Mass center of system (airplane and helicopter crash)

JJBladester

Gold Member
286
2
1. The problem statement, all variables and given/known data

A small airplane of mass 1500 kg and a helicopter of mass 3000 kg flying at an altitude of 1200 m are observed to collide directly above a tower located at O in a wooded area. Four minutes earlier the helicopter had been sighted 8.4km due west of the tower and the airplane 16 km west and 12 km north of the tower. As a result of the collision the helicopter was split into two pieces, H1 and H2, of mass m1 = 1000 kg and m2 = 2000 kg, respectively; the airplane remained in one piece as it fell to the ground. Knowing that the two fragments of the helicopter were located at points H1 (500 m, -100 m) and H2 (600 m, -500 m), respectively, and assuming that all pieces hit the ground at the same time, determine the coordinates of the point A where the wreckage of the airplane will be found.

Answer:
(1004 m)i - (48.7 m)j

airplane%20helicopter.jpg


2. Relevant equations

[tex]m\mathbf{\overline{r}}=\sum_{i=1}^{n}m_i\mathbf{r_i}[/tex]

[itex]\mathbf{\overline{r}}[/itex] = position vector of mass center of system

"Problems of this type can be solved by writing the equation of motion of the mass center of the system in vectorial form and expressing its position vector before and after the collision in terms of the position vectors of the relevant bodies. You can then rewrite the vector equation as two or three scalar equations and solve these equations for an equivalent number of unknowns."

3. The attempt at a solution

[tex]m_A=1500kg[/tex]

[tex]m_H=3000kg[/tex]

[tex]t=4\, min=240\, s[/tex]

[tex]\mathbf{r_A}=-16000\mathbf{i}+12000\mathbf{j}+1200\mathbf{k}[/tex]

[tex]\mathbf{r_H}=-8400\mathbf{i}+0\mathbf{j}+1200\mathbf{k}[/tex]

[tex]\mathbf{\overline{r}}=\frac{m_A\mathbf{r_A}+m_H\mathbf{r_H}}{m_A+m_H}[/tex]

[tex]\mathbf{\overline{r}}=\frac{(1500kg)(-16000\mathbf{i}+12000\mathbf{j}+1200\mathbf{k})+(3000kg)(-8400\mathbf{i}+0\mathbf{j}+1200\mathbf{k})}{1500kg+3000kg}[/tex]

[tex]\mathbf{\overline{r}}=-10933\mathbf{i}+4000\mathbf{j}+1200\mathbf{k}[/tex]

Airplane motion before collision (not in freefall):
[tex]\mathbf{v_0}=\frac{16000m}{240s}\mathbf{i}+\frac{12000m}{240s}\mathbf{j}+0\mathbf{k}=66.7\mathbf{i}+50\mathbf{j}+0\mathbf{k}[/tex]

Helicopter motion before collision (not in freefall):
[tex]\mathbf{v_0}=\frac{8400m}{240s}\mathbf{i}+0\mathbf{j}+0\mathbf{k}=35\mathbf{i}+0\mathbf{j}+0\mathbf{k}[/tex]

Airplane motion after collision (in freefall):
Not sure... But I know the velocities in the x- and y-directions do not undergo acceleration. The velocity in the k-direction undergoes acceleration equal to gravity (g).

Helicopter motion after collision (in freefall):
Not sure... But I know the velocities in the x- and y-directions do not undergo acceleration. The velocity in the k-direction undergoes acceleration equal to gravity (g).

Final center of mass position vector:
[tex]\mathbf{\overline{r}}=\frac{m_A\mathbf{r_{A}}+m_1\mathbf{r_{H_1}}+m_2\mathbf{r_{H_2}}}{m_A+m_1+m_2}[/tex]

I know the goal is to find the final i and j coordinates of the airplane (we know the k coordinate is zero (all pieces fall on the ground [height = 0]). I know the theory behind this problem is that even though the collision occurs and we end up with pieces everywhere, the pieces should follow the trajectory of the mass center.

I know that at the moment of the collision, the plane and helicopter pieces undergo projectile motion in free fall and are accelerated downward at the rate of 9.81m/s2.

However, I am finding it hard to put all of the pieces together.
 
Last edited:

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