Question

A man weighing 680 n and a woman weighing 470 n have the same momentum. what is the ratio of the man's kinetic energy km to that of the woman kw? view available hint(s)

Answer 1

Answer:

47/68

Explanation:

They have the same momentum, so:

(680/g) v₁ = (470/g) v₂

680 v₁ = 470 v₂

The man's kinetic energy is:

KE = ½ mv²

KE₁ = ½ (680/g) v₁²

The woman's kinetic energy is:

KE = ½ mv²

KE₂ = ½ (470/g) v₂²

The ratio is:

KE₁ / KE₂ = [½ (680/g) v₁²] / [½ (470/g) v₂²]

KE₁ / KE₂ = (680 v₁²) / (470 v₂²)

KE₁ / KE₂ = (680² v₁²) / (680 × 470 v₂²)

KE₁ / KE₂ = (680 v₁)² / (680 × 470 v₂²)

KE₁ / KE₂ = (470 v₂)² / (680 × 470 v₂²)

KE₁ / KE₂ = 470 / 680

KE₁ / KE₂ = 47 / 68