Question

An object at rest explodes into three fragments. The figure shows the momentum vectors of two of the fragments.

What is p_x of the third fragment?

What is p_y of the third fragment?

answer in "kg m/s" please!

"HINT" the answer IS NOT p3_x=0 & p3_y=-2

Answer 1

Concepts and reason

The concepts required for this problem are Linear momentum conservation, vector addition and equality of vectors.

First find the x and y- component of the velocities of two fragment and then use the conservation of linear momentum to find the horizontal and vertical component of velocity.

Fundamentals

Momentum: momentum is a vector quantity which is equal to the product of mass and velocity of the body. This can be written as follows;

$\vec P = m\vec v$

Here,$\vec P$is the momentum,$\vec v$is the velocity of the mass and $m$is the mass of the body.

Law of conservation of momentum states that total momentum of the system remains conserved if there is no external force acting on the system.

It is represented by the expression as follows;

${p_{\rm{i}}} = {p_{\rm{f}}}$

Here, ${p_{\rm{i}}}$ is initial momentum and ${p_{\rm{f}}}$ is final momentum.

(a)

Calculate the x and y- components of the momentum of two fragments as follows:

From the graph, the horizontal component of momentum of first fragment is as follows:

${\vec P_{1x}} = \left( { - 2{\rm{ }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}}$

The y- component of momentum of first fragment is as follows:

${\vec P_{1y}} = \left( {2{\rm{ }}{\bf{\hat j}}} \right){\rm{ }}{\rm{kg}} \cdot {\rm{m/s}}$

The horizontal component of momentum of second fragment is as follows:

${\vec P_{2x}} = \left( {{\rm{3 }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}}$

The y- component of momentum of first fragment is as follows:

${\vec P_{1y}} = 0$

Calculate the horizontal component of momentum for the third particle as follows:

Since the object was initially at rest therefore the initial horizontal component of the system is zero. This can be written as follows:

${\vec P_{{\rm{i}}x}} = 0$

The final horizontal component of the momentum of the system is as follows:

${P_{{\rm{f}}x}} = {\vec P_{{\rm{1}}x}} + {\vec P_{{\rm{2}}x}} + {\vec P_{{\rm{3}}x}}$

Substitute $\left( { - 2{\rm{ }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}}$for${\vec P_{1x}}$ and $\left( {{\rm{3 }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}}$for${\vec P_{2x}}$in the equation${P_{{\rm{f}}x}} = {\vec P_{{\rm{1}}x}} + {\vec P_{{\rm{2}}x}} + {\vec P_{{\rm{3}}x}}$to calculate final horizontal component of the momentum of the system.

${\vec P_{{\rm{f}}x}} = \left( { - 2{\rm{ }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}} + \left( {{\rm{3 }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}} + {\vec P_{3x}}$

According to the law of conservation of momentum,

${\vec P_{{\rm{f}}x}} = {\vec P_{{\rm{i}}x}}$

Substitute $\left( { - 2{\rm{ }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}} + \left( {{\rm{3 }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}} + {\vec P_{3x}}$for${\vec P_{fx}}$and$0$for${\vec P_{ix}}$in the equation${\vec P_{{\rm{f}}x}} = {\vec P_{{\rm{i}}x}}$ and solve for${\vec P_{3x}}$.

$\begin{array}{c}\\\left( { - 2{\rm{ }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}} + \left( {{\rm{3 }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}} + {{\vec P}_{3x}} = 0\\\\{{\vec P}_{3x}} = \left( { - {\rm{ }}{\bf{\hat i}}} \right){\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

Hence, ${P_x}$of the third fragment is$- 1{\rm{ kg}} \cdot {\rm{m/s}}$.

(b)

Calculate the y-component of third fragment as follows:

Since the object was initially at rest therefore the initial horizontal component of the system is as zero. This can be written as follows;

${\vec P_{{\rm{i}}y}} = 0$

The final y-component of the momentum of the system is as follows:

${P_{{\rm{f}}y}} = {\vec P_{{\rm{1}}y}} + {\vec P_{{\rm{2}}y}} + {\vec P_{{\rm{3}}y}}$

Substitute $\left( {2{\rm{ }}{\bf{\hat j}}} \right){\rm{ kg}} \cdot {\rm{m/s}}$for${\vec P_{1y}}$ and $0$for${\vec P_{2y}}$in the equation${P_{{\rm{f}}y}} = {\vec P_{{\rm{1}}y}} + {\vec P_{{\rm{2}}y}} + {\vec P_{{\rm{3}}y}}$to calculate the final horizontal component of the momentum of the system.

${\vec P_{{\rm{f}}y}} = \left( {2{\rm{ }}{\bf{\hat j}}} \right){\rm{ kg}} \cdot {\rm{m/s}} + 0 + {\vec P_{3y}}$

According to the law of conservation of momentum,

${\vec P_{{\rm{f}}y}} = {\vec P_{{\rm{i}}y}}$

Substitute $\left( {2{\rm{ }}{\bf{\hat j}}} \right){\rm{ kg}} \cdot {\rm{m/s}} + 0 + {\vec P_{3y}}$for${\vec P_{{\rm{f}}y}}$ and $0$for${\vec P_{{\rm{i}}y}}$in the equation ${\vec P_{{\rm{f}}y}} = {\vec P_{{\rm{i}}y}}$ and solve for${\vec P_{3y}}$.

$\begin{array}{c}\\\left( {2{\rm{ }}{\bf{\hat j}}} \right){\rm{ kg}} \cdot {\rm{m/s}} + 0 + {{\vec P}_{3y}} = 0\\\\{{\vec P}_{3y}} = \left( { - 2{\rm{ }}{\bf{\hat j}}} \right){\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

Hence, ${P_y}$of the third fragment is$- 2{\rm{ kg}} \cdot {\rm{m/s}}$.

Ans: Part a

The ${P_x}$ of the third fragment is$- 1{\rm{ kg}} \cdot {\rm{m/s}}$.

Part b

The ${P_y}$ of the third fragment is$- 2{\rm{ kg}} \cdot {\rm{m/s}}$.