anonymous
  • 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!
Physics
jamiebookeater
  • jamiebookeater
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Michele_Laino
  • Michele_Laino
here we have to apply the conservation of total momentum
anonymous
  • anonymous
ok!
Michele_Laino
  • 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
  • anonymous
ok! what do we plug in? :O
Michele_Laino
  • 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
  • anonymous
ohh okay!
Michele_Laino
  • 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
  • anonymous
ok!
Michele_Laino
  • Michele_Laino
since before the explosion, the object was at rest
Michele_Laino
  • 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
  • anonymous
ok! how do we solve that? :/
Michele_Laino
  • Michele_Laino
it is simple you have to replace M with 0.15 Kg
Michele_Laino
  • Michele_Laino
you should get two values: V_x and V_y respectively
anonymous
  • anonymous
ohh okay! so we get V_x=8 and V_y = 5.33333 ?
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • 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
  • anonymous
so we get this? 9.6129 ?
Michele_Laino
  • Michele_Laino
please wait, I have made an error
Michele_Laino
  • Michele_Laino
you have to replace M with 0.05 Kg, not with 0.15 Kg
Michele_Laino
  • 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
  • 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
  • Michele_Laino
yes! correct! Sorry for my error
Michele_Laino
  • Michele_Laino
\[\Large V = \sqrt {V_x^2 + V_y^2} = \sqrt {{{24}^2} + {{16}^2}} = ...\]
anonymous
  • anonymous
yay!! and it is okay! :) thank you!
Michele_Laino
  • Michele_Laino
\[V = \sqrt {V_x^2 + V_y^2} = \sqrt {{{24}^2} + {{16}^2}} \cong 29m/\sec \]
Michele_Laino
  • Michele_Laino
thank you!! :)

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