In a police ballistics test, 2.00-g bullet traveling at 700 m/s suddenly hits and becomes embedded in a stationary 5.00-kg wood block. Does the bullet gain or lose momentum?

The bullet will lose momentum

Explanation:

mass of the bullet, m₁ = 2.0 g = 0.002 kg

initial velocity of the bullet, u₁ = 700 m/s

mass of the wood block, m₂ = 5 kg

initial velocity of the wood block, u₂ = 0

let the final velocity of the bullet-block system after collision = v

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(0.002 x 700) + (5 x 0) = v(0.002 + 5)

1.4 = 5.002 v

v = 1.4 / 5.002

v = 0.28 m/s

Change in momentum of the bullet is given by;

ΔP = m₁v - m₁u₁

ΔP = m₁(v - u₁)

ΔP = 0.002( 0.28 - 700)

ΔP = -1.4 kg.m/s

Therefore, the bullet will lose momentum

the maxium height is 34m/s