A 6.6 g bullet is stopped in a block of wood
(mw = 6 kg). The speed of the bullet-pluswood
combination immediately after the collision
is 0.8 m/s.
What was the original speed of the bullet?
Answer in units of m/s.
see the problem you posted under another name. Why do students do that?
To find the original speed of the bullet, we need to use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity.
Let's assign variables to the given values:
- Mass of the bullet = mb = 6.6 g = 0.0066 kg
- Mass of the wood block = mw = 6 kg
- Final velocity of the bullet-plus-wood combination = vf = 0.8 m/s
Since the block of wood was initially at rest, the initial velocity of the bullet-plus-wood combination would be the same as the original speed of the bullet.
Let's denote this original speed of the bullet as vb.
Now applying the conservation of momentum, we can write the equation:
(mb + mw) * vb = (mb + mw) * vf
Since "mb + mw" is common on both sides of the equation, we can cancel it out:
vb = vf
Therefore, the original speed of the bullet, vb, is equal to the final velocity of the bullet-plus-wood combination, vf, which is 0.8 m/s.
So, the original speed of the bullet is 0.8 m/s.