Question

Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each other. If m1 is pulled up and released from rest at a height of 1.00 m, what are the maximum heights achieved by (a) m1 and (b) m2 after an elastic collision?

Answer 1

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Step-by-step explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

`mgh_(max) = (1)/(2)mv_(max)^2`

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

`mgh_(max) = (1)/(2)mv_(max)^2 \\\\gh_(max) = (1)/(2)v_(max)^2\\\\ h_(max) = (1)/(2g)v_(max)^2\\\\ h_(max) = (1)/(2*9.8)(1.477^2)\\\\ h_(max) = 0.11 \ m`

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

`mgh_(max) = (1)/(2)mv_(max)^2 \\\\gh_(max) = (1)/(2)v_(max)^2\\\\ h_(max) = (1)/(2g)v_(max)^2\\\\ h_(max) = (1)/(2*9.8)(2.95^2)\\\\ h_(max) = 0.44 \ m`