Answer :
The lighter fragment slide to a distance of 289.1m.
The distance between two points is referred to as its length.
7m in mass for the heavier fragment
The lighter fragment's velocity is equal to V1.
The velocity of larger pieces is equal to V2.
mv1 + 7mv2 = 0
V1 = -7V2
The frictional force on lighter fragment: f1 = mg
Friction force on a heavier piece is equal to f2 = 7 mg.
The lighter fragment's deceleration equals a1.
frictional force = ma1 = - f1
ma1 = - µmg
a1 = - µg
Deceleration of the heavier piece equals a2.
7ma2 = -f2
7ma2 = - 7µmg
a2 = - µg
V₃²= V₁² + 2a₁d₁
0 = (-7v₂)² + 2(-µg) d₁
d₁ = 49v₂²/ 2µg
d₁ = 49d₂ = 49 ˣ 5.9 = 289.1m
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