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A car with mass mc = 1225 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck of mass mt = 1654 kg traveling south at vt = 8.6 m/s fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficient of friction of μk = 0.5.
A) Write an expression for the velocity of the system after the collision, in terms of the variables given in the problem statement and the unit vectors i and j.
B) How far, in meters, will the vehicles slide after the collision?

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

onyebuchinnaji

(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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