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AP Review Sheets AP Physics
Chapter 9: Linear Momentum & Collisions
Created by: Chandler Woo
Background / Summary Deep into the realm of classical mechanics, we reach a section focusing on the conservation of momentum. This unit covers the concepts of the law of conservation of momentum for an isolated system, impulse for a non-‐isolated system, elastic, inelastic, and perfectly inelastic collisions, 2-‐D collisions, and center of mass.
Key Words:
Impulse is the change in momentum, which can be calculated by the area under the curve of a force-‐time graph
Elastic Collision is when both momentum and kinetic energy are conserved. No energy is lost to heat (i.e. billiards)
Inelastic Collision is when kinetic energy is not conserved. There is heat lost in the collision, and deformation occurs (i.e. car crash)
Perfectly Inelastic Collision is when the objects collide and stick perfectly together. Heat is lost, but both travel at the same velocity (i.e. bullet in block)
“Glancing” occurs when object 1 collides with an initially stationary object 2 and both travel off at an angle with respect to the horizontal (i.e. hitting a billiard ball on its side)
Center of Mass is the weighted average position of the system’s mass
Important Formulae:
𝐾! + 𝐾! = 𝐾!! + 𝐾!!
𝑝! + 𝑝! = 𝑝!! + 𝑝!!
𝑚!𝑣! + 𝑚!𝑣! = 𝑚!𝑣!! + 𝑚!𝑣!!
12𝑚!𝑣!! +
12𝑚!𝑣!! =
12𝑚!𝑣!! ! +
12𝑚!𝑣!! !
𝑚!𝑣! + 𝑚!𝑣! = (𝑚! +𝑚!)𝑣!!
𝐽 = 𝐹 𝑑𝑡!!
!!= ∆𝑝
𝑥!" = 𝑚!𝑥! +𝑚!𝑥! +⋯𝑚!𝑥!
𝑚! +𝑚! +⋯𝑚!=
𝑚!𝑥!𝑚!
𝑟!" =1𝑀
𝑟 𝑑𝑚
Problem types you may encounter:
AP Review Sheets AP Physics
PRACTICE PROBLEMS: 1. [Center of Mass]
A flat piece of metal of uniform density has the shape and dimensions shown above. Find the center of mass for the piece of metal (both the x-‐ and y-‐components). 2. [Collision] A 250 gram rubber bullet is fired horizontally at a 900 gram block of wood sitting on a flat surface. The coefficient of kinetic friction between the block of wood and surface is .55, and the block slides 80 cm before coming to rest. If the bullet collides elastically with the block, what are its initial and final velocities? 3. [Momentum/Impulse] A 5.00 kg gun with a 90 cm long barrel fires a 70-‐gram bullet with a velocity of 550 m/s. What is the: recoil velocity of the gun? Impulse on a bullet? Time the bullet accelerated? Average acceleration of the bullet? And force applied to the gun by the bullet?
40cm 30cm 20cm 10cm
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AP Review Sheets AP Physics
Solutions:
1. The center of mass, for both the x-‐ and y-‐directions, can be calculated using the equation:
𝑥!" = !!!!!!
. There are several ways to think about this problem. You can calculate the 13 little
squares individually that comprise the piece of iron, but you can simplify it down to three symmetric pieces, as seen below:
𝑥!" =𝑚!𝑥! +𝑚!𝑥! +𝑚!𝑥!
𝑚! +𝑚! +𝑚!=3𝑚 15𝑐𝑚 + 2𝑚 10𝑐𝑚 + 8𝑚 20𝑐𝑚
13𝑚=22513
𝑐𝑚
𝑦!" =𝑚!𝑦! +𝑚!𝑦! +𝑚!𝑦!
𝑚! +𝑚! +𝑚!=3𝑚 35𝑐𝑚 + 2𝑚 25𝑐𝑚 + 8𝑚 10𝑐𝑚
13𝑚=23513
𝑐𝑚
The center of mass for the piece of iron is: (!!"!"𝑐𝑚, !"#
!"𝑐𝑚).
2. When solving collision problems, always make sure you know which type of collision it is, and from there choose the appropriate equation. Because it is elastic, we are going to be using two equations:
𝑚!𝑣! + 𝑚!𝑣! = 𝑚!𝑣!! + 𝑚!𝑣!! 12𝑚!𝑣!! +
12𝑚!𝑣!! =
12𝑚!𝑣!! ! +
12𝑚!𝑣!! !
𝑚!𝑣! + 𝑚!𝑣! = 𝑚!𝑣!! + 𝑚!𝑣!! 12𝑚!𝑣!! +
12𝑚!𝑣!! =
12𝑚!𝑣!! ! +
12𝑚!𝑣!! !
𝑚! 𝑣!! − 𝑣!! ! = 𝑚! 𝑣!! − 𝑣!! ! 𝑚! 𝑣! − 𝑣!! 𝑣! + 𝑣!! = 𝑚! 𝑣! − 𝑣!! 𝑣! + 𝑣!! 𝑚! 𝑣! − 𝑣!! = 𝑚! 𝑣! − 𝑣!! 𝒗𝟏 + 𝒗𝟏! = 𝒗𝟐 + 𝒗𝟐!
AP Review Sheets AP Physics (Solution on next page)
𝐹!"# = 𝑚𝑎; 𝐹! = 𝑚𝑎 ; 𝜇𝑚𝑔 = 𝑚𝑎; 𝑎 = . 55 9.80 = 5.39 m/s2 (of block)
𝑣!! = 𝑣!! + 2𝑎∆𝑥; 𝑣! = 2𝑎∆𝑥 = 2 5.39 . 8 = 2.94 m/s (of block) 𝑚!𝑣! + 𝑚!𝑣! = 𝑚!𝑣!! + 𝑚!𝑣!!
. 25 𝑣! + 0 = . 25 𝑣!! + (.9)(2.94) à 𝒗𝟏 = 𝒗𝟏! + 𝟏𝟎.𝟓𝟗
𝑣! + 𝑣!! = 𝑣! + 𝑣!! 𝒗𝟏 + 𝒗𝟏! = 𝟐.𝟗𝟒 + 𝟎
Combine equations: 𝑣!! = −3.825 𝑚/𝑠; 𝑣! = 6.765 𝑚/𝑠
3. a) Recoil velocity of gun
𝑚!𝑣! + 𝑚!𝑣! = 𝑚!𝑣!! + 𝑚!𝑣!!
0 = 5𝑘𝑔 𝑣!! + . 070𝑘𝑔 (550𝑚/𝑠)
𝑣!! = −7.7 𝑚/𝑠
b) Impulse on bullet
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡 = 𝑚∆𝑣
= . 070𝑘𝑔 550 − 0 = 38.5 𝑘𝑔 ∙𝑚/𝑠
c) Time the bullet accelerated
𝑡 =𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑠𝑝𝑒𝑒𝑑
= . 9𝑚
550𝑚/𝑠= .00164 𝑠
d) Average acceleration of bullet
𝑎 =𝑣! − 𝑣!
𝑡=550 − 0. 00164
= 3.35𝑒5 𝑚/𝑠!
e) Force applied to the gun by the bullet
𝐹 = 𝑚𝑎 = . 070𝑘𝑔 3.35𝑒5 𝑚/𝑠!
𝐹 = 23450 𝑁
*Credit for these problems go to Mr. White. I changed up the variables, but the problem setup was his.