Question

A 3.0-kilogram object initially at rest explodes and splits into three fragments. One fragment has a mass of 0.50 kg and flies off along the negative x axis at a speed of 2.8 m/s, and another has a mass of 1.3 kg and flies off along the negative y axis at a speed of 1.5 m/s.
Required:
What are the speed and direction of the third fragment?

Answer 1

Answer:

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.

Explanation:

The object is not under the influence of any external force, meaning that Principle of Momentum Conservation to calculation of the velocity of the third fragment:

$(m_{1}+m_{2}+m_{3})\cdot \vec {v}_{o} = m_{1}\cdot \vec v_{1} + m_{2}\cdot \vec v_{2} + m_{3}\cdot \vec v_{3}$ (1)

Where:

$m_{1}$, $m_{2}$, $m_{3}$ - Masses of the first, second and third fragments, in kilograms.

$\vec v_{o}$ - Initial velocity of the object, in meters per second.

$\vec v_{1}$, $\vec v_{2}$, $\vec v_{3}$ - Velocities of the first, second and third fragments, in meters per second.

If we know that $m_{1} = 0.5\,kg$, $m_{2} = 1.3\,kg$, $m_{3} = 1.2\,kg$, $\vec v_{o} = (0,0)\,\left[\frac{m}{s} \right]$, $\vec v_{1} = \left(-2.8, 0\right)\,\left[\frac{m}{s} \right]$ and $\vec v_{2} = \left(0,-1.5\right)\,\left[\frac{m}{s} \right]$, the velocity of the third fragment is:

$(-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)$

$1.2\cdot \vec v_{3} = (1.4,1.95)$

$\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right]$

The speed of the third fragment is the magnitude of the result found above:

$v_{3} = 2\,\frac{m}{s}$

And the direction of the third fragment is:

$\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)$

$\theta_{3} \approx 54.316^{\circ}$

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.