Answer:
a) F = 30 N, b) I = 12 N s
, c) I = -12 N s
, d) ΔI = 0 N s
Explanation:
This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved
Initial, before the explosion
p₀ = m v
The speed can be found by kinematics
v = v₀ - g t
v = 0 - 10 0.4
v = -4.0 m / s
Final after division
pf = m₁ v₁f + m₂ v₂f
p₀ = pf
M v = m₁ v₁f + m₂ v₂f
Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )
a) before the explosion the only force acting on the body is gravity
F = mg
F = 3 10 = 30 N
b) The expression for momentum is
I = Ft
Before the explosion the only force that acts is the weight
I = mg t
I = 3 10 0.4
I = 12 N s
c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2
M v = m₁ v₁f + m₂ v₂f
v₂f = (M v - m₁ v₁f) / m₂
v₂f = (3 (-4) - 1 6) / 2
v₂f = - 9 m / 2
The negative sign indicates that body 2 (botton) is descending
Now we can use the momentum and momentum relationship for the body during the explosion
I = F t = Dp
F t = pf –po)
F t= [m₁ v₁f + m₂ v₂f]
I = [1 6 + 2 (-9) -0]
I = -12 N s
This is the impulse during the explosion the negative sign indicates that it is headed down
d) impulse change
I₀ = Mv
I₀ = 3 *4
I₀ =-12 N s
ΔI =If – I₀
ΔI = - 12 – (-12)
ΔI = -0 N s
