Question

An experiment is performed aboard the International Space Station to verify that linear momentum is conserved during collisions in a zero‑gravity environment. The experiment involves a three‑dimensional, completely inelastic collision of three drops of honey. At the moment just before they all collide, the masses m and velocities → v of the drops are m 1 = 35.5 g → v 1 = ( 13.1 m / s ) ^ x m 2 = 52.3 g → v 2 = ( 14.5 m / s ) ^ y m 3 = 75.7 g → v 3 = ( 18.3 m / s ) ^ z What is your prediction for the speed, V , of the combined honey drop after the collision?

Answer 1

Answer:

$v=15.9554\ m.s^{-1}$

Explanation:

Given:

• mass of the honey drop 1, $m_1=35.5\times 10^{-3}\ kg$

• velocity of the honey drop 1, $v_1=13.1\ m.s^{-1}$

• mass of the honey drop 2, $m_1=52.3\times 10^{-3}\ kg$

• velocity of the honey drop 2, $v_2=14.5\ m.s^{-1}$

• mass of the honey drop 3, $m_1=75.7\times 10^{-3}\ kg$

• velocity of the honey drop 3, $v_3=18.3\ m.s^{-1}$

In ISS there is zero gravity an the collision is completely inelastic.

So, applying the law of conservation of momentum:

$m_1.v_1+m_2.v_2+m_3.v_3=(m_1+m_2+m_3).v$

$35.5\times 10^{-3}\times 13.1+52.3\times 10^{-3}\times 14.5+75.7\times 10^{-3}\times 18.3=(35.5+52.3+75.7)\times 10^{-3}\times v$

$v=15.9554\ m.s^{-1}$