Conservation of Momentum II

by Renee "Princess" Wald and Kate "Bitchin" Tanski

Table of Contents

• Quantities
• Previous Formulas
• New Formulas
• General Problem Solving Strategy
• Example Problem #1
• Example Problem #2
• Example Problem #3
• Go back to Tutorial Page

Here's the quantities you can know when working with Conservation of Momentum:

• X Displacement (meters: m)
• t Time (seconds: s)
• V_o Initial velocity (meters per second: m/s)
• V_a Average Velocity (meters per second: m/s)
• V Final Velocity (meters per second: m/s)
• a Acceleration (meters per second per second: m/s/s)
• g Acceleration of Gravity on Earth (meters per second per second: 9.80 m/s/s)
• m Mass (kilograms: kg)
• F Force (Newtons: N = kg m/s/s)
• F_N Normal Force (Newtons: N = kg m/s/s)
• Mu Coefficient of Friction
• p Momentum (Bergevins: B = kg m/s)

Go back to: Table of Contents

Here are formulas from previous lessons that we will use:

Linear Kinematics

1. X = V_at
2. V = V_o + at
3. V_a = (V_o + V)/₂
4. X = V_ot + ¹/₂at²
5. 2aX = V² - V_o²

Dynamics

1. F = ma
2. F = mg
3. F = Mu(F_N)
4. F = -kX

1. p = mV
2. Ft = mV
3. F = p/t

1. Read the problem.
2. Draw a cartoon that shows the relationship of every object at different moments throughout the problem
3. Express the momentum of every separate object for each frame of your cartoon
4. Since momentum must be conserved, set frames or parts of frames total momentum equal to each other and solve for the unknowns.

Go back to: Table of Contents

109 Kg Thor and his 5.26 Kg hammer are at rest on top of the first of two uncoupled frictionless 89.7 Kg carts that are next to each other. (ok--it's a physics word problem)   Thor runs and jumps from one cart to the other and lands on it.  (Still holding the hammer) He, the hammer, and the cart are going +1.56 m/s in the end.

A) What must be the velocity of the other cart (Cart 1)?

B) What was Thor's velocity after he left the first cart, but before he landed on the second?

C) With what velocity must he throw his hammer to give himself and the cart he lands on (Not the one he starts on) a velocity of 2.00 m/s

D) If instead of throwing the hammer in C), he were to jump from the second cart in such a way that he was not moving horizontally with respect to the ground, what would be the velocity of the second cart?
 

Here's our initial table of knowns and unknowns:

• m_(Cart-1)= 89.7 kg
• m_(Cart-2)= 89.7 kg
• _{m(Thor)= 109 kg}
• _{m(hammer)= 5.26 kg}
• _{Vo(Thor + hammer + Cart-2)= 1.56 m/s}
• _{Vo(Cart 1, hammer, Thor)= 0 m/s (it's at rest)}
• _{Vo(Cart 2)= 0 m/s (it's at rest)}
• _{V(Cart-1)= ? m/s}
• _{V(Thor)= ? m/s}
• _{V(hammer)= ? m/s}
• _{V(Cart-2)= ? m/s}

_{First let's find the momentum of Cart 2 after Thor jumps onto it.}
 

_{p = mV       (Thor + hammer + Cart 2)}

_{m(Thor + hammer + Cart-2)=203.96 kg}

_{Vo(Thor + hammer + Cart-2)= 1.56 m/s}
 

_{in numbers:}

_{p = (203.96 kg) * (1.56 m/s) = 318.17 (no units)}
 

_{Since total momentum is zero, the momentum of Cart 2 must equal -318.17.}

m_(Cart-2)= 89.7 kg       Therefore the Velocity of Cart 2 equals (-318.17) / (89.7 kg) = -3.55 m/s

Part B

_{The momentum of Thor and the hammer before they reach the second cart is the same as the momentum they will have after they reach the cart (318.17).}

_{m(Thor + hammer)=114.26 kg}

_{V(Thor + hammer )= p/m}

_{in numbers:}

_{(318.17) / (114.26 kg) = 2.78 m/s}

Part C

_{First we have to find the momentum of Thor and Cart 2.}

_{p = mV}

_{m(Thor + Cart-2)= 198.7 kg}

_{V(Thor + Cart-2)= 2.00  m/s}

_Therefore

_{p(Thor + Cart-2)= (198.7 kg) * (2.00 m/s) = 397.4}

_{ptotal= 0 = p(Thor + Cart-2) + p(Cart 1)+ p(hammer)}

_{p(hammer)= -(p(Thor + Cart-2) + p(Cart 1) )}

_{p(hammer)= - (397.4  - 318.17) = -79.22}

_{V(hammer)= p(hammer / m(hammer) = .79.22 / 5.26 kg = -15.1 m/s}

Part D

_{p = mV}

_{V(Thor + hammer)= 0  m/s}

_{m(Thor + hammer)= 114.26 kg}

_{p(Thor +hammer)= (114.26 kg) (0 m/s) = 0}

_{ptotal= 0 = p( Cart-2) + p(Cart 1)+ p(Thor + hammer)}

_{p(Cart 2)= -(p(Thor + hammer) + p(Cart 1) )}

_{p(Cart 2)= - (0 - 318.17) = 318.17}

_{V(Cart 2r)= p(Cart 2)  / m(Cart 2r) = 318.17 / 89.7 kg = 3.55 m/s}
 
 

_{Go back to: Table of Contents}

_{Priscilla (65.4 Kg) is at rest on the first of two 82.0 Kg carts which are also at rest on a frictionless level surface.  She jumps from the first cart to the second, and then to the ground. Following this maneuver, the first cart has a velocity of -1.45 m/s, and the second a velocity of 2.26 m/s.}

_{A) What was her velocity between the first and second cart?}

_{B) What was her velocity between the second cart and the ground?}

_{C) What would have been the velocity of her and the second cart if she had remained on the second cart instead of jumping off?}

_{D) If her interaction with the first cart took 2.6 seconds, what force did she exert on it?}

_{E) What would have been the velocity of the second cart if she was motionless after leaving it?}
 
 

_{Here what the question gives us:}

• _{m(Pricilla)= 65.4 kg}
• _{m(Cart-1)= 82.0 kg}
• _{m(Cart-2)= 82.0 kg}
• _{Vo(Pricilla)= 0 m/s}
• _{Vo(Cart-1)= 0 m/s}
• _{Vo(Cart-2)= 0 m/s}
• _{V(Cart-1)= -1.45 m/s}
• _{V(Cart-2)= 2.26 m/s}
• _{V(Between Cart 1 and Cart 2)= ? m/s}
• _{V(Between Cart 2 and the ground)= ? m/s}
• _{V( Cart 2 and Pricilla)= ? m/s}
•  

First we find the momentum of Cart 1 after Pricilla jumps off.

_{p(Cart-1)= m(Cart-1)*V(Cart-1)= (82.0 kg) * (-1.45 m/s) = -118.9}

_{ptotal= 0 = p( Cart-1) + p(Pricilla)}

_{p(Pricilla)= -p( Cart-1) = 118.9}

_{V(Pricilla)= p(Pricilla) / m(Pricillar) = 118.9 / 65.4 kg = 1.82 m/s}

Part B

First we find the momentum of Cart 2 after Pricilla jumps off.

_{p(Cart-2)= m(Cart-2)*V(Cart-2)= (82.0 kg) * (2.26 m/s) = 185.32}

_{ptotal= 0 = p( Cart-2) + p(Pricilla)+ p(Cart 1)}

_{p(Pricilla)= -(p( Cart-1) + p(Cart 2))= - ( -118.9 + 185.32) = -66.42}

_{V(Pricilla)= p(Pricilla) / m(Pricillar) = -66.42 / 65.4 kg = -1.02 m/s}

Part C

_{ptotal= 0 = p( Cart-1) + p(Cart 2 + Pricilla)}

_{p(Cart 2 + Pricilla)= -p( Cart-1) = 118.9}

_{V(Cart 2 +Pricilla)= p( Cart 2 + Pricilla) / m(Cart 2 + Pricillar) = 118.9 / (82.0 kg + 65.4 kg) = .807 m/s}

Part D

t = 2.6 s

_{F =} Dp / t

_{F = p(Cart 1) / 2.6 s = -45.7 N}

Part E

_{p(Pricilla)= m(Pricilla) * V(Pricilla)= (65.2 kg) * ( 0 m/s) = 0}

_{ptotal= 0 = p( Cart-1) + p(Pricilla)+ p(Cart 2)}

_{p(Cart 2)= -(p( Cart-1) + p(Pricilla) ) = - ( -118.9 + 0 ) = 118.9}

_{V(Cart 2)= p(Cart 2) / m(Cart 2r) = 118.9 / 82.0 kg = 1.45 m/s}
 

_{Go back to: Table of Contents}
 

Two 746 g blocks of wood are at rest on that by now familer frictionless surface. A 123 g bullet is shot through the first and sticks in the second.  Following this the first block is moving 12.5 m/s, and the second with the bullet stuck in it is moving 34.6 m/s.  Neither the bullet nor the blocks lose any pieces.

A) What was the bullet's velocity between the blocks?

B) What was the Bullet's velocity before it hit the first block?

C) Suppose the bullet's interaction with the first block had taken .0200 seconds.  What force would it have exerted on the block?

D) If in actuality there had been a frictional force of about .50 N on the second block with the bullet stuck in it, over what time would it have been brought to rest?

E) Suppose the bullet had stuck in the first block, causing it to slide into and stick to the second block.  What would have been the velocity of the bullet and the two blocks?
 
 

_{The Initial Data Table:}

• _{m(bullet)= 65.4 kg}
• _{m(block-1)= 82.0 kg}
• _{m(block-2)= 82.0 kg}
• _{Vo(bullet)= 0 m/s}
• _{Vo(block-1)= 0 m/s}
• _{Vo(block-2)= 0 m/s}
• _{V(block-1)= 12.5 m/s}
• _{V(block-2)= 34.6 m/s}
• _{V(Between block 1 and block 2)= ? m/s}
• _{V(before it hits block 1)= ? m/s}
• _{V(bullet + block 1 + block 2)= ? m/s}
• _{F(bullet on block 1)= ? N}
• _{t (to bring block 2 to rest)= ? s}

_{ptotal= p( block-1) + p(block 2 + bullet)}

_{p( block-1) = m( block-1) * V(block 1) =( .746 kg) * (12.5 m/s) = 9.325}

_{p(block 2 + bullet) = (.746 kg + .123 kg) * (34.6 m/s) = 30.0674}

_{ptotal= (9.325) + ( 30.0674) = 39.3924}

_{p(bullet)= p(total) - p(block 1)= (39.3924 - 9.325) = 30.0674}

_{V(bullet)= p(bullet) / m(bullet) = 30.0674 / .123 kg = 244 m/s}

Part B

_{p(bullet)= p(total) = 39.3924}

_{V(bullet)= p(bullet) / m(bullet) = 39.3924 / .123 kg = 320 m/s}

Part C

t =.0200 s

_{F =} Dp / t

_{F = p(block 1) / .0200 s = 466 N}

Part D

The momentum would be brought from 30.0674 to 0, so Dp = 30.0674 since

_{F =} Dp / t ,    t = Dp / F = (30.0674) / .50 N = 60.1 s

Part E

_{p(total) = 39.3924}

_{m(bullet + block 1 + block 2)= 1.615 kg}

_{V(bullet + block 1 + block 2)= p(total) / m(bullet + block 1 + block 2)= ( 39.3924) / (1.615 kg) = 24.4 m/s}
 
 
 
 

_{Go back to: Table of Contents}