An 7.95 g bullet is fired horizontally into a 9.03 kg block of wood on an air table and is embedded in it. After the collision, the block and bullet slide along the frictionless surface together with a speed of 11.6 cm/s. What was the initial speed of the bullet?
Use the law of conservation of momentum to relate the velocity V' of the block (with embedded bullet) after impact, and the velocity of the bullet before impact, V.
0.00795 V = (9.03 + 0.0008)*V'
You already know V'. Solve for V.
To find the initial speed of the bullet, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity. In this case, we need to calculate the momentum of the bullet before the collision and the momentum of the block and bullet together after the collision.
Let's use the following variables:
- m_bullet: mass of the bullet
- v_bullet: initial velocity of the bullet
- m_block: mass of the wood block
- v_final: final velocity of the block and bullet together after the collision
Given information:
- m_bullet = 7.95 g = 0.00795 kg
- m_block = 9.03 kg
- v_final = 11.6 cm/s = 0.116 m/s
Now we can apply the conservation of momentum principle:
Initial momentum = Final momentum
m_bullet * v_bullet = (m_bullet + m_block) * v_final
Substituting the given values:
0.00795 kg * v_bullet = (0.00795 kg + 9.03 kg) * 0.116 m/s
Simplifying the equation:
v_bullet = ((0.00795 kg + 9.03 kg) * 0.116 m/s) / 0.00795 kg
v_bullet = (9.03895 kg * 0.116 m/s) / 0.00795 kg
v_bullet ≈ 0.131 m/s
Therefore, the initial speed of the bullet was approximately 0.131 m/s.