A frictionless track contains a loop of radius R=0.5m. Situated on top of the track lies a small ball of mass m=2kg at a height h. It is then dropped and collides with another ball (of negligible size) of mass M=5kg.
<image_1>
Let h be the minimum height that m was dropped such that M would be able to move all the way around the loop. The coefficient of restitution for this collision is given as e=21.
Now consider a different scenario. Assume that the balls can now collide perfectly inelastically, which means that they stick to each other instantaneously after collision for the rest of the motion. If m was dropped from a height 3R, find the minimum value of Mm such that the combined mass can fully move all the way around the loop. Let this minimum value be k. Compute α=h2k2. (Note that this question is only asking for α but you need to find h to find α ). Assume the balls are point masses (neglect rotational effects).
Given constants:
- Proton mass, mp=1.67⋅10−27kg
- Neutron mass, mn=1.67⋅10−27kg
- Electron mass, me=9.11⋅10−31kg
- Avogadro's constant, N0=6.02⋅1023mol−1
- Universal gas constant, R=8.31J/(mol⋅K)
- Boltzmann's constant, kB=1.38⋅10−23J/K
- Electron charge magnitude, e=1.60⋅10−19C
- 1 electron volt, 1eV=1.60⋅10−19J
- Speed of light, c=3.00⋅108m/s
- Universal Gravitational constant,
G=6.67⋅10−11(N⋅m2)/kg2
- Solar Mass
M⊙=1.988⋅1030kg
- Acceleration due to gravity, g=9.8m/s2
- 1 unified atomic mass unit,
1u=1.66⋅10−27kg=931MeV/c2
- Planck's constant,
h=6.63⋅10−34J⋅s=4.41⋅10−15eV⋅s
- Permittivity of free space,
ϵ0=8.85⋅10−12C2/(N⋅m2)
- Coulomb's law constant,
k=4πϵ01=8.99⋅109(N⋅m2)/C2
- Permeability of free space,
μ0=4π⋅10−7T⋅m/A
- Magnetic constant,
4πμ0=1⋅10−7(T⋅m)/A
- 1 atmospheric pressure,
1atm=1.01⋅105N/m2=1.01⋅105Pa
- Wien's displacement constant, b=2.9. 10−3m⋅K
- Stefan-Boltzmann constant,
σ=5.67⋅10−8W/m2/K4