A 0.940 kg block is attached to a horizontal spring with spring constant 2550 N/m. The block is at rest on a frictionless surface. A 7.90 g bullet is fired into the block, in the face opposite the spring, and sticks.

What was the bullet's speed if the subsequent oscillations have an amplitude of 10.2 cm?

The spring's potential energy at maximum compression (when X is the amplitude of viration) equals the initial kinetic energy at zero deflection, after the bullet hits.

Use that fact to get the initial velocity of the vibrating Bullet-Block combo.

Then use conservation of momentum to get the velocity of the bullet before impact.

To find the bullet's speed, we can use the principle of conservation of momentum. The momentum before the collision should be equal to the momentum after the collision.

Let's consider the block and bullet as a system. Initially, the block is at rest, so its momentum is zero. The momentum of the bullet before the collision is given by:

Momentum before = mass of bullet × initial velocity of the bullet

After the collision, the bullet and block move together with the same velocity, so their combined momentum is given by:

Momentum after = (mass of bullet + mass of block) × final velocity

According to the conservation of momentum:

Momentum before = Momentum after

mass of bullet × initial velocity of the bullet = (mass of bullet + mass of block) × final velocity

Now we can substitute the given values:

(mass of bullet) × initial velocity of the bullet = (mass of bullet + mass of block) × final velocity

(0.00790 kg) × initial velocity of the bullet = (0.00790 kg + 0.940 kg) × final velocity

0.00790 kg × initial velocity of the bullet = 0.9479 kg × final velocity

Solving for the initial velocity of the bullet, we get:

initial velocity of the bullet = (0.9479 kg × final velocity) / 0.00790 kg

Now, we need to find the final velocity of the bullet-block system when it reaches the maximum amplitude during oscillation.

The maximum amplitude is given as 10.2 cm. In simple harmonic motion, the maximum displacement is equal to the amplitude, which is given by:

maximum displacement = amplitude = A

Using the formula for the maximum displacement of a mass-spring system:

A = (mass of bullet + mass of block) × 9.8 m/s^2 / spring constant

Here, the mass of bullet + mass of block is 0.00790 kg + 0.940 kg = 0.9479 kg, and the spring constant is 2550 N/m.

Substituting the values, we have:

10.2 cm = 0.9479 kg × 9.8 m/s^2 / 2550 N/m

Solving for the maximum displacement, we get:

0.102 m = 0.9479 × 9.8 / 2550

Now, we can solve for the final velocity of the bullet-block system using the equation of motion for simple harmonic motion:

final velocity = √(2 × spring constant × maximum displacement / (mass of bullet + mass of block))

final velocity = √(2 × 2550 N/m × 0.102 m / 0.9479 kg)

Finally, substitute the calculated values into the equation we derived earlier to find the initial velocity of the bullet:

initial velocity of the bullet = (0.9479 kg × final velocity) / 0.00790 kg

After solving these equations, we find the bullet's initial velocity.