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.

Answer

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

finding the launch speed of bullet

Velocity of wooden block + bullet just after impact

= [\tex]\sqrt{2gh}[/tex]

=[\tex]\sqrt{2\times 9.8 \times 0.6}[/tex]

= 3.43 m/s

v₁ be the launch velocity

Applying law of conservation of momentum

0.03 x v₂ = 0.530 x 3.43

v₂ = 60.6 m /s

if v₁ be initial velocity

v₂² = v₁² + 2 g h

v₁² = v₂² - 2 gh

v₁² = 60.6 ² - 2 x (-9.8 )x 0.4

v₁ = 60.65 m /s this is launch speed

i think they are all correct

explanation:

they seem good

answer; equivalent to a one(1) horsepower motor;