A. The bullet's speed is [ ( M + m ) / m ] √ ( 2 μ g d )
B. The initial speed of the 9.0 g bullet is about 610 m/s
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Let's recall Impulse formula as follows:
[tex]\boxed {I = \Sigma F \times t}[/tex]
where:
I = impulse on the object ( kg m/s )
∑F = net force acting on object ( kg m /s² = Newton )
t = elapsed time ( s )
Let us now tackle the problem!
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Given:
mass of bullet = m = 9.0 g = 9.0 × 10⁻³ kg
mass of block = M = 12 kg
sliding distance = d = 5.4 cm = 5.4 × 10⁻² m
coefficient of kinetic friction = k = 0.20
Asked:
initial bullet's speed = u₁ = ?
Solution:
Firstly, we will use Conservation of Energy formula to find the speed of the block:
[tex]W = \Delta Ek[/tex]
[tex]fd = \frac{1}{2}(M+m)v^2[/tex]
[tex]\mu N d = \frac{1}{2}(M+m)v^2[/tex]
[tex]\mu (M + m)g d = \frac{1}{2}(M+m)v^2[/tex]
[tex]\mu g d = \frac{1}{2} v^2[/tex]
[tex]\boxed {v = \sqrt{2 \mu g d}}[/tex] → Equation A
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Next, we will use Conservation of Momentum formula to find the initial speed of the bullet:
[tex]\texttt{Total Momentum Before Collision = Total Momentum After Collision}[/tex]
[tex]m u_1 + M u_2 = ( m + M ) v[/tex]
[tex]m u_1 + M (0) = ( m + M ) v[/tex]
[tex]m u_1 = ( m + M ) v[/tex]
[tex]m u_1 = ( m + M ) \sqrt { 2\mu g d}[/tex] ← Equation A
[tex]\boxed {u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}}[/tex]
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[tex]u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}[/tex]
[tex]u_1 = \frac { 9.0 \times 10^{-3} + 12 }{ 9.0 \times 10^{-3} } \sqrt { 2 \times 0.20 \times 9.80 \times 5.4 \times 10^{-2}}[/tex]
[tex]\boxed{u_1 \approx 610 \texttt{ m/s}}[/tex]
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Grade: High School
Subject: Physics
Chapter: Dynamics
(a) The expression for the speed of the bullet is [tex]u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}[/tex]
(b) The speed of the bullet at the given parameters is 613.78 m/s.
The given parameters;
Apply the principle of conservation of linear momentum to determine the initial speed of the bullet;
[tex]m_1 u_1 + m_2 u_2 = V(m_1 + m_2)\\\\m_1 u_1 + 0 = V(m_1 + m_2)\\\\u_1 = \frac{V(m_1 + m_2)}{m_1}[/tex]
Apply the principle of work-energy theorem to determine the speed of the bullet-block system;
[tex]K.E - P.E = W_f\\\\\frac{1}{2} MV^2 - 0 = \mu_k (Mg)d\\\\V^2 = 2\mu_k gd\\\\V = \sqrt{2\mu_k gd[/tex]
The expression for the speed of the bullet is written as;
[tex]u_1 = \frac{V(m_1 + m_2)}{m_1} \\\\u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}[/tex]
The speed of the bullet at the given parameters is calculated as follows;
[tex]u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1} \\\\u_1 = \frac{(0.009 + 12) \sqrt{2\times 0.2 \times 9.8 \times 0.054} }{0.009} \\\\u_1 = 613.78 \ m/s[/tex]
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