Answer:
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
