Question

A watermelon is blown into three pieces by a large firecracker. Two pieces of equal mass m fly away perpendicular to one another, one in the x direction another in the y direction. Both of these pieces fly away with a speed of V = 35 m/s. The third piece has three times the mass of the other two pieces. (a) Write an expression for the speed of the larger piece, that is in terms of only the variable V.
No Attempt No Attempt 33% Part (b) What is the numeric value for the speed of the larger piece, in meters per second?
show answer No Attempt 33% Part (c) At what angle does the largest piece travel with respect to the -y axis, in degrees?

Answer 1

Answer:

(a) Speed of the third piece in terms of $V$: $\displaystyle \frac{\sqrt{2}}{3}\; V$.

(b) Speed of the third piece in meters per second: approximately $\rm 16.5\; m \cdot s^{-1}$.

(c) Refer to the diagram attached to this answer. The third piece travels at $\rm 45^{\circ}$ to the left of the negative y-axis.

Explanation:

(a)

Momentum shall conserve if this explosion is isolated from the surroundings. The vector sum of the momentum of the three pieces should be equal to zero.

Start by expressing the velocities of the first two pieces as vectors.

The first piece flies in the direction of the positive x-axis. If its velocity is $1$, its velocity vector would be $\displaystyle \left[ \begin{array}{c}1 \\ 0\end{array}\right]$. (The first component is the speed in the x-direction, and the second the speed in the y-direction.) However, since it is moving $V$ times as fast, its actual velocity vector will be $\vec{v}_1 =\displaystyle V\times \left[ \begin{array}{c}1 \\ 0\end{array}\right] = \left[ \begin{array}{c}V \\ 0\end{array}\right]$.

Similarly, the second piece flies in the direction of the positive y-axis. If its velocity is $1$, its velocity vector would be $\displaystyle = \left[ \begin{array}{c}0 \\ 1\end{array}\right]$. However, since it is moving $V$ times as fast, its actual velocity vector will be $\vec{v}_2 = \displaystyle V\times \left[ \begin{array}{c}0 \\ 1\end{array}\right] = \left[ \begin{array}{c}0 \\ V\end{array}\right]$.

The momentum $\vec{p}$ (a vector) of an object is equal to its mass (a scalar) times its velocity (also a vector.)

The question states that the mass of both the first and the second pieces are equal to $m$. The momentum of the first piece will be thus equal to $\displaystyle p_1 = m \cdot \left[ \begin{array}{c}V \\ 0\end{array}\right] = \left[ \begin{array}{c}m\cdot V \\ 0\end{array}\right]$.

Similarly, the momentum of the second piece will be equal to $\displaystyle p_2 = m \cdot \left[ \begin{array}{c}0 \\ V\end{array}\right] = \left[ \begin{array}{c}0 \\ m \cdot V\end{array}\right]$.

Let the momentum of the third piece be $p_3$. If momentum is indeed conserved,

$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = \vec{0}$.

That is:

$\displaystyle \vec{p}_3 + \left[ \begin{array}{c}m \cdot V \\ 0\end{array}\right] + \left[ \begin{array}{c}0 \\ m \cdot V\end{array}\right] = \left[\begin{array}{c}0 \\ 0 \end{array}\right]$.

Substract $\displaystyle \left(\left[ \begin{array}{c}m \cdot V \\ 0\end{array}\right] + \left[ \begin{array}{c}0 \\ m \cdot V\end{array}\right]\right) = \left[\begin{array}{cc} m\cdot V \\ m \cdot V \end{array}\right]$ from both sides of the equation:

$\displaystyle \vec{p}_3 = \left[\begin{array}{c}-m\cdot V \\ -m\cdot V \end{array}\right]$.

Both the momentum (a vector) and the mass (a scalar) of the third object are now known. The velocity (also a vector) of this object can be found by multiplying $\displaystyle \frac{1}{m_3}$ (also a scalar) to $\vec{p}_3$.

$\displaystyle \vec{v}_3 = \frac{1}{m_3} \cdot \vec{p}_3 = \frac{1}{3\;m} \cdot \vec{p}_3 = \left[\begin{array}{c}\displaystyle -\frac{m \cdot V}{3\;m}\\[1em] \displaystyle -\frac{m \cdot V}{3\; m}\end{array}\right] = \left[\begin{array}{c}\displaystyle -\frac{V}{3}\\[1em] \displaystyle -\frac{V}{3}\end{array}\right]$.

The speed of an object is the magnitude of its velocity.

$\begin{aligned} v_3 &= ||\vec{v}_3|| \\&= \sqrt{\left(-\frac{V}{3}\right)^{2} + \left(-\frac{V}{3}\right)^{2}} \\ &= \sqrt{2\left(-\frac{V}{3}\right)^{2}}\\&= \left|-\frac{V}{3}\right|\sqrt{2}\\&= \frac{V}{3}\sqrt{2} = \frac{\sqrt{2}}{3} \;V\end{aligned}$.

(b)

Evaluate the expression $v_3 = \displaystyle \frac{\sqrt{2}}{3} \;V$ for $\rm V = 35\; m \cdot s^{-1}$ to obtain $v_3 \approx \rm 16.5\; m \cdot s^{-1}$.

(c)

Evaluate the expression $\displaystyle \vec{v}_3 = \left[\begin{array}{c}\displaystyle -\frac{V}{3}\\[1em] \displaystyle -\frac{V}{3}\end{array}\right]$ for $V = \rm 35\; m \cdot s^{-1}$:

$\displaystyle \vec{v}_3 =\left[\begin{array}{c}\displaystyle -\frac{35}{3}\\[1em] \displaystyle -\frac{35}{3}\end{array}\right]$.

Plot this vector on a cartesian plane. The exact value of the angle (shown in blue) between this vector and the negative y-axis will be equal to

$\displaystyle \arctan{\left(\frac{\text{opposite}}{\text{adjacent}}\right)} = \arctan{\left(\frac{35/3}{35/3}\right)} = \arctan(1) = 45^{\circ}$.

Answer 2

The motion of the watermelon are as follows:

(a) $The \ speed \ of \ the \ third \ piece, \, is\ \mathbf{ \dfrac{v \cdot \sqrt{2} }{3}}$

(b) The numeric value of the speed of the larger piece is approximately 16.5 m/s

(c) The direction of travel of the larger piece with respect to the y-axis is 45°

The reasons why the above values are  correct are given as follows:

Known parameters:

The number of pieces into which the watermelon is blown = Three pieces

The mass of two of the pieces are equal; m₁ = m₂

The direction to which the two pieces fly are; m₁ in the x-direction, and m₂ in the y-direction

The speed of the two equal mass pieces. v = 35 m/s

The mass of the third piece = 3 × The mass of each of the other two pieces

Therefore, the mass of the third piece = 3·m₁ = 3·m₂ = 3·m

(a) According to the law of conservation of momentum, we have;

m·v·i + m·v·j + 3·m·v₃ = 0

3v₃ = -v·i - v·j

$v_3 = \dfrac{-v\cdot i - v \cdot j}{3} = \dfrac{v\cdot(- i - j)}{3} = \dfrac{v \cdot \sqrt{(-1)^2 + (-1)^2} }{3} =\dfrac{v \cdot \sqrt{2} }{3}$

$The \ speed \ of \ the \ third \ piece, \ v_3 = \dfrac{v \cdot \sqrt{2} }{3}$

(b) The numeric value of the speed of the larger third piece is given by plugging in v = 35 m/s in the equation for v₃ as follows;

$The \ numeric \ value \ for \ the \ speed \ of \ v_3 = \dfrac{35 \ m/s\times \sqrt{2} }{3} \approx 16.5 \ m/s$

(c) Given that v₃ = v₁·i + v₂·j, we have;

The vector components of v₃ are  v₁·i + v₂·j, where  v₁ = v₂ = 35 m/s, we have;

$tan(\theta) = \dfrac{v_2 \cdot j}{v_1 \cdot i}} = \dfrac{35 \times j}{35 \times i}} = \dfrac{ j}{i}} = \dfrac{ 1}{1}} =1$

θ = arctan(1) = 45°

The angle the largest piece travels with respect to the y-axis is 45°

Learn more about the law of conservation of linear momentum here:

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