Question

Two skaters, one with mass 75 kg and the other with mass 60 kg, stand on an ice rink holding a pole of length 14 m and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the 60 kg skater move

Answer 1

Answer: 70m

Explanation:

Given, 2 skaters, one with a mass of 60kg.

Another 75kg.

Distance is 14m.

Assuming the system centre if mass does not change, both skaters would meet at the centre of the mass

let the 60kg skater be xm away from the centre, then, the 75kg skater would be (14 - x)m away from the centre too

m1x + m2(14 - x) = 0

m1x + 14m2 - m2x = 0

60x + 14*75 - 75x = 0

60x - 75x + 1050 = 0

-15x + 1050 = 0

15x = 1050

x = 70

Thus, the distance the 65kg skater moves is 70m

Answer 2

Answer:

The 60 kg skater move 7.8 m

Explanation:

given information:

first skater's mass, m₁ = 75 kg

second skater's mass, m₂ = 60 kg

pole length, x = 14 m

to calculate how far the the second skater moved, we can use the following equation:

$x_{com}$ = $\frac{1}{M}$Σ$m_{i} x_{i}$

where

M = ∑$m_{i}$

   = m₁ + m₂

   = 75 + 60

   = 135 kg

$x_{com}$ = $\frac{1}{M}$Σ$m_{i} x_{i}$

$x_{com}$ = $\frac{1}{135}$ (m₁x₁ + m₂x₂), $x_{com}$ = 0 since both skater meet

now let assume the the second skater moved x₂ = a m, so the first skater moved x₁ = 14 - a

so,

$x_{com}$ = $\frac{1}{135}$ (m₁x₁ + m₂x₂)

0 = $\frac{1}{135}$ [(75(14-a) + 60a)]

0 = $\frac{1}{135}$(1050 - 75a + 60(-a))

0 = (1050 - 135a)

135a = 1050

a = 1050/135

  = 7.8 m