The Laws of Collisions in Physics
A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball.
a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball?
Answer:
a) the large sphere has 3 times the mass of the small sphere
b) What final height, in terms of H, does the small ball reach?
Answer:
b) y = 4H
Explanation:
We must start this problem by calculating the speed with which the spheres reach the floor
vf² = vo² - 2g y
As the spheres are released v₀ = 0
vf² = - 2g H
vf = √ (2g H)
The two spheres arrive with the same speed to the floor.
The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres
V₁₀ = √2gH
The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation
Let's call vh = √2gH
Small sphere m₂ and v₂o = - √2gH = -vh
Large sphere m₁ and v₁o = √ 2gh = vh
Before crash
p₀ = m₁ v₁₀ + m₂ v₂₀
After the crash
pf = m₁ v₁f + m₂ v₂f
po = pf
m₁ v₁₀ + m₂ v₂₀ = m₁ v₁f + m₂ v₂f
The conservation of kinetic energy
Ko = ½ m₁ v₁₀² + ½ m₂ v₂₀²
Kf = ½ m₁ v₁f² + ½ m₂ v₂f²
Ko = KF
½ m₁ v1₁₀² + ½ m₂ v₂₀² = ½ m₁ v₁f² + ½ m₂ v₂f²
Let's write the values
-m₁ vh + m₂ vh = m₁ v₁f + m₂ v₂f
m₁ vh² + m₂ vh² = m₁ v₁f² + m₂ v₂f²
The solution to this system of equations is
mt = m₁ + m₂
v1f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂
v₂f = 2m₁ /mt v₁₀ + (m₂-m₁) / mt v₂
The large sphere is labeled 1, we are asked for the mass so that v1f = 0, let's clear the equation
v₁f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂₀
0 = (m₁-m₂) / mt vh + 2 m₂ / mt (-vh)
(m₁-m₂) / mt vh = 2 m₂ / mt vh
(m₁-m₂) = 2m₂
m₁ = 3 m₂
The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.
b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)
v₂f = 2m₁ / mt v₁₀ + (m₂-m₁) / mt v₂₀
v₂f = 2 m₁ / mt vh + (m₂-m₁) mt (-vh)
In addition, we know that m₁ = 3 m₂
mt = 3m2 + m2
mt= 4m2
v₂f = 2 3m₂ / 4m₂ vh - (m₂-3m₂) 4m₂ vh
v₂f = 3/2 vh +1/2 vh
v₂f = 2 vh
v₂f = 2 √ 2gh
This is the rate of rise of sphere 2 (small. At the highest point its zero velocity vf = 0
V² = v₂f² - 2 g Y
0 = (2√2gh)² - 2gy
2gy = 4 (2gH)
y = 4H
What will be the mass of the sphere and height covered by the small ball?
a) The large sphere has 3 times the mass of the small sphere
b) The final height at which small ball reach y = 4H