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A 22. 0 g bullet strikes a 0. 85 kg block attached to a fixed horizontal spring whose spring constant is 5. 30×10 3 N/m and sets it into vibration with an amplitude of 20 cm. What was the speed of the bullet before impact if the two objects move together after impact?

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Answer: To solve this problem, we can use the principle of conservation of momentum and conservation of mechanical energy.

Step 1: Conservation of Momentum

Before the impact, the bullet is moving with speed "v" towards the block. After the impact, the combined system of the bullet and the block move together with a common velocity "V". According to the conservation of momentum:

Initial momentum of the system = Final momentum of the system

m_bullet * v = (m_bullet + m_block) * V

Step 2: Conservation of Mechanical Energy

When the bullet and the block move together after impact, their maximum displacement occurs when all the initial kinetic energy of the bullet is converted into the elastic potential energy stored in the spring at the maximum displacement. Using conservation of mechanical energy:

Initial kinetic energy of the bullet = Maximum elastic potential energy in the spring

(1/2) * m_bullet * v^2 = (1/2) * k * A^2

where "k" is the spring constant, and "A" is the amplitude of vibration.

Now, let's plug in the given values:

m_bullet = 22.0 g = 0.0220 kg

m_block = 0.85 kg

k = 5.30 × 10^3 N/m

A = 20 cm = 0.20 m

Step 3: Solve for "v" using the equations from Step 1 and Step 2:

From Step 1:

0.0220 kg * v = (0.0220 kg + 0.85 kg) * V

v = 0.87234 * V ............(i)

From Step 2:

(1/2) * 0.0220 kg * v^2 = (1/2) * 5.30 × 10^3 N/m * (0.20 m)^2

0.5 * 0.0220 kg * v^2 = 530 * 0.04

v^2 = 530 * 0.04 / 0.0220 kg

v^2 = 960

v = √960

v = 31.0485 m/s ...........(ii)

Step 4: Substitute the value of "v" from equation (ii) into equation (i) to find "V":

v = 0.87234 * V

31.0485 m/s = 0.87234 * V

V = 31.0485 m/s / 0.87234

V = 35.6 m/s

Therefore, the speed of the bullet before impact was approximately 35.6 m/s.

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