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A bullet of mass 4.00 g is fired horizontally into a wooden block of mass 1.30 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.170. The bullet remains embedded in the block, which is observed to slide a distance 0.240 m along the surface before stopping. Part A What was the initial speed of the bullet

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boffeemadrid

Answer:

[tex]291.67\ \text{m/s}[/tex]

Explanation:

[tex]m_1[/tex] = Mass of bullet = 4 g

[tex]m_2[/tex] = Mass of block = 1.3 kg

[tex]\mu[/tex] = Coefficient of friction = 0.17

[tex]s[/tex] = Displacement of block = 0.24 m

[tex]v_1[/tex] = Velocity of bullet

[tex]v[/tex] = Velocity of combined mass

[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

The energy balance of the system is given by

[tex]\dfrac{1}{2}(m_1+m_2)v^2=\mu(m_1+m_2)gs\\\Rightarrow v=\sqrt{2\mu gs}[/tex]

As the momentum is conserved in the system we have

[tex]m_1v_1=(m_1+m_2)v\\\Rightarrow m_1v_1=(m_1+m_2)\sqrt{2\mu gs}\\\Rightarrow v_1=\dfrac{(m_1+m_2)\sqrt{2\mu gs}}{m_1}\\\Rightarrow v_1=\dfrac{(4\times 10^{-3}+1.3)\times \sqrt{2\times 0.17\times 9.81\times 0.24}}{4\times 10^{-3}}\\\Rightarrow v_1=291.67\ \text{m/s}[/tex]

The initial speed of the bullet is [tex]291.67\ \text{m/s}[/tex].

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