A fireworks rocket is launched vertically (+y-direction) into the night sky with

Question

A fireworks rocket is launched vertically (+y-direction) into the night sky with an initial speed of 44.3 m/s. The rocket coasts after being launched, then explodes and breaks into two pieces of equal mass 2.41 s later.
(a) If each piece follows a trajectory that is initially at 45.6° to the vertical, what was their speed immediately after the explosion?
m/s

(b) What is the velocity of the rocket's center of mass before and after the explosion?
( m/s) y-direction

(c) What is the acceleration of the rocket's center of mass before and after the explosion?
( m/s2) y-direction

Explanation / Answer

initial velocity= 44.3 m/s

final velocity (V at t=2.41sec)= 44.3-9.8*2.41 (from v=u-at)

= 20.682 m/s in upward direction

(a) since two pices are divided in equal parts so their wight will be m/2. and their direction are at 45.6° so their speed will be same.

since no external force is working at the moment of exposion so net momentum will conserve, hence

momeentum in horizontal direction

(m/2)*v1*sin(45.6°)= (m/2)*v2*sin(45.6°)

so v1=v2

hence, momentum in vertical direction

m.20.682={2*v*m/2}cos(45.6)

v= 29.56m/s at 45.6° to the vertical

(B) since no external force is applied for expolsion so momentum will conserve.

before explosion v=20.682m/s in vertically upward direction

as mass loss is zero velocity of center of mass will be remain same as before = 20.682m/s in vertically upward direction.

(C) since no external force is working change in acceleration of center of mass will be zero.

initial acceleration= 9.8m/s^2 in vertically downdord direction

so final acceleration= 9.8m/s^2 in vertically downdord direction.

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