Respuesta :
Answer:
The speed of each fragment is as follows
mA = 16.55 m/s
mB = 71.73 m/s.
Explanation:
To solve the question, we note that
Maximum height of fireworks rocket = 70.0 m
mA = 1.30 kg
mB = 0.300 kg
Energy converted = 950 J
Since the available energy to be converted = 950 J
we have Kinetic energy, K.E. = 950 J
However K.E. = [tex]\frac{1}{2}\cdot m\cdot v^{2}[/tex] and the rocket breaks into two pieces mA and mB which constitute the mass in the K.E. equation.
That is K.E. = [tex]\frac{1}{2}[/tex] ×mA×v[tex]_A[/tex]² +
To find v[tex]_A[/tex] and v[tex]_B[/tex], we note that from the principle of conservation of linear momentum we have at the time of explosion;
mA×v[tex]_A[/tex] = mA×v[tex]_B[/tex]
1.30 kg×v[tex]_A[/tex] = 0.300 kg×v[tex]_B[/tex]
That is
v[tex]_A[/tex] = [tex]\frac{3}{13}[/tex]×v[tex]_B[/tex]
Therefore
K.E. = [tex]\frac{1}{2}[/tex] ×mA×v[tex]_A[/tex]² +
950 J = [tex]\frac{1}{2}[/tex] ×1.3 kg×v[tex]_A[/tex]² +
= [tex]\frac{1}{2}[/tex] ×1.3 kg×([tex]\frac{3}{13}[/tex]×v[tex]_B[/tex])² +
= [tex]\frac{9}{260}[/tex]×v[tex]_B[/tex]² + [tex]\frac{3}{20}[/tex]×v
∴ v[tex]_B[/tex]² 950 J × [tex]\frac{65}{12}[/tex] =5145.83 m²/s²
v[tex]_B[/tex] = [tex]\sqrt{5145.83 m^2/s^2}[/tex] = 71.73 m/s
Therefore, since v[tex]_A[/tex] = [tex]\frac{3}{13}[/tex]×v[tex]_B[/tex]
v[tex]_A[/tex] = [tex]\frac{3}{13}[/tex] × 71.73 m/s = 16.55 m/s.
