A 14.0-g wad of sticky clay is hurled horizontally at a 90-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact

the speed of the clay immediately before impact =72.58m/s

Explanation:

Given that  

mass of the stick clay, M₁= 14.0 g = 0.014 kg

mass of the block ,M₂= 90 g = 0.09 kg

Therefore the total mass= (M₁+M₂) = 104g = 0.104 kg

Also, distance, s = 7.50 m

coefficient of friction μ= 0.650

Acceleration due to gravity ,g = 9.8 m/s²

Using the Work- Energy theorem,

change in kinetic energy =  work done

final kinetic energy(K₂) - initial  kinetic energy(K₁) =   force, F x coefficient of friction, μ x distance,s

The final kinetic energy is zero  because after the impact,  the block with the clay comes to a stop after 7.50m

kinetic energy =Work done

0.5 x m x v²=coefficient of friction,  μ x force(F)  x  distance,s(Since force = m g )

0.5 x m x v²= μ x m x g x s

0.5 x 0.104 x v² = 0.650 x 0.104x 9.8 x 7.5

v²= 0.650 x 0.104x 9.8 x 7.5 / 0.5 x 0.104

v²==95.55

V = 9.77 m/s

Using the  conservation of momentum formulae where

M₁ V₁ + M₂ V₂ = (M₁ + M₂ ) V

Since V₂  which is the velocity of block  is zero as the  block is initially at rest, We now have that

M₁ V₁ = (M₁ + M₂ ) V

0.014 kg x V₁ = 0.104 x 9.77

V₁=0.104 x 9.77 / 0.014

V=72.58m/s

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