A 9.0 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 (in m/s) immediately before impact?

107.58 m/s

Explanation:

from the question we are given the following:

mass of the stick (M₁) = 9 g = 0.009 kg

mass of the block (M₂) = 90 g = 0.09 kg

total mass (M₁+M₂) = 99 g = 0.099 kg

distance (s) = 7.5 m

coefficient of friction (Uk) = 0.650

acceleration due to gravity (g) = 9.8 m/s^{2}

apply the equation below

change in kinetic energy = net work done

final kinetic energy (K₂) - final kinetic energy (K₁) = Uk x force (F) x distance (s)

final kinetic energy is zero because the clay and the wood comes to a stop

kinetic energy = 0.5 x m x v^{2} and

force = m x g so the equation now becomes

0.5 x m x v^{2} = Uk x m x g x s

0.5 x 0.099 x v^{2} = 0.650 x 0.099 x 9.8 x 7.5

V = 9.78 m/s

from the conservation of momentum

M₁ V₁ + M₂ V₂ = (M₁ + M₂) V

V₂ is zero since the block is initially at rest, so our equation becomes

M₁ V₁ = (M₁ + M₂) V

0.009 x V₁ = 0.099 x 9.78

V₁ = 107.58 m/s

the speed of the clay before impact is 107.58 m/s