Question

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?

Answer 1

In a police ballistics test, when a bullet traveling at 700 m/s hits and becomes embedded in a stationary wood block, the bullet loses momentum.

Step-by-step explanation:

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

Momentum is defined as the product of an object's mass and velocity. When the bullet hits the wood block and becomes embedded in it, the system consisting of the bullet and the block gains momentum but the bullet itself loses momentum.

This is because momentum is conserved in an isolated system, meaning that the total momentum before the collision is equal to the total momentum after the collision. Since the wood block is initially at rest, its momentum is zero. Therefore, the bullet's momentum must be equal in magnitude and opposite in direction to the momentum of the bullet and the block together after the collision.

Answer 2

The bullet will lose momentum

Step-by-step 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