A 10.0-g bullet is fired horizontally into a 101-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 155 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 90.0 cm, what was the speed of the bullet at impact with the block?

To find the speed of the bullet at impact with the block, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.

Let's break down the problem step by step:

Step 1: Find the momentum before the collision.
The momentum of an object is given by the product of its mass and velocity. Since the bullet is fired horizontally, its initial vertical velocity is 0 m/s. Therefore, the momentum of the bullet before the collision is:

Momentum of bullet = mass of bullet × velocity of bullet

Given:
Mass of bullet = 10.0 g = 0.010 kg (convert grams to kilograms)

Since the bullet is fired horizontally into the block, the entire momentum of the bullet is transferred to the bullet-block system. Hence, the momentum before the collision is given by:

Momentum before collision = Momentum of bullet

Step 2: Find the velocity of the bullet-block system after the collision.
After the collision, the bullet becomes embedded in the block, and they move together with a common final velocity. Let's assume this final velocity as 'v'.

The momentum after the collision is given by:

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

Given:
Mass of bullet = 0.010 kg (as calculated earlier)
Mass of block = 101 g = 0.101 kg (convert grams to kilograms)

Now, the momentum after the collision is given by:

Momentum after collision = (0.101 kg + 0.010 kg) × v
= 0.111 kg × v

Step 3: Apply the conservation of linear momentum.
According to the principle of conservation of linear momentum:

Momentum before collision = Momentum after collision

Therefore,

Momentum of bullet = Momentum after collision
(mass of bullet × velocity of bullet) = (0.111 kg × v)

Step 4: Solve for the velocity of the bullet-block system.
Now, we can solve the equation for 'v', which is the velocity of the bullet-block system after the collision:

(0.010 kg × velocity of bullet) = (0.111 kg × v)

Rearranging the equation:

velocity of bullet = (0.111 kg × v) / 0.010 kg

Step 5: Find the maximum compression of the spring.
The maximum compression of the spring can be found using Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it is given by:

Force (F) = spring constant (k) × displacement (x)

In this case, the spring constant is given as 155 N/m, and the maximum compression is given as 90.0 cm = 0.90 m.

Therefore, the force exerted by the spring is:

Force (F) = 155 N/m × 0.90 m

Step 6: Find the work done by the spring.
The work done by the spring is equal to the energy stored in the spring. It can be calculated using the formula:

Work (W) = (1/2) × spring constant (k) × (displacement)^2

In this case, the work done by the spring is:

Work (W) = (1/2) × 155 N/m × (0.90 m)^2

Step 7: Relate work done by the spring to the kinetic energy of the bullet-block system.
According to the principle of conservation of energy, the work done by the spring is equal to the change in kinetic energy of the bullet-block system:

Work done by the spring = Kinetic energy after collision - Kinetic energy before collision

The bullet-block system is initially at rest, so the kinetic energy before the collision is zero. Therefore, the equation becomes:

Work done by the spring = Kinetic energy after collision

Substituting the value of work done by the spring calculated in Step 6:

(1/2) × 155 N/m × (0.90 m)^2 = (1/2) × (mass of bullet + mass of block) × (velocity after collision)^2

Given:
Mass of bullet = 0.010 kg (as calculated earlier)
Mass of block = 0.101 kg (as calculated earlier)

Now, we can solve the equation for the velocity after the collision.

Step 8: Calculate the final velocity of the bullet-block system.
Solve the equation from Step 7 for 'v', which is the velocity of the bullet-block system after the collision.

Step 9: Calculate the speed of the bullet.
As the bullet and the block move together after the collision, the final velocity of the bullet-block system is also the final velocity of the bullet.

Therefore, the speed of the bullet at impact with the block is equal to the absolute value of the final velocity.