Respuesta :
Answer:
10,041.7 J
Explanation:
Using Newton's second law of motion of motion; the linear momentum in each direction will be conserved.
For each direction,
Momentum before collision = momentum after collision
In the x- direction, there was initially no x-component for velocity before collision for the two bodies,
0 = m₁v₁ₓ + m₂v₂ₓ
0 = (89.3)(7.43) + (57.7)(v₂ₓ)
v₂ₓ = (-89.3×7.43)/57.7
v₂ₓ = -11.50 m/s (minus indicating that the motion is in the opposite direction the first diver's motion).
In the y- direction, there was initially no y-component for velocity before collision for the two bodies,
0 = m₁v₁ᵧ + m₂v₂ᵧ
0 = (89.3)(5.75) + (57.7)(v₂ᵧ)
v₂ᵧ = (-89.3×5.75)/57.7
v₂ᵧ = -8.90 m/s (minus indicating that the motion is in the opposite direction the first diver's motion).
And the velocity in the z-axis for both has been given to be 55.9 m/s
Kinetic energy of the system before leaving each other = (1/2)(89.3+57.7)(55.9²) = 229673.535 J
Kinetic energy of skydiver 1 after collision
To do this, we first find the magnitude of the velocity
v₁ = √(7.43² + 5.75² + 55.9²) = 56.684 m/s
Kinetic energy = (1/2)(89.3)(56.684²) = 143463.84 J
Kinetic energy of skydiver 2 after collision
To do this, we first find the magnitude of the velocity
v₂ = √[(-11.50)² + (-8.90)² + 55.9²] = 57.76 m/s
Kinetic energy = (1/2)(57.7)(57.76²) = 96251.39 J
Change in kinetic energy of the system = (final kinetic energy) - (initial kinetic energy)
Change in kinetic energy of the system = 143463.84 + 96251.39 - 229673.535 = 10041.695 J = 10041.7 J
Answer:
Change in K.E=10040.583 J
Explanation:
Given Data:
Initial Velocity=[tex]V_i= 55.9 m/s[/tex]
[tex]m_1=89.3 kg\\m_2=57.7 kg[/tex]
1st skydiver velocities:
[tex]V_{1,x} = 7.43 m/s,\ \ \ \ V_{1,y} = 5.75 m/s \ \ \ V_{1,z} = 55.9 m/s[/tex]
Now, we will calculate velocities for 2nd diver:
Along x-axis:
[tex]m_1v_{1x}+m_2v_{2x}=0\\v_{2x}=-\frac{m_1*v_{1x}}{m_2}\\ v_{2x}=-\frac{89.3*7.43}{57.7}\\v_{2x}=-11.499\ m/s[/tex]
Along y-axis:
[tex]m_1v_{1y}+m_2v_{2y}=0\\v_{2y}=-\frac{m_1*v_{1y}}{m_2}\\ v_{2y}=-\frac{89.3*5.75}{57.7}\\v_{2y}=-8.899\ m/s[/tex]
Initial Kinetic Energy:
[tex]K.E=\frac{1}{2}(m_1+m_2)(v_i)^2\\ K.E=\frac{1}{2}(89.3+57.7)(55.9)^2\\\\K.E=229673.535 J[/tex]
Final Kinetic Energy:
[tex]K.E=\frac{1}{2}(m_1)(v_1)^2+K.E=\frac{1}{2}(m_2)(v_2)^2\\K.E=\frac{1}{2}(89.3)[(7.43)^2+(5.75)^2+ (55.9)^2]+frac{1}{2}(57.7)[(-11.499)^2+(-8.899)^2+ (55.9)^2]\\\\K.E=143463.9056+96250.2124\\K.E=239714.118 J[/tex]
Change in K.E= Final Kinetic Energy - Initial Kinetic Energy
Change in K.E= 239714.118-229673.535
Change in K.E=10040.583 J
