Question

In a physics demonstration, two rubber balls are dropped down a tube towards the ground, with the lighter ball behind, and not quite in contact with, the heavier ball. The heavier ball collides with the ground and bounces, then collides with the still descending lighter ball. Assume that all the collisions are elastic. In one such demonstration, the heavier ball has a mass of 500 grams, the lighter one a mass of 100 grams, and they are released 1.00 m above the ground. How high does the lighter ball rise after the two collisions

Answer 1

h = 5.45 m

Step-by-step explanation:

Let's perform this exercise in parts, let's look for the speed of the balls when they reach the ground, for this we use the concepts of energy conservation

Initial, at the point where they are released

Em₀ = U = m gh

Fianl on reaching the ground

`Em_(f)` = ½ m v²

Em₀ =
`Em_(f)`

mgh = ½ m v²

v = √ 2gh

We see that the speed of the two balls is the same when they reach the ground, because it does not depend on the mass

v = √ (2 9.8 1)

v = 4,427 m / s

Now let's use the conservation of the amount of movement to find the velocity of the heaviest ball when it bounces on the ground.

Before crash

p₀ = Mv

After it bounces

`p_(f)` = -M v

p₀ =
`p_(f)`

-M v = M
`v_(f)`

`v_(f)` = -v

The ball bounces with the same speed, but in the opposite direction

Now let's study the clash of the two balls, suppose it has the same small diameter, the magnitude of the velocity is the same, but the large bullet has an upward direction (positive)

Initial before the crash

p₀ = M v - m v

After the crash

`p_(f)` = m
`v_(1f)` + M
`v_(2f)`

p₀ =
`p_(f)`

(M-m) v = m
`v_(1f)` + M
`v_(2f)`

As the shock is elastic the kinetic energy is conserved

K₀ = ½ M v² + ½ m v²

K₀ = ½ (M + m) v²

`K_(f)` = ½ m
`v_(1f)`² + ½ M
`v_(2f)`²

K₀ =
`K_(f)`

(M + m) v² = m
`v_(1f)`² + v
`v_(2f)`²

Let's write and solve our system of equations

(M-m) v = m v1f + M v2f

(M + m) v² = m
`v_(1f)`² +M
`v_(2f)`²

Let's replace

0.4 4.427 = 0.1 v1f +0.5 v2f

0.6 4.427² = 0.1
`v_(1f)`² + 0.5
`v_(2f)`²

1.77 = 0.1 v1f +0.5 v2f

11.76 = 0.1
`v_(1f)`² + 0.5
`v_(2f)`²

`v_(2f)` = (1.77 - 0.1
`v_(1f)`) / 0.5

I substitute in the other equation

11.76 = 0.1
`v_(1f)`² + (1.77 -0.1
`v_(1f)`)² / 0.5

11.76= 0.1
`v_(1f)`² + 6.2658 - 0.708
`v_(1f)` + 0.02
`v_(1f)`²

0.12
`v_(1f)`² - 0.708
`v_(1f)` -5.4942 = 0

`v_(1f)`² - 5.9
`v_(1f)`- 45. 785 = 0

Let's solve the second degree equation

`v_(1f)` = [5.9 ±√(5.9 2 + 4 45,785)] / 2 = [5.9 + - 14.76] / 2

Results

`v_(1f)` = 10.3315 m / s

`v_(1f)` = -4.43 m / s

The correct result is the positive, since the negative result the lighter ball should follow its trajectory, which is impossible since the heaviest is going up

`v_(1f)` = 10. 3315 m / s

With this speed we can use energy to calculate until the lightest ball rises

Initial

Em₀ = K = ½ m
`v_(1f)`²

Final

`Em_(f)` = U = m g h

Em₀ =
`Em_(f)`

½ m
`v_(1f)`² = m g h

h = ½
`v_(1f)`² / g

h = ½ 10.3315² / 9.8

h = 5.45 m