Apyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. at the same instant the firecracker begins to explode into two pieces, "top" and "bottom", with masses mtop = 1 kg and mbottom=2 kg. at the end of the explosion (t = 0.6 s), the top piece is moving upward with speed 6 m/s the mass of the explosive substance is negligible in comparison to the mass of the two pieces.

a) determine the magnitude of the net force on the firecracker system before the explosion. (use g=10). explain.

i said f=m*a=(3)(10)= 30n

b) determine the magnitude of the net force on the firecracker system during the explosion.

i said the net force would still be 30 n but i'm not sure.

c) determine the magnitude and reaction of the net impulse on the firecracker system during the explosion (from t=0.4 to t=0.6). explain.

i did p=m*v=(30)(0.2)=6 but i'm not sure if this is right

d) use the impulse momentum theorem to determine the change in magnitude and direction of momentum of the firecracker system during the explosion.

draw a vector diagram for initial, change, and final momentum of the top, bottom, and system as a whole.

A) The magnitude of the net force on the firecracker system before the explosion is; F_net = 30 N

B) The magnitude of the net force on the firecracker system during the explosion is; F_net = 30 N

C) The net impulse on the firecracker system during the explosion from t=0.4 to t=0.6 s is; I = -12 N.s

D) The change in magnitude and direction of momentum of the firecracker system during the explosion is; ΔP = 0 N.s

We are given;

Mass before explosion; m = 3kg

Speed before explosion; v = -4 m/s (because it is descending)

Time before explosion; t = 0,4 s

Mass of top piece after explosion; m₁ = 1 kg

Mass of bottom piece after explosion; m₂ = 2 kg

speed of top piece; v₁ = 6 m/s

Time after explosion; t = 0.6 s

A) The magnitude of the net force on the firecracker system before the explosion is; F_net = mg

F_net = 3 × 10

F_net = 30 N

B) The net force during collision will be equal to the net force before collision as a result of newton's third law of motion.

C) From conservation of momentum, we can say that;

mv = m₁v₁ + m₂v₂

where;

v₁ is speed of top piece

v₂ is speed of bottom piece

Plugging in the relevant values gives;

3 × -4 = (1 × 6) + (2v₂)

2v₂ = -18

v₂ = -18/2

v₂ = -9 m/s

net impulse on the firecracker system during the explosion from t=0.4 to t=0.6 is;

I = m₁v₁ + m₂v₂

I = (1 × 6) + (2 × -9)

I = -12 N.s

D) From the  impulse momentum theorem, the change in magnitude and direction of momentum of the firecracker system during the explosion is;

ΔP = ΔI = I - I₀

where I₀ = mv = 3 * -4

I₀ = -12 N.s

Thus;

ΔP = -12 -(-12)

ΔP = 0 N.s

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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