# A gun is fired vertically into a 1.55 kg block of wood at rest directly above it. If the bullet has a mass of 21.0 g and a speed of 190 m/s, how high will the block rise into the air after the bullet becomes embedded in it?

i don't even know where to begin... would it have something to do with 1/2mv^2=mgh?

Yes! The kinetic energy of the block and embedded bullet gets converted to potential energy when in reaches its maximum height. Do you know how to get the kinetic energy of the block + bullet?

i don't think so, could you add the masses for m?

This is conservation of momentum. The momentum of the bullet before it hits the block has to equal the momentum of the bullet + block. This will give you the velocity of the bullet + block. Then, you can calulate its kinetic energy!!

thanks!

9 months ago

## 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.