Respuesta :
Answer:
Explanation:
mass of car, m1 = 1170 kg
mass of truck, m2 = 9800 kg
initial velocity of car, u1 = 25 m/s
initial velocity of truck, u2 = 20 m/s
final velocity of car, v1 = 18 m/s
Let the final velocity of truck is v2.
(a) Use the conservation of momentum
m1 x u1 + m2 x u2 + m1 x v1 + m2 x v2
1170 x 25 + 9800 x 20 = 1170 x 18 + 9800 x v2
29250 + 196000 = 21060 + 9800 v2
v2 = 20.84 m/s
(b) Total kinetic energy before collision,
Ki = 0.5 x 1170 x 25 x 25 + 0.5 x 9800 x 20 x 20
Ki = 365625 + 1960000 = 2325625 J
Total final kinetic energy
Kf = 0.5 x 1170 x 18 x 18 + 0.5 x 9800 x 20.84 x 20.84
Kf = 189540 + 2128097.44
Kf = 2317637.44 J
Change in energy = 2325625 - 2317637.44 = 7987.56 J
(c) It is due to the heat energy
(a) he velocity of the truck right after the collision is 30.24 m/s.
(b) The change in the mechanical energy of the car-truck system in the collision is 29,698 J.
(c) The change in the mechanical energy is due to friction.
The given parameters;
- mass of the car, m₁ = 1170 kg
- initial velocity of the car, u₁ = 25 m/s
- mass of the truck, m₂ = 800 kg
- initial velocity of the truck, u₂ = 20 m/s
- velocity of the car after collision, v₁ = 18 m/s
The velocity of the truck after the collision is calculated by applying the principle of conservation of linear momentum;
[tex]m_1 u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\1170(25) \ + \ 800(20)= 1170(18) + 800v_2\\\\800v_2 = 24,190\\\\v_2 = \frac{24,190}{800} \\\\v_2 = 30.24 \ m/s[/tex]
The change in the mechanical energy of the car-truck system in the collision is calculated as;
[tex]\Delta K.E = K.E_f - K.E_i\\\\\Delta K.E = (\frac{1}{2}m_1v_1^2 + \frac{1}{2} m_2v_2^2 )-(\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 ) \\\\\Delta K.E = [\frac{1}{2}(1170)(18)^2 + \frac{1}{2} (800)(30.24)^2 ]-[\frac{1}{2}(1170)(25) + \frac{1}{2}(800)(20)^2 ]\\\\\Delta K.E = 29,698 \ J[/tex]
The change in the mechanical energy is due to friction.
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