Answer:
a) v = 10.3 m/s at 17.1° south of west
b)ΔE = 147,964.5 J
Explanation:
Let us consider west as positive x-axis and the south as positive y-axis
Given
Mass of the first car [tex]m_{1}[/tex] = 1150 kg
Mass of the second car [tex]m_{2}[/tex] =750 kg
Speed of the first car [tex]v_{1}[/tex] = 5.00 m/s in y axis
Speed of the second car [tex]v_{2}[/tex] = 25.0 m/s in x axis
Solution
Since there are no external forces the momentum is conserved
in X-axis
[tex]m_{1}\times0+m_{2} u_{2} =(m_{1}+m_{2})v_{x} \\v_{x} =\frac{m_{2} u_{2}}{m_{1}+m_{2}} \\v_{x} =\frac{750\times25}{1150+750}\\v_{x} =\frac{18750}{1900}\\v_{x} = 9.87 m/s[/tex]
in Y-Axis
[tex]m_{1}u_{1}+m_{2} \times0} =(m_{1}+m_{2})v_{y} \\v_{y} =\frac{m_{1} u_{1}}{m_{1}+m_{2}} \\v_{y} =\frac{1150\times5}{1150+750}\\v_{y} =\frac{5750}{1900}\\v_{y} =3.03 m/s[/tex]
Final velocity
[tex]v = \sqrt{v_{x} ^{2} +v_{y} ^{2} } \\v = \sqrt{9.87 ^{2} +3.03 ^{2} } \\v = 10.3 m/s[/tex]
Direction
[tex]\theta = acrtan(\frac{v_{y} }{v_{x} } )\\\theta = acrtan(\frac{3.03 }{9.87 } )\\\theta = 17.1^{o}[/tex]
Energy Lost = Initial kinetic energy - final kinetic energy
[tex]\Delta E = [\frac{1}{2}m_{1}u_{1}^2 +\frac{1}{2}m_{2}u_{2}^2 ] - \frac{1}{2}(m_{1}+m_{2})v^2\\\Delta E = [\frac{1}{2}\times1150\times5}^2 +\frac{1}{2}\times750 \times 25^2 ] - \frac{1}{2}(1150+750})10.3^2\\\Delta E = 147,964.5 J[/tex]
The type of collision experienced by the two cars is a perfectly inelastic
collision.
(a) The final velocity of the two cars, is approximately 10.32 m/s.
(b) The kinetic energy lost in the collision is approximately 147,532.89 J.
Reasons:
Mass of the first car, m₁ = 1150 kg
The mass of the second, m₂ = 750 kg
Velocity of the first car, v₁ = -5·j m/s
Velocity of the second car, v₂ = 25·i m/s
By conservation of linear momentum, we have;
1150 × -5.j + 750 × 25·i = (1150 + 750)×v
[tex]v = \dfrac{1150 \times (-5)\cdot \mathbf {i} + 750 \times 25\cdot \mathbf {j} }{1150 + 750} = -\dfrac{115}{38} \cdot \mathbf {i} +\dfrac{375}{38} \cdot \mathbf {j}[/tex]
[tex]v = \underline{ -\dfrac{115}{38} \cdot \mathbf {i} +\dfrac{375}{38} \cdot \mathbf {j}}[/tex]
Magnitude of the velocity is therefore;
[tex]|v| = \sqrt{\left (-\dfrac{115}{38} \right)^2 +\left (\dfrac{375}{38} \right)^2 } = \dfrac{5 \cdot \sqrt{6154} }{38}[/tex]
[tex]\mathrm{The \ final \ velocity \ of \ the \ two \ cars}, v = \dfrac{5 \cdot \sqrt{6154} }{38} \ m/s \approx \underline{ 10.32 \ m/s}[/tex]
(b) The total initial kinetic energy, K.E.₁ = 0.5·m₁·v₁² + 0.5·m₂·v₂²
∴ K.E.₁ = 0.5×1150 ×5² + 0.5 × 750 × 25² = 248,750
The initial kinetic energy, K.E.₁ = 248,750 J
The final kinetic energy, K.E.₂ = 0.5·(m₁ + m₂)·v²
[tex]K.E._2 = \dfrac{1}{2} \times (1150 + 750) \times \left(\dfrac{5 \cdot \sqrt{6154} }{38} \right)^2 = 101,217.11 \ m^2[/tex]
Kinetic energy lost in the collision (energy of deformation), E = -ΔK.E.
Change in kinetic energy, ΔK.E. = 101,217.11 J - 248,750 J = -147,532.89 J
Kinetic energy lost, E = 147,532.89 J
Learn more here:
https://brainly.com/question/12904920
