A bullet with mass (m) is fired vertically into a block of wood with mass (M) that is hanging from a rope at rest. If the billet has an initial speed Vi (initial velocity) find the height (h) that the block will swing into the air after the bullet becomes embedded in it

m=0.021 kg
M=1.4 kg
Vi(intial)= 210 m/s

To find the height that the block will swing into the air after the bullet becomes embedded in it, we can apply the principle of conservation of momentum.

1. Calculate the momentum of the bullet before it hits the block:
Momentum of the bullet = mass of the bullet x velocity of the bullet
Momentum of the bullet = m x Vi

2. The bullet becomes embedded in the block, so the total mass after the bullet is embedded is the sum of the block mass and the bullet mass:
Total mass after embedding = M + m

3. Apply the principle of conservation of momentum:
Momentum before = Momentum after
m x Vi = (M + m) x Vf
where Vf is the final velocity of the block after the bullet is embedded.

4. Rearrange the equation to solve for Vf:
Vf = (m x Vi) / (M + m)

5. Apply the principle of conservation of energy to find the maximum height the block reaches:
The potential energy gained by the block is equal to the kinetic energy lost due to the decrease in velocity.
mgh = 0.5 x (M + m) x Vf^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2).

6. Solve the equation for h which represents the maximum height:
h = (0.5 x (M + m) x Vf^2) / (mg)

Now, let's plug in the values and calculate the height:

m = 0.021 kg
M = 1.4 kg
Vi = 210 m/s

1. Calculate the momentum of the bullet before it hits the block:
Momentum of the bullet = m x Vi = 0.021 kg x 210 m/s = 4.41 kg·m/s

2. Calculate the total mass after embedding:
Total mass after embedding = M + m = 1.4 kg + 0.021 kg = 1.421 kg

3. Calculate the final velocity of the block:
Vf = (m x Vi) / (M + m) = (0.021 kg x 210 m/s) / (1.4 kg + 0.021 kg) = 0.003 kg·m/s

4. Calculate the maximum height:
h = (0.5 x (M + m) x Vf^2) / (mg) = (0.5 x 1.421 kg x (0.003 kg·m/s)^2) / (0.021 kg x 9.8 m/s^2) = 0.002456 m or 2.456 cm

Therefore, the block will swing into the air to a maximum height of approximately 2.456 cm.

To find the height (h) that the block will swing into the air after the bullet becomes embedded in it, we need to apply the law of conservation of momentum and the law of conservation of energy.

First, let's find the initial momentum of the bullet-block system before the collision. The momentum (p) is defined as the product of mass and velocity:

p_initial = m * Vi

Next, since the bullet becomes embedded in the block, we have a new system with combined mass (M + m). Therefore, we can write the law of conservation of momentum:

p_initial = (M + m) * Vf

Where Vf is the final velocity of the combined system after the bullet is embedded. Note that the velocity here is the velocity of the block and bullet moving together.

Now, let's find the final velocity of the combined system using the law of conservation of energy. The initial energy (E_initial) is the kinetic energy of the bullet, given by:

E_initial = (1/2) * m * Vi^2

The final energy (E_final) is the potential energy of the block at its highest point, given by:

E_final = (M + m) * g * h

Here, g is the acceleration due to gravity (9.8 m/s^2), and h is the height to be found.

Since energy is conserved, we can equate the initial energy (E_initial) to the final energy (E_final):

(1/2) * m * Vi^2 = (M + m) * g * h

Now, solve this equation for h:

h = (1/2) * m * Vi^2 / [(M + m) * g]

Plugging in the given values:

m = 0.021 kg
M = 1.4 kg
Vi = 210 m/s
g = 9.8 m/s^2

h = (1/2) * 0.021 kg * (210 m/s)^2 / [(1.4 kg + 0.021 kg) * 9.8 m/s^2]

Simplifying the equation, we can calculate the value of h.