Question

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 1

Answer:

94 ms⁻¹

Explanation:

$V$ = Speed of the bullet-block combination after collision

$d$ = distance traveled by the combination before coming to stop = 6.5 m

$\mu$ = Coefficient of kinetic friction = 0.750

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

$a = - \mu g = - (0.750) (9.8) = - 7.35 ms^{-2}$

$V_{f}$ = Final speed of the combination = 0 m/s

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

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

$M$ = mass of the wooden block = 110 g

$m$ = mass of the bullet = 12.8 g

$v$ = Speed of the bullet before collision

Using conservation of momentum for inelastic collision , we have

$m v = (m + M) V\\(12.8) v = (12.8 + 110) (9.8)\\v = 94 ms^{-1}$