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A bullet with a mass m = 12.5 g and speed v = 97.4 m/s is fired into a wooden block with M = 113 g which is initially at rest on a horizontal surface. The bullet is embedded into the block. The block-bullet combination slides across the surface for a distance d before stopping due to friction between the block and surface. The coefficients of friction are µs = 0.753 and µk = 0.659.(a) What is the speed (m/s) of the block-bullet combination immediately after the collision?(b) What is the distance d (m)?

Respuesta :

Answer:

Explanation:

[tex]m_1=12.5 gm[/tex]

initial speed [tex]v_1=97.4 m/s[/tex]

Mass of block [tex]m_2=113 gm[/tex]

Initial momentum [tex]P_i=m_1\cdot v_1[/tex]

Final Momentum [tex]P_f=(m_1+m_2)\cdot v_2[/tex]

Conserving momentum

[tex]P_i=P_f[/tex]

[tex]m_1\times v_1=(m_1+m_2)v_2[/tex]

[tex]v_2=\frac{m_1}{m_1+m_2}\times v_1[/tex]

[tex]v_2=\frac{12.5}{12.5+113}\times 97.4[/tex]

[tex]v_2=9.701 m/s[/tex]

(b)Frictional Force on combined mass

[tex]F_{r}=\mu _kN[/tex]

[tex]F_{r}=0.659\times (m_1+m_2)\cdot g[/tex]

[tex]F_r=0.659\times (0.0125+0.113)\cdot 9.8[/tex]

[tex]F_r=0.81 N[/tex]

acceleration a=\frac{F_r}{m_1+m_2}[/tex]

[tex]a=\frac{0.81}{0.1255}[/tex]

[tex]a=6.45 m/s^2[/tex] (deceleration)

using [tex]v^2-u^2=2 as[/tex]

[tex]0-(9.701)^2=2\cdot (-6.45)\cdot s[/tex]

[tex]s=\frac{94.109}{12.9}[/tex]

[tex]s=7.295 m[/tex]

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