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Mila [183]
3 years ago
12

Un soldado de 1.80 m de altura ha sido herido por una bala, cuya masa es de 100 g. Deciden colocar al soldado en una centrifugad

ora gigante, con el objeto de producir el desplazamiento deseado de la bala. Se coloca al hombre radialmente en la máquina, con sus pies a 20.0 cm del centro. Si la máxima fuerza de fricción entre la bala y el cerebro es 5.00 N ¿Cuál debería ser la rapidez angular de rotación de la máquina para lograr que la bala se desplace de la forma requerida?
Physics
1 answer:
maks197457 [2]3 years ago
3 0

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

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Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E
Leviafan [203]
We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\
Because both carts have the same mass we can cancel those out:
3v=v_A+2v_B
Now we have our system of equation that we have to solve:
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B
Part A
We need to solve our system for v_a. We will solve second equation for v_b and then plug that in the first equation.
3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}
Now we have to plug this in the first equation:
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\
We will multiply the first equation with 2 and divide by m:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\
Now we plug in the second equation into first one:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0
We end up with quadratic equation that we have to solve, I won't solve it by hand. 
Coefficients are:
a=1\\&#10;b=-6v\\&#10;c=v^2-\frac{E}{m}
Solutions are:
v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}
Part B
We do the same thing here, but we must express v_a from momentum equation:
3v=v_A+2v_B\\&#10;v_A=3v-2v_B
Now we plug this into our energy conservation equation:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\&#10;3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\&#10;3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\&#10;3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\&#10;3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\&#10;3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0&#10;
Again we end up with quadratic equation. Coefficients are:
a=3\\&#10;b=-6v\\&#10;c=6v^2-\frac{3E}{2m}
Solutions are:
v_B=\frac{6v+\sqrt{-36v^2+\frac{18E}{m}}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+\frac{18E}{m}}}{6}



8 0
3 years ago
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