a bullet of mass 0.05 kg strikes a wooden block of mass 5 kg and becomes embedded within the block. The block and bullet then fly off at 10 m/s after the collision. What was the original velocity of the bullet if the block was initially at rest?

To solve this problem, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.

Let's assume the initial velocity of the bullet is "v" (to be determined) and the final velocity of the combined block and bullet is 10 m/s. The mass of the bullet is 0.05 kg and the mass of the block is 5 kg.

Before the collision:

momentum of bullet = mass of bullet * velocity of bullet = 0.05 kg * v
momentum of block = mass of block * velocity of block = 5 kg * 0 (since the block is initially at rest)

After the collision, the bullet and block become a combined system moving together, so the combined momentum will be the sum of their individual momenta:

momentum of bullet and block = (mass of bullet + mass of block) * velocity of block and bullet = (0.05 kg + 5 kg) * 10 m/s = 5.05 kg * 10 m/s = 50.5 kg·m/s

Since the total momentum before the collision is equal to the total momentum after the collision, we can equate these two values:

0.05 kg * v = 50.5 kg·m/s

Solving for v:

v = 50.5 kg·m/s / 0.05 kg = 1010 m/s

Therefore, the original velocity of the bullet was 1010 m/s.

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

The principle of conservation of momentum states that the total momentum of a closed system remains constant before and after a collision, as long as no external forces are acting on it.

In this case, before the collision, only the bullet is in motion, and the block is at rest. After the collision, both the bullet and the block move together.

Let's denote:
m_bullet = mass of the bullet = 0.05 kg
m_block = mass of the block = 5 kg
v_bullet_initial = original velocity of the bullet
v_after_collision = velocity of the bullet and block after the collision = 10 m/s

According to the conservation of momentum, the total momentum before the collision (which is only from the bullet) is equal to the total momentum after the collision (which is the combination of the bullet and the block):

(m_bullet * v_bullet_initial) + (m_block * 0) = (m_bullet + m_block) * v_after_collision

Since the block is initially at rest, its initial velocity is zero (0).

Applying the equation and solving for v_bullet_initial:

(0.05 kg * v_bullet_initial) + (5 kg * 0) = (0.05 kg + 5 kg) * 10 m/s

0.05 kg * v_bullet_initial = 50.25 kg*m/s

Dividing both sides by 0.05 kg:

v_bullet_initial = 50.25 kg*m/s / 0.05 kg

v_bullet_initial ≈ 1005 m/s

Therefore, the original velocity of the bullet was approximately 1005 m/s.

momentum (as usual) is conserved

.05 * v = (5 + .05) * 10