Chapter 9: Linear Momentum

9.1: Linear Momentum and Its Conservation 9.2: Impulse and Momentum 9.3: Collisions in One Dimension: collision, elastic collision, inelastic collision 9.4: Two-Dimensional Collisions

9.5: The Center of Mass 9.6: Motion of a System of Particles: velocity of the center of mass, total momentum of a system of particles, acceleration of the center of mass, Newton's second law for a system of particles 9.7: Rocket Propulsion

9.1: Linear Momentum and Its Conservation

Our book states that in isolated systems momentum is conserved. Another way to state this is that we can account for all the parts involved in the problem.

ΣF = ma + m2a2 = 0 à ΣF = m dv/dt + m2 dv2 / dt = 0 à ΣF = d (mv + m2v2) / dt = 0 ΣF = d (mv) /dt ΣF = dp/ dt pbefore = pafter à à à

If two particles strike each with equal and opposite force then And since sum is constant...the total momentum in a system must also be constant Where a is in the opposite direction of a2 and a = dv/dt So, what are the units of momentum p = mv has the units of kg m/s... A couple of interesting notes: The derivative with respect to time is zero... thus the sum must be constant or conserved

Demo: Bi-Ball: ME-N-BB

9.2: Impulse and Momentum

Let's play with units... kg m/s kg m/s (s/s) kg m/s (s/s) = Newton second

Multiply both sides by seconds / seconds So this gives us a hint we need to integrate with respect to time.

F = m a F = m dv/dt F dt = m dv ∫ F dt = ∫m dv FΔt = m Δv FΔt = Δmv FΔt = Δp

Let’s assume F is independent of time Impulse (F dt) = change of momentum (Δp = Δmv)

Demo: Half-Liter Paper Rocket: ME-N-PR Estimate the distance of the deformed tip of the rocket. Use d = ½at^2 to determine time of impact. Calculate Force exerted on Rocket during this time.

9.3: Collisions in One Dimension

BIG SUV…small car (BUG) Smash small car on grill of big SUV pbefore = pafter when à à à

If we account for all parts (objects) then momentum is conserved during the collision (this is a closed system)

Let’s say a 7000 kg SUV (totally loaded) is traveling North at 90 mph (about 40 m/s) and a little car, let’s call it a bug, and is traveling to the south at 65 mph (about 30 m /s). These are both controlled remotely. We ram them together. What happens? pbefore = msuvvsuv + mbugvbug; pafter = msuvvsuv-f + mbugvbug-f pbefore = pafter msuvvsuv + mbugvbug = (msuv + mbug) vf but bug is squashed on grill, thus final velocity is identical 7000(40) + 1000(-30) = (7000 + 1000) vf 250000 = 8000 vf vf = 33 1/3 m/s

The final velocity is 33 1/3 m/s. The Massive SUV goes from 40 to 33 1/3 m/s for a change of 6 2/3 m/s. That’s like hitting a concrete wall doing about 14 to 15 mph. The SUV is wrecked badly.

The bug is traveling in the opposite direction…thus a negative. So the bug starts at -30 m/s and end up at 33 1/3 m/s…or a change of 63 1/3 m/s or about 140 mph directly in a concrete wall.

Like we said earlier…the bug is squash on the grill of the SUV.

Demo: Newton’s Cradle: ME-N-NC

9.4: Two-Dimensional Collisions

Σpxi = constant Σpxf = constant KEi = KEf

First off, this is an artificial situation. 3D is a must; we're looking at this as a 2D equation for simplicity.

A 2D Look at a Neutron Moderator

Initial

Final

px = (m) 3.5 x 10^5   +   0 py = 0 Kinitial = ½(m) (3.5 x 10^5)^2   +   0

px = (m) v1x + (m) v2x px = (m) v1 cos 37° + (m) v2x cos φ py = (m) v1y + (m) v2y py = (m) v1 sin 37° + (m) v2 sin φ Kfinal = ½(m) v1^2 + ½(m) v2^2

Kinetic energy isn't broke into components (v^2, a scaler). Kinetic energy is conserved in elastic collisions only; where Ktotalbefore = Ktotalafter

px à (m) 3.5 x 10^5 = (m) v1 cos 37° + (m) v2x cos φ Eq (1) px à 3.5 x 10^5 = v1f cos 37° + v2f cos φ Eq (2) py à 0     = v1f sin 37° + v2f sin φ Eq (3) K à (3.5 x 10^5)^2 = v1f^2 + v2f^2

v1f = 2.80 x 10^5 m/s v2f = 2.11 x 10^5 m/s φ = 53.0°

Solve these 3 equations simultaneously

9.5: The Center of Mass

Demo: Leaning Tower: ME-J-LT

If 3D then CMx = Σmi*xi / mTotal CMx = 1 /mT ∫x dm

CMy = Σmi*yi / mT CMy = 1 /mT ∫y dm

CMz = Σmi*zi / mT CMz = 1 /mT ∫z dm

Where is the center of mass in a rod is λ = mass / length How do we find the center of mass of a homogeneous symmetric object (Like solid 2D square or the figure to the right)? IT's AT THE CENTER

Just as obvious is the center of mass of the above bar bell set, where the center of mass is marked by the “X”. This time…let’s write the expression for the center of mass. à à à

m1x1 + m2x2 xCM =    ——————                m1 + m2

Now we just extend this principle to many more objects for the x, y, and z axis.

And rCM = xCM i + yCM j + yCM k

Σi mixi xCM = ———— Σi mi

Σi miyi yCM =    ———— Σi mi

Σi mizi zCM =    ————  Σi mi

Σi miri rCM =    ————  Σi mi

Example 9.14: Where is the center of mass of a right triangle?

We need mass / unit area of the triangle (total mass / total area) dm = mass / unit area * area of green strip dm = ( M / ½ a b) * y dx

dm = 2M/ab * y dx dm = 2M/ab * y dx dm = 2M/ab * (x*b/a) dx dm = 2M/ab * (b/a) xdx dm = 2M/a² ( x dx)

xCM = ^1/M ∫ x dm xCM = ^1/M ∫ x (2M/a² * xdx) combine like terms xCM = 2/a² ∫ x² dx       integrate from 0 to a xCM = ^2/3 a

Expression for equivalent proportions ^x/y = ^a/b à y = x*b/a

Example 3: Center of Mass for varying linear density

λ is mass per unit length (g / m) λ = 2 + 40x² What is its mass if it's 30 cm long? m = ∫ λ dx = m = ∫ (2 + 40x²)dx = m = 2x + 40x³/3 from 0 to 0.3 m m = 0.6 g + 0.36 g m = 0.96 grams

Where is its center of mass from x = 0? xCM = ∫ x dm / mT xCM = 1 / mT ∫ x λ dx xCM = 1 / mT ∫ x (2 + 40x²) dx xCM = 1 / mT ∫ 2xdx + 40∫ x³dx xCM = 1 / mT (x² + 10x^4) from 0 to 0.3 xCM = 1/0.96 (0.09 + 0.081) xCM = 0.178 m past 0 m

Demo: Walking the Spool: ME-K-WS Demo: Center of Mass O.C Ruler: ME-J-CE Demo: Center of Mass, Irregular Object: ME-J-CI

9.6: Motion of a System of Particles

A rocket is fired vertically upward. At the instant it reaches an altitude of 1000 meters and a speed of 300 m/s it explodes into three fragments. The 1^st fragment has twice the mass of either the 2^nd or 3^rd fragments and continues to move upward with a speed of 200 m/s. The second fragment has a speed of 240 m/s and is moving east right after the explosion. What is the velocity of the 3 fragment?

Whole = 1^st frag + 2^nd frag + 3^rd frag 4 m = 2 m + m + m

Vertical components pbefore = pafter 4m(300) = 2m(v1) + m(v3y) (v1 is all y-comp) 4m(300) = 2m(v1) + m(v3y) 1200 = 2(200) + (v3y) v3y = 800 m/s

Horizontal components pbefore = pafter 0 = 2m(v2) + m(v3x) (v2 is all x-comp) 0 = m(v2) + m(v3x) 0 = (240) + (v3x) v3x = -240 m/s

v3rd = (240^2 + 800^2)^½ v3rd = 835.2 m/s @ 106.7°

q3rd = tan^-1(800/240) q3rd = 73.3° We want 180 - q3rd

9.7: Rocket Propulsion

pbefore = pafter procket&fuel = procket + pfuel (M+Dm)v = M(v + Dv) + Dm(v - vexhaust) Mv +Dmv = Mv + MDv + Dmv - Dmvexh 0 = MDv - Dmvexh MDv = Dm vexh (the velocity of the Mdv = vexh dm exhaust is a constant) (limit of m à 0) Mdv = -vexh dM òdv = -vexh òdM / M Dv = vexh ln (Mi / Mf)

M is the mass of the rocket (which may also containing some remaining fuel) We know downward momentum of the exhaust opposes the upward momentum of the rocket… thus vexhaust dm = -Dv dM We know that the dm = -dM dM is the mass of the rocket with the remaining fuel and dm is the spent fuel. As more and more fuel (+dm) exits the rocket (-dM, mass of rocket) is reduced (-) accordingly

FThrust = m a FThrust = m dv/dt FThrust = vexh dM/dt