A 0.0155 kg bullet hits a 1.00 kg block of wood that is attached to wires (lengths equals 1.75 m) so that it can swing like a pendulum...?

As the block of wood and the embedded bullet swings to a maximum height of 0.141 meter, their velocity decreases from their initial velocity to 0 m/s at the rate of 9.8 m/s each second. Use the following equation to determine their initial velocity.

vf^2 = vi^2 + 2 * a * d
0 = vi^2 + 2 * -9.8 * 0.141
vi = √2.7636

This is approximately 1.66 m/s. Let’s use conservation of momentum to determine the initial speed of the bullet.

Initial momentum = 0.0155 * v
Final momentum = 1.0155 * √2.7636
0.0155 * v = 1.0155 * √2.7636
v = 1.0155 * √2.7636 ÷ 0.0155
This is approximately 108.9 m/s.

Take g = 9.81 m/s²

Loss in kinetic energy of the bullet (in J)
= (1/2)mv²
= (1/2) × 0.0155 × v²

Gain in potential energy of the block and the bullet (in J)
= (m + M)gh
= (0.0155 + 1.00) × 9.81 × 0.141

According to the law of conservation of mechanical energy :
Loss in kinetic energy of the bullet = Gain in potential energy of the block and the bullet
(1/2) × 0.0155 × v² = (0.0155 + 1.00) × 9.81 × 0.141
v = 13.5 (m/s)

Initial speed of the bullet = 13.5 m/s

Momentum is conserved when the bullet hits the block, giving the block + bullet kinetic energy.

Energy is conserved as the block swings. The kinetic energy turns into potential energy.

So work the calculations backward.
1. Take the final potential energy mgh and set it equal to (1/2)mv^2, the original kinetic energy of the block + bullet. Solve for v.

2. Use that v to calculate the original momentum mv of the block + bullet and set it equal to mv for the bullet (different m, different v). Solve for v of the bullet.