A 0.0382-kg bullet is fired horizontally into a 3.78-kg wooden block attached to one end of a massless, horizontal spring (k = 833 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.190 m. What is the speed of the bullet?

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Respuesta :

JemdetNasr JemdetNasr · Answer 1 · 2019-11-02T12:34:55+01:00

Answer:

[tex]280.87[/tex] ms⁻¹

Explanation:

Consider the motion of the bullet-block combination after collision

[tex]m[/tex] = mass of the bullet = 0.0382 kg

[tex]M[/tex] = mass of wooden block = 3.78 kg

[tex]V[/tex] = velocity of the bullet-block combination after collision

[tex]k[/tex] = spring constant of the spring = 833 N m⁻¹

[tex]A[/tex] = Amplitude of oscillation = 0.190 m

Using conservation of energy

Kinetic energy of bullet-block combination after collision = Spring potential energy gained due to compression of spring

[tex](0.5)(m + M)V^{2} = (0.5)kA^{2}[/tex]

[tex](0.0382 + 3.78)V^{2} = (833)(0.190)^{2}[/tex]

[tex]V = 2.81[/tex] ms⁻¹

[tex]v_{o}[/tex] = initial velocity of the bullet before striking the block

Using conservation of momentum for the collision between bullet and block

[tex]m v_{o} = (m + M) V[/tex]

[tex](0.0382) v_{o} = (0.0382 + 3.78) (2.81)[/tex]

[tex]v_{o} = 280.87[/tex] ms⁻¹