Respuesta :
Answer:
The third piece moves at 6.36 m/s at an angle of 65° below the horizon
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 [tex]\vec v[/tex], the linear momentum is computed by
[tex]\displaystyle \vec{P}=m.\vec{v}[/tex]
a)
The momentum of the board before the explosion is
[tex]\displaystyle \vec{P}_{t1}=m_t\ \vec{v}_o[/tex]
Since the board was initially at rest
[tex]\displaystyle \vec{P}_{t1}=<0,0>[/tex]
After the explosion, 3 pieces are propelled in different directions and velocities, and the total momentum is
[tex]\displaystyle \vec{P}_{t2}=m_1\ \vec{v}_1\ +\ m_2\ \vec{v}_2+m_3\ \vec{v}_3[/tex]
The first piece of 2 kg moves at 10 m/s in a 60° direction
[tex]\displaystyle \vec{v}_1=(10\ m/s,60^o)[/tex]
We find the components of that velocity
[tex]\displaystyle \vec{v}_1=<10\ cos60^o,10\ sin60^o>[/tex]
[tex]\displaystyle \vec{v}_1=<5,5\sqrt{3}>m/s[/tex]
The second piece of 1.2 kg goes at 15 m/s in a 180° direction
[tex]\displaystyle \vec{v}_2=(15,180^o)[/tex]
Its components are computed
[tex]\displaystyle \vec{v}_2=(15\ cos180^o,15\ sin180^o)[/tex]
[tex]\displaystyle \vec{v}_2=(-15,0)\ m/s[/tex]
The total momentum becomes
[tex]\displaystyle P_{t2}=2<5,5\sqrt{3}>+1.2<-15,0>+m_3\ \vec{v}_3[/tex]
Operating
[tex]\displaystyle P_{t2}=<10,10\sqrt{3}>+<-18,0>+m_3\ \vec{v}_3[/tex]
Knowing the total momentum equals the initial momentum
[tex]\displaystyle P_{t2}=<-8,10\sqrt{3}>+m_3\ \vec{v}_3=0[/tex]
Rearranging
[tex]\displaystyle m_3\ \vec{v}_3=<8,-10\sqrt{3}>[/tex]
Calculating
[tex]\displaystyle m_3\ \vec{v}_3=<8,-17.32>[/tex]
This is the momentum of the third piece
b)
From the above equation, we solve for [tex]\vec v_3[/tex]:
[tex]\displaystyle \vec{v}_3=\frac{1}{3}<8,-17.32>[/tex]
[tex]\displaystyle \vec{v}_3=<2.67,-5.77>m/s[/tex]
The magnitude of the velocity is
[tex]\displaystyle \vec{v}_3|=\sqrt{2.67^2+(-5.77)^2}=6.36[/tex]
And the angle is
[tex]\displaystyle tan\theta =\frac{-5.77}{2.67}=-2.161[/tex]
[tex]\displaystyle \theta =-65.17^o[/tex]
The third piece moves at 6.36 m/s at an angle of 65° below the horizon