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A bullet with a mass my = 11.9 g is fired into a block of wood at velocity = 245 m/s. The block is attached to a spring that has a spring constant k 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 m, of the wooden block.___________kg

User Feuda
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Final answer:

To find the mass of the wooden block, we can use the principle of conservation of momentum and the elastic potential energy stored in the spring. By using these principles and solving the equations, we find that the mass of the wooden block is 0.588 kg.

Step-by-step explanation:

To find the mass of the wooden block, we can use the principle of conservation of momentum. Since the bullet and the block are initially at rest, the momentum before the collision is zero. After the collision, the bullet and the block move together. Let's call the mass of the wooden block m. The bullet has a mass of 11.9 g = 0.0119 kg and a velocity of 245 m/s. The total momentum after the collision is (0.0119 kg + m) * V, where V is the common velocity of the bullet and the block. According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:

0 = (0.0119 kg + m) * V

Solving for V, we get:

V = -0.0119 kg * 245 m/s / (0.0119 kg + m)

Next, we can use the elastic potential energy stored in the spring to find the mass of the wooden block. The elastic potential energy is given by:

0.5 * k * x^2 = 0.5 * 205 N/m * (0.35 m)^2 = 12.06 J

where k is the spring constant and x is the compression of the spring. The elastic potential energy is equal to the kinetic energy of the block and the bullet:

0.5 * (0.0119 kg + m + m) * V^2 = 12.06 J

Substituting the expression for V from the momentum equation, we get:

0.5 * (0.0119 kg + m + m) * (-0.0119 kg * 245 m/s / (0.0119 kg + m))^2 = 12.06 J

Simplifying and solving for m, we find:

m = 0.588 kg

User Dean Coakley
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