Rodney is trying out one of Santa's new games that consists of three pieces that blow apart if the wrong key is inserted into the door. The elf picks the wrong key and the board explodes. One piece of mass 2 kg goes off at an angle of 60° from the horizon at 10 m/s. A second piece of mass 1.2 kg travels at an angle of 180° and is moving at 15 m/s.

a) find the momentum of the third piece.

b) find the velocity of the third piece if it's mass is 3 kg.

1 Answer

  • Answer:

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


    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]


    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]


    [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]


    [tex]\displaystyle m_3\ \vec{v}_3=<8,-10\sqrt{3}>[/tex]


    [tex]\displaystyle m_3\ \vec{v}_3=<8,-17.32>[/tex]

    This is the momentum of the third piece


    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