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A 12-g wad of sticky clay is hurled horizontally at a 100-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.5 m before coming to rest. If the coefficient of friction between the block and the surface is 0.65, what was the speed of the clay immediately before the impact?

Answer :

Answer:

91.5 m/s

Explanation:

m = mass of clay = 12 g = 0.012 kg

M = mass of wooden block = 100 g = 0.1 kg

d = distance traveled by the combination before coming to rest = 7.5 m

μ = Coefficient of friction = 0.65

V = speed of the combination of clay and lock just after collision

V' = final speed of the combination after coming to rest = 0 m/s

acceleration caused due to friction is given as

a = - μ g

a = - (0.65) (9.8)

a = - 6.37 m/s²

Using the kinematics equation

V'² = V² + 2 a d

0² = V² + 2(- 6.37) (7.5)

V = 9.8 m/s²

v = speed of clay just before collision

Using conservation of momentum

m v = (m + M) V

(0.012) v = (0.012 + 0.100) (9.8)

v = 91.5 m/s

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