The magnitude of the block's velocity just after impact is about 2.60 m/s.
The initial speed of the bullet is about 325 m/s
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Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !
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Given:
mass of bullet = m₁ = 8.00 gram = 8 × 10⁻³ kg
mass of block = m₂ = 0.992 kg
initial speed of block = u₂ = 0 m/s
compression of spring = x = 15.0 cm = 0.15 m
spring constant = k = 0.750 ÷ (2.5 ×10⁻³) = 300 N/m
Asked:
final speed of block = v₂ = ?
initial speed of bullet = u₁ = ?
Solution:
Let:
final speed of the block just after impact = final speed of the bullet just after impact = v
Firstly , we will use Conservation of Energy as follows:
[tex]Ek = Ep[/tex]
[tex]\frac{1}{2}(m_1 + m_2)v^2 = \frac{1}{2} k x^2[/tex]
[tex](m_1 + m_2) v^2 = k x^2[/tex]
[tex](0.008 + 0.992) v^2 = 300 \times 0.15^2[/tex]
[tex]v^2 = 6.75[/tex]
[tex]v = \frac{3}{2}\sqrt{3} \texttt{ m/s}[/tex]
[tex]v \approx 2.60 \texttt{ m/s}[/tex]
The magnitude of the block's velocity just after impact is about 2.60 m/s
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Next , we will use Conservation of Momentum Law as follows:
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]
[tex]0.008u_1 + 0.992(0) = 0.008(\frac{3}{2}\sqrt{3}) + 0.992(\frac{3}{2}\sqrt{3})[/tex]
[tex]0.008u_1 = \frac{3}{2}\sqrt{3}[/tex]
[tex]u_1 \approx 325 \texttt{ m/s}[/tex]
The initial speed of the bullet is about 325 m/s
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Grade: High School
Subject: Physics
Chapter: Dynamics
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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
