Answer: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
