A 12 g bullet is stopped in a block of wood

(mw = 5.9 kg). The speed of the bullet-
plus-wood combination immediately after the
collision is 0.95 m/s.
What was the original speed of the bullet?
Answer in units of m/s

12

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

The principle of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision, provided that there are no external forces involved.

In this case, the momentum before the collision is equal to the momentum after the collision.

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

The momentum before the collision, p_initial, can be calculated as follows:

p_initial = m_bullet * v_initial

Where:
m_bullet is the mass of the bullet
v_initial is the initial velocity of the bullet

The momentum after the collision, p_final, can be calculated as follows:

p_final = (m_bullet + m_wood) * v_final

Where:
m_wood is the mass of the wood block
v_final is the final velocity of the bullet-plus-wood combination

Since the mass of the wood block is given as 5.9 kg, the mass of the bullet is 0.012 kg, and the final velocity is given as 0.95 m/s, we can rewrite the equations as:

m_bullet * v_initial = (m_bullet + m_wood) * v_final

Now we can solve for v_initial, the original velocity of the bullet.

v_initial = ((m_bullet + m_wood) * v_final) / m_bullet

Substituting the given values, we get:

v_initial = ((0.012 kg + 5.9 kg) * 0.95 m/s) / 0.012 kg

Simplifying the expression:

v_initial = (5.912 kg * 0.95 m/s) / 0.012 kg

Calculating the result:

v_initial ≈ 466.67 m/s (rounded to two decimal places)

Therefore, the original speed of the bullet is approximately 466.67 m/s.