Respuesta :
Answer:
a) v(f) = -4i - 5j
b) 4.18 m
Explanation:
The equation to be used for this question is
v(c)m(c) + v(t)m(t) = [m(c) + m(t)] v(f)
if we rearrange and make v(f) subject of formula, then
v(f) = v(c)m(c) + v(t)m(t) / [m(c) + m(t)]
One vehicle is headed towards south and the other vehicle, west when they collide they will travel together in a southwestern direction. This means that both vehicles are traveling in the negative direction taking a standard frame of reference. Thus, we can write the equation in component form by substituting the values as
v(f) = 1225(-9.5i) + 1654(-8.6j) / 1225 + 1654
v(f) = -11637.5i - 14224.4j / 2879
v(f) = -4i - 5j m/s
From the answer,
v(f) = √(4² + 5²)
v(f) = √41
v(f) = 6.4 m/s
And we know that
KE = ½mv²
Fd = umgd
And, KE = Fd, so
½mv² = umgd
½v² = ugd
Making d the subject of formula,
d = v²/2ug
d = 6.4² / 2 * 0.5 * 9.8
d = 41 / 9.8
d = 4.18 m
(a) The velocity of the system after collision is 4.04 i + 4.9 j.
(b)The distance traveled by the vehicles after collision is 1.73 m.
The given parameters;
- mass of the car, Mc = 1225 kg
- velocity of the car, Vc = 9.5 m/s
- mass of the truck, Mt = 1654 kg
- velocity of the truck, Vt = 8.6 m/s
Apply the principle of conservation of linear momentum to determine the velocity of the system after collision;
[tex]m_1u_x_1 + m_2 u_y_2 = V(m_1 + m_2)\\\\V = \frac{(1225\times 9.5)_x \ + \ (1654 \times 8.6)_y}{m_1 + m_2} \\\\V = \frac{(1225\times 9.5)_x \ + \ (1654 \times 8.6)_y}{1225+ 1654} \\\\V= \frac{(11,637.5)_x \ + \ (14,224.4)_y}{2879} \\\\V = 4.04x \ + 4.94y\\\\V = 4.04i \ + 4.9 j[/tex]
The magnitude of the final velocity of the system is calculated as;
[tex]V = \sqrt{v_x^2 + v_y^2} \\\\V = \sqrt{(4.04)^2 + (4.9)^2} \\\\V = 6.35 \ m/s[/tex]
The change in the mechanical energy of the system;
[tex]\Delta K.E = K.E_f - K.E_i\\\\[/tex]
The initial kinetic energy of the cars before collision is calculated as;
[tex]K.E_i = \frac{1}{2} m_1u_1_x^2 \ + \frac{1}{2} m_1u_2_y^2 \\\\K.E_i = \frac{1}{2} (1225)(9.5)^2\ + \frac{1}{2} (1654)(8.6)^2\\\\K.E_i = 55,278.13_x \ + \ 61,164.92_y\\\\K.E_i = \sqrt{55,278.13^2 \ + \ 61,164.92^2} \\\\K.E_i = \sqrt{6,796,819,094.9} \\\\K.E_i = 82,442.82 \ J[/tex]
The final kinetic energy of the system;
[tex]K.E_f = \frac{1}{2} (m_1 + m_2)V^2\\\\K.E_f = \frac{1}{2} (1225 + 1654)(6.35)^2\\\\K.E_f = 58,044.24 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f -K.E_i\\\\\Delta K.E= (58,044.24) - (82,442.82)\\\\\Delta K.E = -24,398.58 \ J[/tex]
Apply the principle of work-energy theorem, to determine the distance traveled by the vehicles after collision;
[tex]W = \Delta K.E\\\\- \mu Fd = - 24,398.58\\\\\mu mgd= 24,398.58\\\\d = \frac{24,398.58}{\mu mg} \\\\d = \frac{24,398.58}{0.5 \times 9.8(1225 + 1654)} \\\\d = 1.73 \ m[/tex]
Thus, the distance traveled by the vehicles after collision is 1.73 m.
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