Question

A gadget of mass 21.85 kg floats in space without motion. Because of some internal malfunction, the gadget violently breaks up into 3 fragments flying away from each other. The first fragment has mass m1 = 6.42 kg and speed v1 = 6.8 m/s while the second fragment has mass m2 = 8.26 kg and speed v2 = 3.54 m/s. The angle between the velocity vectors ~v1 and ~v2 is θ12 = 64 ◦ . What is the speed v3 of the third fragment? Answer in units of m/s.

Answer 1

To solve this problem we will apply the concepts related to the conservation of momentum. For this purpose we will determine the velocities in the three body in the vertical and horizontal components. Once the system of equations is obtained, we will proceed to find the angle and the speed at which the third fragment is directed.

Mass of third part is

$m_3 = m-(m_1+m_2)$

$m_3= 21.85-6.42-8.26$

$m_3 =7.17 kg$

Assume that $m_1$ is along X-axis we have that $m_2$ makes an angle is 64 degrees with x-axis and $m_3$ makes an angle $\theta$  with x-axis.

Using law of conservation of momentum along X-axis

$0 = (6.42*6.8)+(8.26*3.54*cos(64))+(7.71v_3 cos\theta)$

$(7.71v_3 cos\theta) = 56.4741$

$v_3 cos\theta = 7.3247$ $\rightarrow \text{Equation 1}$

Now applying the same through the Y-axis.

$0=0+8.26*3.54*sin(64\°) + 7.71*v_3*sin\theta$

$-8.26*3.54*sin(64\°)=7.71*v_3*sin\theta$

$v_3*sin\theta = -3.409$ $\rightarrow \text{Equation 2}$

If we divide the equation 1 with the equation 2 we have that

$\frac{v_3cos\theta}{v_3 sin\theta } = \frac{7.3247}{-3.409}$

$tan\theta = \frac{7.3247}{-3.409}$

$\theta = tan^{-1} (\frac{7.3247}{-3.409})$

$\theta = -65.04\°$

Using this angle in the second equation we have that velocity 3 is,

$v_3 = \frac{-3.409}{sin(-65.04)}$

$v_3 = 3.7601m/s$

Therefore the speed of the third fragment is $3.7601\frac{m}{s} \angle -65.04\°$