Answer:
The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.
Explanation:
Given;
mass of the bullet, m₁ = 2.47 g = 0.00247 kg
mass of the wooden block, m₂ = 2.43 kg
initial velocity of the wooden block, u₂ = 0
height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m
let the initial velocity of the bullet on leaving the gun's barrel = v₁
let final velocity of the bullet-wooden block system after collision = v₂
Apply the principle of conservation of linear momentum;
Total initial momentum = Total final momentum
m₁v₁ + m₂u₂ = v₂(m₁ + m₂)
0.00247v₁ + 2.43 x 0 = v₂(2.43 + 0.00247)
0.00247v₁ = 2.4325v₂ -------(1)
The kinetic energy of the bullet-block system after collision;
K.E = ¹/₂(m₁ + m₂)v₂²
K.E = ¹/₂ (2.4325)v₂²
The potential energy of the bullet-block system after collision;
P.E = mgh
P.E = (2.4325)(9.8)(0.00295)
P.E = 0.07032
Apply the principle of conservation of mechanical energy;
K.E = P.E
¹/₂ (2.4325)v₂² = 0.07032
1.21625 v₂² = 0.07032
v₂² = 0.07032 / 1.21625
v₂² = 0.0578
v₂ = √0.0578
v₂ = 0.24 m/s
Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;
0.00247v₁ = 2.4325v₂
0.00247v₁ = 2.4325 (0.24)
0.00247v₁ = 0.5838
v₁ = 0.5838 / 0.00247
v₁ = 236.36 m/s
Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.
