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

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