A fireworks rocket is fired vertically upward. At its maximum height of 70,0m ,

Question

A fireworks rocket is fired vertically upward. At its maximum height of 70,0m , it explodes and breaks into two pieces, one with mass mA = 1.50kg and the other with mass mB = 0.290kg. In the explosion, 930J of chemical energy is converted to kinetic energy of the two fragments. Part A What is the speed of each fragment just after the explosion? vA, vB = Part B It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored. |xA - xB| =

Explanation / Answer

1)

Let the velocity of mass , mA be vA and that of mb be vB

So, Kinetic energy of the two combined, KE,tot = 0.5*mA*vA^2 + 0.5*mB*vB^2 = 930

At the top of its flight, the speed of the rocket is 0

At this point, we can conserve momentum, to get,

mA*vA + mB*vB = 0

So, vA = -mB*vB/mA

Plugging this value of vA in the Kinetic energy equation, we get,

0.5*mA*(-mB*vB/mA)^2 + 0.5*mB*vB^2 = 930

So, 0.5*1.5*(-0.29*vB/1.5)^2 + 0.5*0.29*vB^2 = 930

Solving this equation, we get,

vB = 73.3 m/s <-------------answer

So, vA = -0.29*73.3/1.5 = 14.2 m/s <-----------answer

2)

Using the equation of motion,

s = ut + 0.5*at^2

In vertical motion after the explosion,

where s = height of the rocket = 70 m

u = initial vertical velocity = 0 m/s

a = 9.8 m/s2

So, 70 = 0*t + 0.5*9.8*t^2

So, t = 3.78 s <------- time taken to reach ground by both

So, distance traveled in horizontal distance by both,

by mA, xA = vA*t = -14.2*3.78 = -53.7 m <------ negative sign for oppite direction to vB

Similarly, xB = vB*t = 73.3*3.78 = 277 m

So,

| xA - xB | = 277+53.7 = 330.7 m <--------------answer

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