A pyro technician releases a firecracker of mass mcracker = 7.0 kg from rest at

Question

A pyro technician releases a firecracker of mass mcracker = 7.0 kg from rest at time t = 0. At t = 0.7 s, the firecracker is moving downward with speed v1. At this same instant, the firecracker begins to explode into two pieces with masses mtop = 2.0 kg and mbottom = 5.0 kg. At the end of the explosion (t = 1.3 s), the top piece is moving upward with speed 8.0 m/s.

a) Assuming that the firecracker is in free-fall from t = 0 s to t = 0.7 s, what is the speed of the firecracker at t = 0.7 s?

b) What is the magnitude of the net force on the firecracker system between t = 0.7 s and t = 1.3 s?

c) What is the magnitude of the net impulse on the firecracker system between t = 0.7 s and t = 1.3 s?

e) What is the speed of the bottom piece of the firecracker at t = 1.3 s?

Explanation / Answer

A) SPEED OF FIRECRACKER AT T =.7 =at = 6.86m/s

b) speed of enr of mass at t = 1.3 = at = 12.74

net force = change in kinetic energy = 1/2m(v^2-u^2) = 1/2*7*( 12.74^2-6.86^2) = 403.368

c) net impulse = force/time = 403.368/.6 = 672.28

d) taking bottom to be positive

conserving momentum

-2*8+5*v = 7*12.74

v =1.636

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