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Novosadov [1.4K]

2 years ago

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A 4 kg body is traveling at 3 m/s with no external force acting on it. At a certain instant an internal explosion occurs, splitt

ing the body into two chunks of 2 kg mass each. The explosion gives the chunks an additional 18 J of kinetic energy. Neither chunk leaves the line of original motion. Determine the speed and direction of motion of each of the chunks after the explosion.

1 answer:

Ann [662]2 years ago

5 0

Answer

given,

mass of body = 4 Kg

travelling at speed = 3 m/s

exploded into two fragments of equal mass i.e. = 2 Kg

energy after collision increases by 18 J

original kinetic energy

$KE = \dfrac{1}{2}MV^2$

$KE = \dfrac{1}{2}\times 4 \times 3^2$

KE = 18 J

energy increases after collision = 18 J

total energy = 36 J

now,

$\dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2 = 36$

$\dfrac{1}{2}v_1^2 + \dfrac{1}{2}v_2^2 = 36$

$v_1^2+v_2^2 = 72$

using conservation of momentum

m₁v₁ + m₂v₂ = MV

2 v₁ + 2 v₂ = 4 x 3

v₁ + v₂ = 6-------------(1)

squaring both side in the above equation

v₁² + v₂² + 2v₁v₂ = 36

72 + 2v₁v₂ = 36

v₁v₂ = -18-------(2)

putting value of equation (1) in (2)

v₁ (-18 - v₁) = 6

-18 v₁ - v₁² = 6

v₁² + 18 v₁ + 6 = 0

on solving the above equation we get

v₁ = -0.34 m/s

putting value of v₁ in equation (1)

v₂ = 6 + 0.34

v₂ = 6.34 m/s

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Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

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(I will assume the positions are in meters)

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$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

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$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

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$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

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which yields:

$r_{CP}^{2}= 30 m^{2}$

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which yields:

$F_{CP}=4.20 GN$

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$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

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$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

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So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

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