Question

Consider the roller coaster pictured. The coaster car (m= 1000 kg) starts from rest at (1) the top of the first hill at height H = 40 m. It then (2) drops, (3) goes over the second hill assume semicircular) of height L, and (4) around the loop of radius R, all frictionless. After finishing the loop (5), it encounteres frictional resistance u = 1 on the track at and (6) hits an armadillo (mA = 5 kg) at distance D = 25 m with a coefficient of restitution e = 0.3. 1 3 4 H AG 2 6 D- 1. <40 m is the max height L can be for the car to make it over 2. The maximum radius R for the car to still round the loop. 3. Find the speed of the car at (4), the top of the loop. 4. What is conserved from states 5-6 pre-collision? Select all that apply. Mass, energy, momentum, velocity, work 5. 17.2 m/s is the speed of the car at (6), just before it hits the armadillo. 6. Find the post-collision speed of the armadillo.

Answer 1

Solution ; H=40m 1. • object FN, Fg R P.E:=0 A Potential Energy (ma is mass of Potential Ene gy at height H = m gH object Applying conservation of energy from 0 to (K.E.), + (P.E.), = (K. E.)2 + (P.E.)2 Potential energy at point a. kinetic energy kinetic energy at point I at point 1. at point 2. Potential energy 0+ mgh I muhto (Us = velocity at ③) £mu? = mgh - Applying conservation of energy at point ® and © K.E.)2 + (P.E.)= (K.E.)+(P.E.)3 mor to & mus + mgh value of L put Us=o. Ź mu? = mg/max 2 To find max. CS Scanned with CamScanner

mglman mgh (:1mus = mgh) Lmax EH H=40m (given) L < 40m 2 Velocity at point > mug? mglmax La oss = m gH = 40g Us=roog mis max. nadius R for the car to still round the loop Newton's law at point 6 » Applying -FN-Fg = mi-a) a a gravitational force (-ve sign indicates centripetal downward direction) acceleration Nonnal force -FN-mg = muy R CS Scanned with CamScanner

Minimum Speed at point U to complete loop (FN=0) trag =travu R max IgRmad at and o- = Applying conservation of Energy (k.e)s + (P.E.)s (K.E.y +(P.E.) Žus to I hop tong Rmax 80g A Roma + grm max 800 Rmax Rmax wo 3 speed at car of the vu 9.8 X 80 16.2 mis Mass is Energy is conserved conserved (But mechanical energy is not conserved due to friction) is not conserved as Frictional force is acting. not conserved. As , it will decrease Momentum Velocity is CS Scanned with CamSaariner

due to frictional force work is conserved. Speed at point S = us=480g = 28 mls to a constant from point frictional force acts on the object ie. fr = MEN 1. mg 1000x9.8 N = 9800 N for ma a retcondation a = for 9800 loo =9.8 mis? we need to find speed at point 6 US- o} thad 6 y distance bhw 0 and 6 D- given V? (28)2 219.8) (25) negative sign at a is in opposity direction uz 6 294 mis? winot v U = 17.2 mis 16 let speed of roller coaster Armadillo )) 1) 4, initial Uz = G, = Final V2 = Speed of roller coaster Armadillo 12 ICS Scanned with CamScanner

coefficient of restitution =e=0.3 Uz-u, e (4₂-UI) armadillo is at rest) U2=0 (initially u, = 17.2 mis Uz-u, = -0.310-17.2) Uz-u, 5.16 mis = 0, Uz-5.16 During collision momentum is conserved > mu,+ malz = miv, tmavz 1000 (17.2) to = 1000 u, tsuz 1000 u, +5uz = 17200 17 Putting value value of u from 1000 (Uz - 5.15) + 5 U2 = 17200 1005 U2 - 5160=17200 speed of armadillo V2 = 22.25 mis post - collision CS Scanned with Coman