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A bullet with a mass mb=11.5 g is fired into a block of wood at velocity ????b=265 m/s. The block is attached to a spring that has a spring constant ???? of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass of the wooden block. From left to right, a bullet of mass m subscript b has a velocity of v subscript b with a velocity vector pointing directly to the right. The vector points toward a block that is attached on its right face to a spring of force constant k. The other end of the spring is attached to a fixed, vertical surface. mass of wooden block: ____________

Respuesta :

davidlcastiblanco

Answer:

[tex]m=0.496 kg[/tex]

Explanation:

Knowing the potential energy in the spring is equal to the initial kinetic energy  so:

[tex]F_s=E_K[/tex]

[tex]\frac{1}{2}*K*x^2=\frac{1}{2}*m_b*v_f^2[/tex]

[tex]\frac{1}{2}*205n/m*(0.35m)^2=\frac{1}{2}*m_b*v_f^2[/tex]

Now using the conservation of momentum

[tex]P_i=P_f[/tex]

[tex]m_b*v_i=(m_b+m_w)*v_f[/tex]

[tex]11.5x10^3kg*265m/s=(11.5x10^3kg+m)*v_f[/tex]

[tex]v_f=\frac{3.0475 kg*m/s}{(11.5x1063kg+m)}[/tex]

Replacing in initial equation so:

[tex]12.5556J=\frac{1}{2}*11.5x10^3kg*(\frac{3.0475 kg*m/s}{11.5x10^3kg+m})^2[/tex]

Solve to m

[tex]m=\frac{6.23}{12.556}[/tex]

[tex]m=0.496 kg[/tex]