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.