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oksano4ka [1.4K]
3 years ago
7

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 th

e 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.
Physics
1 answer:
ch4aika [34]3 years ago
5 0

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 \vec v, the linear momentum is computed by

\displaystyle \vec{P}=m.\vec{v}

a)

The momentum of the board before the explosion is

\displaystyle \vec{P}_{t1}=m_t\ \vec{v}_o

Since the board was initially at rest

\displaystyle \vec{P}_{t1}=

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

\displaystyle \vec{P}_{t2}=m_1\ \vec{v}_1\ +\ m_2\ \vec{v}_2+m_3\ \vec{v}_3

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

\displaystyle \vec{v}_1=(10\ m/s,60^o)

We find the components of that velocity

\displaystyle \vec{v}_1=

\displaystyle \vec{v}_1=m/s

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

\displaystyle \vec{v}_2=(15,180^o)

Its components are computed

\displaystyle \vec{v}_2=(15\ cos180^o,15\ sin180^o)

\displaystyle \vec{v}_2=(-15,0)\ m/s

The total momentum becomes

\displaystyle P_{t2}=2+1.2+m_3\ \vec{v}_3

Operating

\displaystyle P_{t2}=++m_3\ \vec{v}_3

Knowing the total momentum equals the initial momentum

\displaystyle P_{t2}=+m_3\ \vec{v}_3=0

Rearranging

\displaystyle m_3\ \vec{v}_3=

Calculating

\displaystyle m_3\ \vec{v}_3=

This is the momentum of the third piece

b)

From the above equation, we solve for \vec v_3:

\displaystyle \vec{v}_3=\frac{1}{3}

\displaystyle \vec{v}_3=m/s

The magnitude of the velocity is

\displaystyle \vec{v}_3|=\sqrt{2.67^2+(-5.77)^2}=6.36

And the angle is

\displaystyle tan\theta =\frac{-5.77}{2.67}=-2.161

\displaystyle \theta =-65.17^o

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

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