Answer:
Explanation:
Let assume that ball strikes a vertical wall in horizontal direction. The situation can be modelled by the appropriate use of the definition of Moment and Impulse Theorem, that is:
The average force acting on the ball during the collision is:
Naturally we assume that 10000 km/hr is initial velocity (same as being shot from a cannon), and no air resistance. With so high a velocity, the effect of diminishing gravity with increasing radius must be taken into account, so you use an energy solution. M is earth mass, r is earth radius.
KE/m = (9000000/3600)^2/2 = 3858025 J/kg
ΔPE/m = (PE(at height) - PE(at surface))/m = -GM/(r+h) + GM/r
KE/m = ΔPE/m
KE/m - GM/r = -GM/(r+h)
h = -GM / (KE/m - GM/r) - r = 335665.44 m
(Using G = 6.673E-11 Nm^2/kg^2, M = 5.9742E24 kg, r = 6378100 m)
Answer:
1.603 s
Explanation:
Given that
Initial mass, = 0.45 kg
Initial period, = 1.45 s
Initial radius, = 0.14 m
Final mass, = 0.55 kg
Final period, = ?
Final radios, = 0.14 m
Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that
m₁r₁ω₁² = m₂r₂ω2²
Where, ω = 2π/T, on substituting, we have
0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²
0.45 / 1.45² = 0.550 / T₂²
T₂² = 0.550 * 1.45² / 0.45
T₂² = 2.56972
T₂ = √2.56972
T₂ = 1.603 sec
