 m/s/s)


Vo = 8.345 m/s 





Part C


Now that we have found velocity we can put it into the first momentum formula with the mass:


p = mV


.824 kg * 8.345 m/s = 6.87646 kg m/s





And finally using the momentum formula again we can determine the bullets original speed:


p=mV


Vo(bullet) = 6.87646 kg m/s / .024 kg = 286.519 m/s





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5.

 Moe and Joe each have a mass of 89.25 kg and each of them are initially at rest on a different 25.2 kg cart that moves on frictionless wheels. (Really!!) Initially they are at rest, and Joe is 1.2 m to the right of Moe. Joe, however, is holding a 5.00 kg shot put that he throws at a velocity of -1.25 m/s to Moe. (Negative is to the left). Moe catches the shot put. A) What is Joe's final velocity? B) What is Moe's final velocity? C) What velocity (in our frame of reference) must Moe give the shot put when he throws it back to Joe, so that when Joe catches it he is going .1760 m/s? (+.0546 m/s, -.0523 m/s, +2.95 m/s)

Here is what you start with: 
Part A


First calculate the momentum of the shotput:
 

p = mV


p(shot put) = 5 kg * -1.25 m/s = -6.25 kg m/s





The total mass given momentum will be Joe plus his cart


m(cart) + m(Joe) = m


25.2 kg + 89.25 kg = 114.45



Joe's momentum is the opposite of the shotput (he moves to the right), so:


p=mV


V(Joe) = 6.25 kg m/s / 114.45 kg = .05461





Part B


Moe's total mass will be that of himself, his cart, plus the shot put:


m(Moe) + m(cart) + m(shot put) = m


89.25 kg + 25.2 kg + 5 kg = 119.45 kg





Moe's momentum will equal that of the shotput (he moves to the left), so:


p = mV


V(Moe) = -6.25 kg m/s / 119.45 kg = .05232 kg 





Part C


First we want to calculate the final momentum Joe will need:


p(Joe) = 119.45 kg * .1760 m/s = 21.0232 kg m/s





Then we will subtract his current momentum to find how much more he'll need:


21.0232 kg m/s -6.25 kg m/s = 14.7732 kg m/s





Now using the mass of the shot put and the necessary momentum we can find the necessary velocity:


p = mV


V(shot put) = 14.77322 kg m/s / 5 kg = 2.9546 m/s

  

 

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