P = mv (kg•m/s) impulse: j = F

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P = W/t (W)
K = ½mv

m_Av_A + m_Bv_B = m_Av_A’ + m_Bv_B’

Steps

Algebra

start with Newton's Law

multiply both side by t

substitute mv for Ft

substitute v' – v for v

collect like v terms

multiply by -1

F_A = -F_B

F_At = -F_Bt

m_Av_A = -m_Bv_B

m_A(v_A’ – v_A) = -m_B(v_B’ – v_B)

-m_Av_A – m_Bv_B = -m_Av_A’ – m_Bv_B’

m_Av_A + m_Bv_B = m_Av_A’ + m_Bv_B’

2. two particles collide and stick together

a. inelastic collisions

b. m_Av_A + m_Bv_B = (m_A + m_B)v’

Steps

Algebra

start with

substitute v' for v_A' and v_B'

simplify

m_Av_A + m_Bv_B = m_Av_A’ + m_Bv_B’

m_Av_A + m_Bv_B = m_Av’ + m_Bv’

m_Av_A + m_Bv_B = (m_A + m_B)v’

3. two particles collide and bounce off

a. elastic collisions

b. difference in velocity is the same after collision: v_A – v_B = -(v_A’ – v_B’)—proof later

c. solving two equations and two unknowns

fill in v_A and v_B into v_A – v_B = -(v_A' – v_B’)
write expression for v_A’ in terms of v_B'
substitute v_A' expression in equation:

m_Av_A + m_Bv_B = m_Av_A’ + m_Bv_B’

solve for v_B’
solve for v_A’ using the expression for v_A' above

4. collisions in two dimensions

a. p_x is conserved independently of p_y

b. elastic collision

1. m_Av_Ax+m_Bv_Bx=m_Av_Ax’+m_Bv_Bx’

2. m_Av_Ay+m_Bv_By=m_Av_Ay’+m_Bv_By’

3. solve two equations & two unknowns

c. inelastic collision

1. m_Av_Ax+m_Bv_Bx=(m_A+m_B)v_x'

2. m_Av_Ay+m_Bv_By=(m_A+m_B)v_y'

3. solve two equations & two unknowns

c. object explodes into two pieces m_A and m_B

1. (m_A + m_B)v = m_Av_A’ + m_Bv_B’

2. opposite inelastic collision equation

B. Forms of Energy

1. scalar value measured in joules, J = 1 N•m

2. work, W = F_||d (J)

a. Only component of F parallel to d does work



1. F_|| = Fcos  W = (Fcos)d

2. include sign W > 0 when F  d 

b. W_net > 0 (acceleration), W_net < 0 (deceleration)

c. W_net= 0

1. F_net = 0 (lift)

2. F  when d  (orbit)

d. work done by variable force—stretching a spring

1. graph spring force (F_s) vs. position (x)

F_s
	slope = k Area = W	x

2. F_s = kx  slope = F_s/x = k

3. W = F_sx  area = ½bh = ½x(kx) = ½kx² = W

3. power, P = W/t (W)

a. rate that work is done: Watt, W = J/s

b. P = Fv_av, where v is average (W/t = F(d/t) = Fv_av)

c. graphing

1. P = slope of W vs. t graph

2. P = area under F vs. v graph

d. kilowatt-hour is a unit of energy

(1KWh = 3.6 x 10⁶ J)

4. mechanical energy

a. work-energy theorem: work done to an object increases mechanical energy; work done by an object decreases mechanical energy

b. scalar quantity, like work

c. kinetic energy—energy of motion

1. positive only

2. K = ½mv²= p²/2m

Steps	Algebra
start with assume K_o = 0  v_o = 0 solve for ad	v² = v₀² + 2ad v² = 2ad ad = ½v²
start with substitute ma for F substitute ½v² for ad rearrange	K = W = Fd K = (ma)d = m(ad) K = m(½v²) K = ½mv²
start with square both sides divide both sides by 2m substitute K for ½mv²	mv = p m²v² = p² m²v²/2m = p²/2m K = p²/2m

d. potential energy—energy of relative position

1. gravitational potential energy

a. based on arbitrary zero

(usually closest or farthest apart)

b. U_g = mgh (near the Earth's surface)

Steps

Algebra

start with

substitute mg for F

substitute h for d

U_g = W = Fd

U_g = (mg)d

U_g = mgh

c. U_g = -GMm/r (orbiting system)

1. G = 6.67 x 10^-11 N•m²/kg²

2. r = distance from center to center

3. U_g = 0 when r is U_g < 0 for all values of r because positive work is needed reach U_g = 0

2. spring (elastic) potential energy, U_s = ½kx²

a. U_s = W to stretch the spring

b. see work by a variable force above

C. Conservation of Energy

1. work done on object A by a "nonconservative" force (push or pull, friction) results in the gain in mechanical energy for object A equal to the loss of energy by the source of the nonconservative force

2. work done on object A by a "conservative" force (gravity, spring) results in the change in form of mechanical energy (U  K) for object A, but no loss in energy

a. conservative forces (F_g and F_s)

1. F_g  d : U_g  K, F_g  d : K  U_g

2. F_s  d : U_s  K, F_s  d : K  U_s

b. process isn't 100 % efficient

1. friction (W = F_fd) reduces mechanical energy

2. mechanical energy is converted into random kinetic energy of the object's atoms and the temperature increases = heat energy—Q

3. total energy is still conserved

3. work done by object A on object B

a. W = m_Aad (uses up kinetic energy to decelerate)

b. energy loss by object A = energy gain by object B

c. some energy is lost due to friction

4. examples

		Process	Energy
1	Work done to	pull a pendulum bob off center	W  U_g
	Transformation	release pendulum	U_g  K
	Work done by	hit a stationary object	K  W
2	Work done to	throw a ball into the air	W  K
	Transformation	ball rises and falls	K  U_g
	Work done by	falling ball dents ground	K  W
3	Work done to	load a projectile in spring-gun	W  U_s
	Transformation	release projectile	U_s  K
	Work done by	projectile penetrates target	K  W

5. elastic collision formula proof (p and K are conserved)

½m_A(v_A² - v_A'²) = ½m_B(v_B'²- v_B²) (v_A²- v_A'²) = (v_B'²- v_B²)

m_A(v_A – v_A') = m_B(v_B' – v_B) (v_A – v_A') = (v_B' – v_B)

(v_A ~~– v~~_A')(v_A + v_A') = (v_B ~~– v~~_B')(v_B + v_B')  v_A - v_B = -(v_A' - v_B')

(v_A ~~– v~~_A') =(v_B~~' – v~~_B)

6. solving conservation of energy problems

determine initial energy of the object, E_o
- if elevated h distance: U_g = mgh
- if accelerated to v velocity: K = ½mv²
- if spring compressed x distance: U_s = ½kx²
determine energy added/subtracted due to an external push or pull: W_p = ±F_||d
determine energy removed from the object by friction: W_f = F_fd = (mgcosd
- d is the distance traveled
-  is the angle of incline (0^o for horizontal)
determine resulting energy, E'= E_o ± W_p – W_f
determine d, h, x or v
- if slides a distance d: 0 = E_o ± W_p – mgcosd'
- if elevates a height h: E' = mgh'
- If compresses a spring x: U_s: E' = ½kx'²
- if accelerated to velocity v: E' = ½mv'²
general equation (not all terms apply for each problem)

K + U_g + U_s ± W_p – W_f = K' + U_g^'+ U_s''

½mv² + mgh + ½kx² ± F_pd – F_fd = ½mv'² + mgh' + ½kx'²

7. solve ballistics problems

	M (v_M = 0)  m v_m bullet collides inelastically with block: mv_m = (M + m)v' block swings or slides (conservation of energy) block swings like a pendulum to height h K = U_g  ½(M + m)v'² = (M + m)gh  h = v'²/2g block slides a distance d along a rough surface K = W_f½(M + m)v'²=(M + m)gdd = v'²/2g

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