20 m/s is the speed of the heavier fragment just after the explosion.
Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
F = ma where, F = Force (Newton)
m= mass
a = acceleration
Given:
mass of rocket = M = 1500 kg
acceleration of rocket = a = 10 m/s²
elapsed time = t = 2.00 s
mass of lighter fragment = m₁ = m = 500 kg
mass of heavier fragment = m₂ = 2m = 1000 kg
maximum height of lighter fragment = h = 530 m
Let's calculate the final speed of the rocket just before the explosion:
v = u + at
v = 0 + 10(2)
v = 20 m/s
Then, we will calculate the height of the rocket just before the explosion:
[tex]h' = ut + \frac{1}{2}at^{2}[/tex]
[tex]h' = 0 + \frac{1}{2} (10)(2.00)^{2}[/tex]
[tex]h' =20m[/tex]
The initial speed of lighter fragment just after the explosion:
[tex]v^{2}[/tex]₁ [tex]= u^{2}[/tex]₁ - [tex]2g[/tex]Δ[tex]h[/tex]
[tex]v^{2}[/tex]₁ [tex]= u^{2}[/tex]₁ - [tex]2g[/tex] [tex](h - h')[/tex]
[tex]0^{2} = u^{2}[/tex]₁[tex]- 2(9.8) (530-20)[/tex]
[tex]u^{2}[/tex]₁[tex]=9996[/tex]
[tex]u[/tex]₁ [tex]=\sqrt{9996} m/s[/tex]
Using Conservation of Momentum Law :
[tex]M v=m[/tex]₁ [tex]u[/tex]₁ [tex]+ m[/tex]₂[tex]u[/tex]₂
[tex]1500 (20)= 500(\sqrt{9996} ) + 1000u[/tex]₂
[tex]u[/tex]₂ ≅ [tex]- 20 m/s[/tex]
Learn more about Newton's law of motion here:
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