Question

In a police rifle test, a 15-g bullet traveling 213 m/s in a vertical direction suddenly buries itself in a 2.4-kg block of wood at rest directly above it. As a result, the bullet-block combination moves vertically upward.

(a) Determine the velocity of the bullet-block combination just after the impact.

(b) Determine the maximum height reached by the bullet/block combination.

(c) Is kinetic energy conserved in this collision?   

Answer 1

given that :

mass of the bullet, m₁ = 15 g = 0.015 kg

mass of the block of wood, m₂ = 2.4 kg

speed of the bullet, v₁ = 213 m/s

(a) the velocity of the bullet-block combination just after the impact which is given as ::

using a conservation of linear momentum,

m₁ v₁ + m₂ v₂ = (m₁ + m₂) V    { eq. 1 }

where, v₂ = 0 m/s

inserting all these values in above eq.

(0.015 kg) (213 m/s) + (2.4 kg) (0 m/s) = [(0.015 kg) + (2.4 kg)] V

(3.195 kg.m/s) + (0) = (2.415 kg) V

V = (3.195 kg.m/s) / (2.415 kg)

V = 1.32 m/s

(b) the maximum height reached by the bullet/block combination is given as ::

using a conservation of energy,

K.E after impact = potential energy after impact

1/2 M V² = M g h

h = (0.5) M V² / M g    { eq. 2 }

where, g = acceleration due to gravity = 9.8 m/s²

inserting the values in eq.2,

h = (0.5) (2.415 kg) (1.32 m/s)² / (2.415 kg) (9.8 m/s²)

h = (0.8712 m/s)² / (9.8 m/s²)

h = 0.088 m

or   h = 8.8 cm

(c) NO, kinetic energy conserved in this collision.

because, kinetic energy before collision : (0.5) m₁ v₁² = (0.5) (0.015 kg) (213 m/s)²

K.E_before = 340.2 J