A 23 g bullet traveling 220 m/s penetrates a 4.2 kg block of wood and emerges cleanly at 140 m/s.

If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

conservation of momentum:

M*0+23*220=M*Vf+23*140
M = 4200 grams solve for Vf

230

140 m/s

To find the speed at which the block moves after the bullet emerges, we can use the principle of conservation of momentum.

The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, the bullet and the block form an isolated system, as there are no external forces acting on them.

The momentum of an object is given by the product of its mass and its velocity. Mathematically, momentum (p) is calculated as:

p = mass (m) × velocity (v)

Using this formula, we can calculate the initial momentum of the bullet (before it hits the block) and the final momentum of the bullet and block system (after the bullet emerges).

Step 1: Calculate the initial momentum of the bullet.
Given:
Mass of the bullet (m1) = 23 g = 0.023 kg
Velocity of the bullet (v1) = 220 m/s

Initial momentum of the bullet (p1) = m1 × v1

Step 2: Calculate the final momentum of the bullet and block system.
Given:
Mass of the block (m2) = 4.2 kg
Velocity of the bullet after emerging (v1') = 140 m/s
Velocity of the block after the bullet emerges (v2') = ?
Since there are no external forces, the total momentum before the collision (p1) should equal the total momentum after the collision (p1' + p2').

p1 = p1' + p2'

Substituting the values:
m1 × v1 = 0.023 × 220 + 4.2 × v2'

Step 3: Calculate the velocity of the block after the bullet emerges.
Rearranging the equation to solve for v2':

v2' = (m1 × v1 - 0.023 × 220) / 4.2

Substituting the values:
v2' = (0.023 × 220 - 0.023 × 220) / 4.2

Simplifying:
v2' = 0 m/s

Therefore, the block does not move at all after the bullet emerges.

Explanation:

The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, since there are no external forces acting on the bullet and block system, their total momentum is conserved.

Before the collision, the bullet has momentum m1 × v1, where m1 is the mass of the bullet and v1 is its velocity. After the bullet emerges, the final momentum of the system is given by the sum of the momentum of the bullet (m1 × v1') and the momentum of the block (m2 × v2').

By applying the conservation of momentum principle, we equate the total momentum before the collision to the total momentum after the collision (p1 = p1' + p2').

Solving the equation, we find that the velocity of the block after the bullet emerges (v2') is 0 m/s. This means that the block does not move at all after the bullet emerges. The bullet imparts its momentum to the block, causing it to come to rest on the frictionless surface.