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- anonymous

A firecracker stuck into a 150 g apple explodes and sends five apple fragments in different directions. The vector sum of momenta 1, 2, 3, and 4 is found from a video of the event to have components p_x = -1.20 kg * m/s, p_y = -0.80 kg * m/s (with no other component). The mass of the fifth fragment is 0.050 kg. What is its velocity right after the explosion?
2.1 m/s , 42 m/s , 18 m/s , or 29 m/s
?? not sure which it would be!

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- anonymous

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- Michele_Laino

here we have to apply the conservation of total momentum

- anonymous

ok!

- Michele_Laino

in other words, we have to write these two components of the momentum of the fifth fragment:
\[\Large \begin{gathered}
{P_x} = M{V_x} \hfill \\
{P_y} = M{V_y} \hfill \\
\end{gathered} \]

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- anonymous

ok! what do we plug in? :O

- Michele_Laino

where V_x and V_y are the components of the velocity of the fifth fragment, whereas M is the mass of the fifth fragment

- anonymous

ohh okay!

- Michele_Laino

so, we can write these two equations:
\[\Large \begin{gathered}
M{V_x} - 1.2 = 0 \hfill \\
M{V_y} - 0.8 = 0 \hfill \\
\end{gathered} \]

- anonymous

ok!

- Michele_Laino

since before the explosion, the object was at rest

- Michele_Laino

then we get:
\[\Large \begin{gathered}
{V_x} = \frac{{1.2}}{M} = ... \hfill \\
\\
{V_y} = \frac{{0.8}}{M} = ... \hfill \\
\end{gathered} \]

- anonymous

ok! how do we solve that? :/

- Michele_Laino

it is simple you have to replace M with 0.15 Kg

- Michele_Laino

you should get two values: V_x and V_y respectively

- anonymous

ohh okay! so we get V_x=8 and V_y = 5.33333 ?

- Michele_Laino

ok!

- Michele_Laino

now, the requested speed V is given by the subsequent formula:
\[\Large V = \sqrt {V_x^2 + V_y^2} = \sqrt {{8^2} + {{5.33}^2}} = ...\]

- anonymous

so we get this?
9.6129 ?

- Michele_Laino

please wait, I have made an error

- Michele_Laino

you have to replace M with 0.05 Kg, not with 0.15 Kg

- Michele_Laino

\[\Large \begin{gathered}
{V_x} = \frac{{1.2}}{M} = \frac{{1.2}}{{0.05}}... \hfill \\
\hfill \\
{V_y} = \frac{{0.8}}{M} = \frac{{0.8}}{{0.05}}... \hfill \\
\end{gathered} \]

- anonymous

ohh okay! so we get thiS?
v_x=24 and V_y=16
so then we put that into sq rt and get this? 28.85? so our answer is choice D? 29 m/s?

- Michele_Laino

yes! correct! Sorry for my error

- Michele_Laino

\[\Large V = \sqrt {V_x^2 + V_y^2} = \sqrt {{{24}^2} + {{16}^2}} = ...\]

- anonymous

yay!! and it is okay! :) thank you!

- Michele_Laino

\[V = \sqrt {V_x^2 + V_y^2} = \sqrt {{{24}^2} + {{16}^2}} \cong 29m/\sec \]

- Michele_Laino

thank you!! :)

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