Question

two arrows are fired horizontally with the same speed of 30.0 m/s. Each arrow has a mass of 0.100kg. One is fired due east and the other due south. Find the magnitude and direction of the total momentum of this two-arrow system. Specify the direction with respect to due east.

Answer 1

|momentum| = (mass) x (speed)

Momentum of each arrow = (0.1 kg) x (30 m/s) = 3 kg-m/s

Everything is equal for the two arrows, except their direction.
So the total momentum vector of the 2-arrow system consists of

-- a 3 kg-m/s component pointing east
plus
-- a 3 kg-m/s component pointing south.

The magnitude of the total momentum is

                √ [  (3 kg-m/s)²  +  (3 kg-m/s)²  ]

            =  √ [  9 kg²-m²/s²  +  9 kg²-m²/s²  ]

            =  √ [  18 kg²-m²/s²  ]

            =  3√2  kg-m/s        (approx  4.243 kg-m/s)

The direction of the total momentum vector is the angle
south of east whose tangent is

                             (3 kg-m/s) / (3 kg-m/s)  

                   =  tan⁻¹ (1)

                   =   45° south of east.

That direction is southeast.  

Answer 2

The magnitude of the total momentum is $\fbox{4.24 kg m/s}$ and the direction is $\bf{south-east}$.

Further Explanation:

The net momentum acting is the sum of all the momentum which is applying on the different bodies considering the direction of action of the momentum. In other words, the net momentum applying on a body is the sum of all the vectors of all the momentum.

The momentum is the vector quantity has some magnitude and unique direction represented in the coordinates system on the x, y and z-axis. It is the related directly to the object mass and object velocity.

Velocity is the vector quantity has some magnitude and some direction represented in the coordinates system on the x, y and z-axis.

Given:

The speed of the two arrows is $30.0 m/s$.

The mass of each arrow is $0.1 kg$.

The first arrow is fired due east and other arrow is fired due south.

Concept:

The expression for the momentum is:

$P=m\vec v$

Here, $p$ is the momentum, $m$ is the mass of the object and $v$ is the velocity of the object.

The velocity can be represented in the vector form.

The first arrow is fired due east can be written as $30.0{\text{ m/s }}\hat i$ and other arrow is fired due south can be written as $- 30.0{\text{ m/s }}\hat j$.

The total momentum can be expressed as:

$\vec P={m_1}{\vec v_1} + {m_2}{\vec v_2}$

Here, $\vec P$ is the total momentum, ${\vec v_1}$ and ${\vec v_2}$ are the velocities of the arrows and ${m_1}$, ${m_2}$ are the masses of the arrow.

Substitute $30.0{\text{ m/s }}\hat i$ for ${\vec v_1}$,  $- 30.0{\text{ m/s }}\hat j$ for ${\vec v_2}$, $0.1 kg$ for ${m_1}$ and $0.1 kg$ for ${m_2}$ in the above equation.

$\begin{aligned}\vec P&=\left( {0.1{\text{ kg}}} \right)\left[ {\left( {30.0{\text{ m/s }}\hat i} \right) + \left( { - 30.0{\text{ m/s }}\hat j} \right)} \right]\\&=\left( {0.1{\text{kg}}}\right)\left[ {30.0{\text{ m/s }}\hat i - 30.0{\text{ m/s }}\hat j} \right] \\&=\left( {3\hat i - 3\hat j} \right){\text{ kg}} \cdot {\text{m/s }} \\\end{aligned}$

Simplify further for the magnitude of the vector.

$\begin{aligned}\left| P \right|&=\sqrt {{3^2} + {3^2}}\\&=\sqrt {9 + 9}\\&=\sqrt {18}\\&=4.24{\text{ kg}}\cdot {\text{m/s}}\\\end{aligned}$

The direction of the vector can be written as:

$\begin{aligned}\theta&={\tan ^{ - 1}}\left( {\frac{{ - 3}}{3}} \right)\\&={\tan ^{ - 1}}\left( { - 1} \right)\\&=- 45^\circ\\\end{aligned}$

Therefore, the magnitude of the total momentum is $\fbox{4.24 kg m/s}$ and the negative sign show the direction is $\bf{south-east}$.

Learn more:

1. State and prove law of conservation of linear momentum https://brainly.in/question/3078233

2. Momentum with numerical. https://brainly.in/question/4130937

3. Impulse momentum. brainly.com/question/11604162

Answer Details:

Grade: Middle school

Subject: Physics

Chapter: Kinematics

Keywords:

Two, arrow, fired, horizontally, same, speed, 30.0 m/s,  ,  , mass, 0.100 kg. One, due, east, south, magnitude, direction, total, momentum, two-arrow system, specify, respect, 4.24 kg m/s, $4.24{\text{ }}\dfrac{{{\text{kg}} \cdot {\text{m}}}}{{\text{s}}}$, south-east.