## No problem! To find the kinetic energy of the bullet + block system, we first need to calculate the velocity of the bullet + block after the collision.

Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be written as:

(m_bullet Ã— v_bullet) + (m_block Ã— 0) = (m_bullet Ã— v_bullet+block) + (m_block Ã— v_block)

The initial velocity of the block is 0 since it is at rest. The velocity of the bullet after the collision will be the same as the velocity of the bullet+block system, denoted as v_bullet+block.

Given:

m_bullet = 21.0 g = 0.021 kg (converted to kilograms)

v_bullet = 190 m/s (Given)

m_block = 1.55 kg (Given)

v_block = velocity of the bullet+block (Unknown)

Plugging in these values, we can solve for v_bullet+block:

(0.021 kg Ã— 190 m/s) + (1.55 kg Ã— 0) = (0.021 kg Ã— v_bullet+block) + (1.55 kg Ã— 0)

(0.021 kg Ã— 190 m/s) = (0.021 kg Ã— v_bullet+block)

0.399 kgÂ·m/s = 0.021 kg Ã— v_bullet+block

v_bullet+block = 0.399 kgÂ·m/s / 0.021 kg

v_bullet+block â‰ˆ 19.0 m/s

So, the velocity of the bullet+block system after the collision is approximately 19.0 m/s.

Now that we have the velocity, we can find the kinetic energy of the bullet+block system. The kinetic energy formula is:

KE = (1/2) Ã— mass Ã— velocity^2

Plugging in the values:

m = m_bullet + m_block = 0.021 kg + 1.55 kg â‰ˆ 1.571 kg

v = v_bullet+block = 19.0 m/s

KE = (1/2) Ã— 1.571 kg Ã— (19.0 m/s)^2

KE â‰ˆ 356 J (Joules)

Now that we have the kinetic energy of the bullet+block system, we can find the potential energy at the maximum height. The potential energy formula is:

PE = mass Ã— gravity Ã— height

Given:

mass = m_block = 1.55 kg (Given)

gravity = 9.8 m/s^2 (acceleration due to gravity)

height = ?

PE = 356 J (calculated)

Plugging in the values:

356 J = 1.55 kg Ã— 9.8 m/s^2 Ã— height

height = 356 J / (1.55 kg Ã— 9.8 m/s^2)

height â‰ˆ 23.2 m

Therefore, the block will rise approximately 23.2 meters into the air after the bullet becomes embedded in it.