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A 110. g wooden block is initially at rest on a rough horizontal surface when a 12.8 g bullet is fired horizontally into (but does not go through) it. After the impact, the block–bullet combination slides 6.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.750, determine the speed of the bullet (in m/s) immediately before impact.

Answer :

Answer:

94 ms⁻¹

Explanation:

[tex]V[/tex] = Speed of the bullet-block combination after collision

[tex]d[/tex] = distance traveled by the combination before coming to stop = 6.5 m

[tex]\mu [/tex] = Coefficient of kinetic friction = 0.750

acceleration due to friction on a flat surface can be given as

[tex]a = - \mu g = - (0.750) (9.8) = - 7.35 ms^{-2}[/tex]

[tex]V_{f}[/tex] = Final speed of the combination = 0 m/s

Based on the above equation , we can use the kinematics equation as

[tex]V_{f}^{2} = V^{2} + 2 a d\\(0)^{2} = V^{2} + 2 (- 7.35) (6.5)\\V = 9.8 ms^{-1}[/tex]

[tex]M[/tex] = mass of the wooden block = 110 g

[tex]m[/tex] = mass of the bullet = 12.8 g

[tex]v[/tex] = Speed of the bullet before collision

Using conservation of momentum for inelastic collision , we have

[tex]m v = (m + M) V\\(12.8) v = (12.8 + 110) (9.8)\\v = 94 ms^{-1}[/tex]

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