Question

A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket has lost 10% of its mass by expelling air at an average velocity of 53.m⋅s−1. What is the velocity of the rocket at that moment

Answer 1

Answer:

$v_r=5.89\ m.s^{-1}$

Explanation:

Given:

• mass of rocket, $m_r=50\ g$

• time of observation, $t=2\ s$

• mass lost by the rocket by expulsion of air, $m_a=10\%\ of m_r=5\ g$

• velocity of air, $v_a=53\ m.s^{-1}$

Now the momentum of air will be equal to the momentum of rocket in the opposite direction: (Using the theory of elastic collision)

$m_a.v_a=(m_r-m_a)\times v_r$

$5\times 53=(50-5)\times v_r$

$v_r=5.89\ m.s^{-1}$