Answer:
The initial speed of the bullet is 169.12 m/s
Explanation:
Given;
mass of bullet, m₁ = 8 g = 0.008 kg
mass of wooden block, m₂ = 1.20-kg
coefficient of kinetic friction, μ = 0.20
distance moved by the block, d = 0.320 m
Step 1:
frictional force on the wooden block after the collision
Fk = μm₂g
Neglect mass of the bullet since it is small compared to mass of the block
Fk = 0.2 x 1.2 x 9.8
Fk = 2.352 N
Step 2:
acceleration of the block after the impact
From Newton's second law of motion;
F = ma
Fk = m₂a
a = Fk / m₂
a = 2.352 / 1.2
a = 1.96 m/s²
Step 3:
velocity of the block after the impact
v² = u² + 2as
where;
s is the distance moved by the block = d
v² = 0² + 2(1.96 x 0.32)
v² = 1.2544
v = √1.2544
v = 1.12 m/s
Step 4:
velocity of the bullet before the collision
Apply the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
u₂ is speed of the wooden block before collision = 0, since it was resting ....
0.008u₁ + 1.2 x 0 = 1.12(0.008 + 1.2)
0.008u₁ = 1.35296
u₁ = 1.35296 / 0.008
u₁ = 169.12 m/s
Therefore, the initial speed of the bullet is 169.12 m/s
The initial speed of the bullet as it was fired horizontally into the wooden block resting on a horizontal surface is 169.12m/s.
Given the data in the question;
Initial speed of the bullet; [tex]v = \ ?[/tex]
First we get the velocity of the Wood. From Conservation of Energy:
[tex]\frac{1}{2}mv^2 = umgs\\\\\frac{1}{2}v^2 = ugs\\\\v = \sqrt{2ugs}[/tex]
Where v is the velocity of the block, μ is the coefficient of kinetic friction, s is the displacement of the bullet and g is acceleration due to gravity( [tex]9.8m/s^2[/tex] ).
We substitute in our values
[tex]v \sqrt{2\ *\ 0.20\ *\ 9.8m/s^2\ *\ 0.320m } \\\\v = \sqrt{1.2544m^2/s^2}\\\\v = 1.12m/s\\\\v_w = 1.12m/s[/tex]
Now, we determine the velocity of the bullet. Using Conservation of Momentum.
[tex]m_bv_b = (m_b + m_w )v_w[/tex]
We substitute our values into the equation
[tex]0.008kg * v_b = ( 0.008kg + 1.20kg) * 1.12m/s\\\\0.008kg * v_b = 1.208kg * 1.12m/s\\\\0.008kg * v_b = 1.35296kg.m/s\\\\v_b = \frac{1.35296kg.m/s}{0.008kg}\\\\v_b = 169.12 m/s[/tex]
Therefore, the initial speed of the bullet as it was fired horizontally into the wooden block resting on a horizontal surface is 169.12m/s.
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