Answer:
d_1 = 396.9 m
Explanation:
Given data:mass of lighter fragment m
Mass of heavier fragment = 7m
Velocity of lighter fragment = V1
Velocity of heavier fragment = V2
As object is rest before explosion occur thus momentum of system is zero
mv1 + 7mv2 = 0
V1 = -7V2
force of Friction on lighter fragment [tex]= f1 = \mu mg[/tex]
force of friction on heavier fragment [tex]= f2 = 7\mu mg[/tex]
Deceleration of lighter fragment = a1
we know that force of friction is given as
ma1 = - f1
[tex]ma1 = -\mu mg[/tex]
[tex]a1 = -\mu g[/tex]
Deceleration of heavier fragment = a2
7ma2 = -f2
[tex]7ma2 = -7\mu mg[/tex]
[tex]a2 = -\mu g[/tex]
Final velocity of lighter fragment = V3 = 0 m/s
Final velocity of heavier fragment = V4 = 0 m/s
Distance traveled by lighter fragment prior to rest position is = d1
Distance traveled by the heavier fragment prior to rest position is = d2 = 8.10 m
[tex]V_4^2 = V_2^2 + 2a_2 d_2
[tex]0^2 = v_2^2 + 2(-\mu g)d_2[/tex]
[tex]d_2 = \frac{v_2^2}{2 \mu g}[/tex]
[tex]V_3^2 =V_1^2 + 2a_1 d_1[/tex]
[tex]0 = (-7v_2)^2 + 2(-\mu g)d_1[/tex]
[tex]d_1 = \frac{49 v_2^2}{2\mu g}[/tex]
[tex]d_1 = 49 d_2[/tex]
[tex]d_1 = 49 \times 8.1 = 396.9 m[/tex]
The lighter fragment slide moves a distance of 396.9 m.
This is defined as the length of the space between two points.
Mass of heavier fragment = 7m
Velocity of lighter fragment = V1
Velocity of heavier fragment = V2
mv1 + 7mv2 = 0
V1 = -7V2
Force of Friction on lighter fragment = f1 = µmg
Force of friction on heavier fragment = f2 = 7µmg
Deceleration of lighter fragment = a1
Force of friction = ma1 = - f1
ma1 = - µmg
a1 = - µg
Deceleration of heavier fragment = a2
7ma2 = -f2
7ma2 = - 7µmg
a2 = - µg
V₃²= V₁² + 2a₁d₁
0 = (-7v₂)² + 2(-µg) d₁
d₁ = 49v₂²/ 2µg
= 49d₂ = 49 ˣ 8.1 = 396.9m
Read more about Distance here https://brainly.com/question/23848540
