A30 gram bullet is shot upward at a wooden block. the bullet is launched at the speed vi. it travels up 0.40 m to strike the wooden block. the wooden block is 20 cm wide and 10 cm high and its thickness gives it a mass of 500 g. the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height. a. what is the launch speed of the bullet? b. how much mechanical energy does the bullet and the block system have before all of the processes? use the surface the block rests on as the reference for where gravitational potential energy is zero. c. how much mechanical energy does the bullet and the block system have after all of the processes? d. how much mechanical energy was lost from beginning to end?

Explanation:

Mass of bullet m = .03 kg

Mass of wooden block M = 0.5 kg

Since the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height

Velocity of wooden block + bullet just after impact = √2gH

=√(2 x 9.8 x 0.6)

= 3.43 m / s

Let the launch velocity of bullet be v₁

If v₂ be the velocity with which bullet hits the block

Applying law of conservation of momentum

.03 x v₂ = .530 x 3.43

v₂ = 60.6 m /s

if v₁ be initial velocity

v₂² = v₁² - 2 gh

v₁² = v₂² + 2 gh

= 60.6 ² + 2 x 9.8 x 0.4

v₁ = 60.65 m /s this is launch speed.

b )

Initial kinetic energy of bullet

= 1/2 m v²

= .5 x .03 x 3680

= 55 J

Potential energy of bullet + block = 0

Total energy = 5 J

c)

Kinetic energy of bullet block system

1/2 m v²

= .5 x .53 x  3.43

= 3.11 J

d )

Loss of energy in the impact =  Total mechanical energy  lost from beginning to end?

3.11 J  - 5

= 1.89 J