A 1.10 kg} block of wood sits at the edge of a table, 0.790 m above the floor. A 1.10×10−2 kg} bullet moving horizontally with a speed of 725 m/s embeds itself within the block.

What horizontal distance does the block cover before hitting the ground?

Use conservation of momentum to find the momentum of the block/bullet after impact, then the velocity. Now, knowing the initial horizonal velocity, you are on your way.

From the height above the floor, find the time it takes to fall.

.79=1/2 g t^2 find t.

Now horiztal distance willbe horizontal veloctiy*time.

Well, that's quite a creative way to see how far the block can travel before hitting the ground! Let's calculate it using some physics magic:

First, we need to find the initial horizontal velocity of the block after the bullet embeds itself. Since we know the momentum is conserved, we can set up an equation:

(mass of bullet * initial bullet velocity) = (mass of block * final block velocity)

Plugging in the values, we have:

(1.10×10−2 kg) * (725 m/s) = (1.10 kg) * V

Solving for V, we find that the final block velocity is 0.727 m/s.

Now that we know the final velocity, we can use some kinematic equations to find the horizontal distance covered by the block before hitting the ground. The equation we can use is:

d = v^2 / (2 * g)

Where d is the distance, v is the final velocity, and g is the acceleration due to gravity.

Plugging in the values, we have:

d = (0.727 m/s)^2 / (2 * 9.8 m/s^2)

After doing the math, we find that the block travels approximately 0.027 meters before hitting the ground.

So, with a little bit of physics and a pinch of magic, the block covers a horizontal distance of about 0.027 meters before hitting the ground. It's not a very impressive distance, but hey, at least the block made it quite far for its size!

To find the horizontal distance the block covers before hitting the ground, we will use the principle of conservation of momentum.

First, we need to find the initial horizontal velocity of the block and the bullet. Since the bullet embeds itself within the block, the total momentum before the collision is equal to the momentum after the collision.

The initial momentum of the bullet is given by:
p_bullet = m_bullet * v_bullet
p_bullet = (1.10 × 10^-2 kg) * (725 m/s)
p_bullet = 7.98 kg⋅m/s

Since the bullet embeds itself within the block, the final horizontal velocity of the combined block and bullet system is the same as the initial velocity of the bullet, i.e., 725 m/s.

Using the principle of conservation of momentum, we can find the initial velocity of the block before the collision. The total momentum after the collision is given by:
p_final = (m_block + m_bullet) * v_final

Since the block and bullet stick together after the collision, their combined mass is 1.10 kg + 1.10 × 10^-2 kg = 1.111 kg.

The total momentum after the collision is:
p_final = (1.111 kg) * (725 m/s)
p_final = 801.575 kg⋅m/s

Since momentum is conserved, we have:
p_initial = p_final

Therefore,
m_block * v_initial = p_final
(1.10 kg) * v_initial = 801.575 kg⋅m/s
v_initial = 801.575 kg⋅m/s / 1.10 kg
v_initial = 728.705 m/s

Now, we can calculate the time the block takes to fall from the table to the ground using the equation:
h = 0.5 * g * t^2

where h is the height (0.790 m) and g is the acceleration due to gravity (9.8 m/s^2).

Solving for t:
t^2 = (2 * h) / g
t^2 = (2 * 0.790 m) / (9.8 m/s^2)
t^2 = 0.1604 m / m/s^2
t^2 = 0.1604 s^2
t ≈ 0.400 s

Finally, we can find the horizontal distance using the equation:
d = v_initial * t

Substituting the values:
d = (728.705 m/s) * (0.400 s)
d ≈ 291.482 m

Therefore, the block covers approximately 291.482 meters horizontally before hitting the ground.

To find the horizontal distance the block covers before hitting the ground, we need to use the principles of conservation of momentum and conservation of energy.

First, let's consider the momentum. The momentum before the collision is given by the equation:

momentum_before = mass_of_bullet * velocity_of_bullet

momentum_before = (1.10 × 10^-2 kg) * (725 m/s)

momentum_before = 7.975 × 10^-2 kg·m/s

Since the bullet embeds itself within the block, the total mass after the collision will be the sum of the mass of the block and the bullet. So the mass after the collision is:

mass_after = mass_of_block + mass_of_bullet

mass_after = (1.10 kg) + (1.10 × 10^-2 kg)

mass_after = 1.11 kg

Now let's consider the conservation of momentum after the collision. The momentum after the collision can be calculated using the equation:

momentum_after = mass_after * velocity_after

momentum_after = (1.11 kg) * velocity_after

Since the block falls straight down after the collision, the velocity after the collision is purely vertical and can be calculated using the equation for conservation of energy:

potential_energy_before = kinetic_energy_after

m * g * h = (1/2) * m * v^2

(1.10 kg) * 9.8 m/s^2 * 0.790 m = (1/2) * (1.11 kg) * v^2

v^2 = (2 * (1.10 kg) * 9.8 m/s^2 * 0.790 m) / (1.11 kg)

v^2 = 13.92 m^2/s^2

v = 3.73 m/s

So, the velocity after the collision is 3.73 m/s.

Now we can substitute this value back into the momentum equation to find the momentum after the collision:

momentum_after = (1.11 kg) * (3.73 m/s)

momentum_after = 4.13 kg·m/s

Now, using the conservation of momentum, we can equate the momentum before the collision to the momentum after the collision:

momentum_before = momentum_after

7.975 × 10^-2 kg·m/s = 4.13 kg·m/s

Finally, we can solve for the horizontal distance the block covers before hitting the ground, which can be calculated using the equation:

horizontal_distance = velocity_of_block * time

Since the block only moves horizontally, the time it takes for the block to hit the ground is the same as the time it takes for the bullet to reach the block. We can calculate this time using the equation:

horizontal_distance = velocity_of_bullet * time

time = horizontal_distance / velocity_of_bullet

horizontal_distance = 7.975 × 10^-2 kg·m/s / (1.10 × 10^-2 kg) * (725 m/s)

horizontal_distance = 0.099 m

Therefore, the block covers a horizontal distance of approximately 0.099 meters before hitting the ground.