Question

IP A fireworks rocket is launched vertically into the night sky with an initial speed of 42.2 m/s . The rocket coasts after being launched, then explodes and breaks into two pieces of equal mass 2.90 s later.

Part A

If each piece follows a trajectory that is initially at 45.0 ∘ to the vertical, what was their speed immediately after the explosion?

Part B

What is the velocity of the rocket's center of mass before the explosion?

Part C

What is the velocity of the rocket's center of mass after the explosion?

Part D

What is the acceleration of the rocket's center of mass before the explosion?

Answer 1

consider the motion of rocket ::

V_i = initial velocity = 42.2 m/s

V_f = final velocity just before explosion

t = time = 2.90 sec

a = acceleration = - 9.81 m/s²

Using the equation

V_f = vi + a t

V_f = 42.2 + (-9.81) (2.90)

V_f = 13.75 m/s verticallu up

a)

let the velocity of each piece = V       making angle 45 with the vertical

m = mass of rocket

m/2 = mass of each piece

using conservation of momentum along vertical direction

m V_f = (m/2) (V cos45) + (m/2) (V cos45)

V_f = V cos45

13.75 = v cos45

v = 19.45 m/s

B)

rockets center of mass ' speed before explosion = V_f = 13.75 m/s vertically upwards

c)

since the momentum is conserved ,

rockets center of mass ' speed after explosion = V_f = 13.75 m/s

d)

acceleration = 9.8 m/s²