Question

A rocket is fired vertically upward. At the instant it reaches some altitude with a speed of 100 m/s. It explodes into three fragments having equal mass. One fragment moves upward with a speed of 150 m/s following the explosion. The second fragment has a speed of 150 m/s and is moving east right after the explosion. What is the velocity of the third fragment immediately after the explosion?

Answer 1

Answer:

The magnitude of the velocity of the third fragment  is 212.13 m/s.

Explanation:

Given that,

Speed = 100 m/s

Vertical speed of first fragment = 150 m/s

Vertical speed of second fragment = 150 m/s

We need to calculate the vertical speed of third fragment

Using conservation of momentum

$P_{iy}=P_{fy}$

$mv=\dfrac{m}{3}{v_{1}}+\dfrac{m}{3}v_{2}+\dfrac{m}{3}v_{3}$

Put the value into the formula

$m\times100=\dfrac{m}{3}\times150+\dfrac{m}{3}\times0+\dfrac{m}{3}\times v_{y}$

$\dfrac{1}{3}v_{y}=100-50$

$v_{y}=150\ m/s$

We need to calculate the horizontal speed of third fragment

Using conservation of momentum

$P_{iy}=P_{fy}$

$mv=\dfrac{m}{3}{v_{1}}+\dfrac{m}{3}v_{2}+\dfrac{m}{3}v_{3}$

Put the value into the formula

$m\times0=\dfrac{m}{3}\times0+\dfrac{m}{3}\times150+\dfrac{m}{3}v_{x}$

$\dfrac{1}{3}v_{x}=-50$

$v_{x}=-150\ m/s$

We need to calculate the magnitude of the velocity of the third fragment

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(-150)^2+(150)^2}$

$v=212.13\ m/s$

Hence, The magnitude of the velocity of the third fragment  is 212.13 m/s.