Answer:
<em>The third piece moves at 6.36 m/s at an angle of 65° below the horizon</em>
Explanation:
Linear Momentum
It's a physical magnitude that measures the product of the velocity by the mass of a moving object. In a system where no external forces are acting, the total momentum remains unchanged regardless of the interactions between the objects in the system.
If the velocity of an object of mass m is
, the linear momentum is computed by

a)
The momentum of the board before the explosion is

Since the board was initially at rest

After the explosion, 3 pieces are propelled in different directions and velocities, and the total momentum is

The first piece of 2 kg moves at 10 m/s in a 60° direction

We find the components of that velocity


The second piece of 1.2 kg goes at 15 m/s in a 180° direction

Its components are computed


The total momentum becomes

Operating

Knowing the total momentum equals the initial momentum

Rearranging

Calculating

This is the momentum of the third piece
b)
From the above equation, we solve for
:


The magnitude of the velocity is

And the angle is


The third piece moves at 6.36 m/s at an angle of 65° below the horizon