Question

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

Answer 1

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

Step-by-step 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