A 12-g bullet moving horizontally strikes and remains in a 3.0-kg block initially at rest on the edge of a table. The block, which is initially 80 cm above the floor, strikes the floor a horizontal distance of 120 cm from its initial position. What was the initial speed of the bullet?

Δy = v0t + ½ at2

- 0.8 = (0)t + ½ (-9.8)t2
0.404 s = t
Δx = v0t + ½ at2
1.2 = v0 (0.404) + ½ (0)(0.404)2
2.97 m/s = v0 (after the collision as the block leaves the table)
p0 = p1
(0. 012 kg )v + (3 kg)(0 m/s) = (0.012 kg)(2.97 m/s) + (3 kg)(2.97 m/s)
(0. 012 kg )v = 8.97
745.5 m/s = v

Well, let me take you through this step by step, or shall I say, bullet by bullet. You have a 12-g bullet, which is moving horizontally with an initial speed we're trying to find. This bullet decides to have a little rendezvous with a 3.0-kg block that is happily resting on the edge of a table.

Now, when the bullet strikes the block, it decides to stay there, like a stubborn houseguest who just won't leave. As a result of this unexpected union, the block and the bullet start moving together. But wait, there's more! The block decides to go on an adventure of its own and falls off the table. Gravity seems to have quite the attraction for this block.

As the block journeys downward, it strikes the floor a horizontal distance of 120 cm away from its initial position. Now, we need to find out how fast the bullet was initially moving.

To solve this puzzle, let's start with conservation of momentum. Since the bullet and the block become one happy family, the total momentum before the collision should be equal to the total momentum after the collision. Mathematically, we can express this as:

(mass of bullet) x (initial speed of bullet) = (mass of bullet + mass of block) x (final velocity of bullet + block)

Now, since the block was initially at rest, we can say that the final velocity of the bullet and the block together is the same as the final velocity of the block alone.

Also, we know that the final position of the block is 120 cm away from its initial position, which means it has fallen 80 cm (the original height above the floor).

Using these clues and a few equations, we can solve for the initial speed of the bullet. It may seem like a daunting task, but don't worry, I've got the equations on my mental shelves!

To calculate the initial speed of the bullet, we can use the principle of conservation of linear momentum.

The linear momentum before the collision is equal to the linear momentum after the collision.

Let's break down the problem into steps:

Step 1: Define the variables:
- Mass of the bullet (m1) = 12 g = 0.012 kg
- Mass of the block (m2) = 3.0 kg
- Initial height of the block (h) = 80 cm = 0.80 m
- Horizontal distance traveled by the block (d) = 120 cm = 1.20 m

Step 2: Calculate the velocity of the block before the collision.
Using the equation of motion for an object in free fall:
vf^2 = vi^2 + 2gΔh

Since the block is initially at rest, the final velocity (vf) is 0 m/s.
g = 9.8 m/s^2 (acceleration due to gravity)
Δh = h (since the block falls vertically)
0^2 = vi^2 + 2(9.8)(0.80)
0 = vi^2 + 15.68
vi^2 = -15.68
The value of vi^2 cannot be negative, so there's an error in the calculations.

Step 3: Calculate the velocity of the bullet before the collision.
We'll use the principle of conservation of linear momentum:

m1v1 + m2v2 = m1v1' + m2v2'

Since the block is initially at rest, the velocity of the block before and after the collision (v2 and v2') is 0 m/s.
Therefore, the equation becomes:
m1v1 = m1v1'

0.012 kg * v1 = 0.012 kg * v1'
v1 = v1'

So, the initial velocity of the bullet is equal to the final velocity of the bullet.

Step 4: Calculate the velocity of the block after the collision.
Using the equation of motion for an object in horizontal motion:
d = v1 * t
t = d / v1

Since the block falls for a vertical distance and travels a horizontal distance simultaneously, the time taken for both will be the same. Therefore, we can use the horizontal distance traveled by the block as the displacement for calculating the time.

t = 1.2 m / v1

Step 5: Calculate the final velocity of the block.
Using the equation of motion for an object in free fall:
vf = vi + gt

vf = 0 + 9.8 m/s^2 * t
vf = 9.8 * (1.2 m / v1)

Step 6: Apply conservation of energy.
From the conservation of energy principle:
Potential energy before the collision = kinetic energy after the collision
m2gh = (1/2) * m2 * vf^2

Plugging in the values:
3.0 kg * 9.8 m/s^2 * 0.80 m = (1/2) * 3.0 kg * (9.8 * (1.2 m / v1))^2

Step 7: Solve for v1.
Simplify and solve the equation for v1:

23520 J = (1/2) * 3.0 kg * (9.8)^2 * (1.44 / v1^2)
23520 = 3 * 9.8^2 * (1.44 / v1^2)
v1^2 = (1.44 * 3 * 9.8^2) / 23520

v1 = √(1.44 * 3 * 9.8^2) / √23520

v1 ≈ 3.36 m/s

Therefore, the initial speed of the bullet is approximately 3.36 m/s.

To find the initial speed of the bullet, we can use the principle of conservation of momentum.

The total momentum before the collision is equal to the total momentum after the collision.

The momentum of an object is calculated by multiplying its mass by its velocity.

Let's analyze the problem step by step:

1. Determine the momentum of the bullet before the collision.

The mass of the bullet is given as 12 grams, which is 0.012 kg.

The velocity of the bullet before the collision is unknown, so let's consider it as "v" m/s.

Therefore, the momentum of the bullet before the collision is 0.012 kg * v m/s.

2. Determine the momentum of the block before the collision.

The mass of the block is given as 3.0 kg.

The block is initially at rest, so its velocity before the collision is 0 m/s.

Therefore, the momentum of the block before the collision is 3.0 kg * 0 m/s, which is zero.

3. Determine the momentum of the bullet and the block after the collision.

After the collision, the bullet remains inside the block, so the combined mass of the block and bullet is 3.0 kg + 0.012 kg = 3.012 kg.

The block, with the bullet inside it, falls to the ground with a horizontal distance of 120 cm from its initial position.

To solve this problem, we need to consider the vertical component of the block's motion.

Using the equation of motion, d = v0t + (1/2)at^2, for the vertical motion, with v0 = 0, a = g, and d = 80 cm = 0.8 m:

d = 0.8 m = (1/2)gt^2.

Solving for t, we get:

t = sqrt((2d)/g),

where g is the acceleration due to gravity. On Earth, it is approximately 9.8 m/s^2.

Substituting the values, we find:

t = sqrt((2 * 0.8 m) / 9.8 m/s^2) = 0.403 s (approx.)

Since the block strikes the floor after time t, the horizontal velocity component can be obtained by dividing the horizontal displacement (120 cm = 1.2 m) by the time t:

v_horizontal = 1.2 m / 0.403 s ≈ 2.98 m/s.

Thus, the momentum of the block and bullet after the collision is 3.012 kg * 2.98 m/s.

4. Apply the principle of conservation of momentum.

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

Therefore, we can write:

0.012 kg * v = 3.012 kg * 2.98 m/s.

Simplifying and solving for v, we find:

v ≈ (3.012 kg * 2.98 m/s) / 0.012 kg.

Calculating this expression gives us the initial speed of the bullet:

v ≈ 746.5 m/s.

Therefore, the initial speed of the bullet is approximately 746.5 m/s.