Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.

Calculate the final velocity of the cars. (Note that since both cars have an initial velocity, you cannot use Equations 7.6a and b. You must look for other simplifying aspects.)

Magnitude (answer in m/s

Direction ° (counterclockwise from west is positive)

BHow much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.)

$m_{1}\times0+m_{2} u_{2} =(m_{1}+m_{2})v_{x} \\v_{x} =\frac{m_{2} u_{2}}{m_{1}+m_{2}} \\v_{x} =\frac{750\times25}{1150+750}\\v_{x} =\frac{18750}{1900}\\v_{x} = 9.87 m/s$

in Y-Axis

$m_{1}u_{1}+m_{2} \times0} =(m_{1}+m_{2})v_{y} \\v_{y} =\frac{m_{1} u_{1}}{m_{1}+m_{2}} \\v_{y} =\frac{1150\times5}{1150+750}\\v_{y} =\frac{5750}{1900}\\v_{y} =3.03 m/s$

Final velocity

$v = \sqrt{v_{x} ^{2} +v_{y} ^{2} } \\v = \sqrt{9.87 ^{2} +3.03 ^{2} } \\v = 10.3 m/s$

Direction

$\theta = acrtan(\frac{v_{y} }{v_{x} } )\\\theta = acrtan(\frac{3.03 }{9.87 } )\\\theta = 17.1^{o}$

Energy Lost = Initial kinetic energy - final kinetic energy

$\Delta E = [\frac{1}{2}m_{1}u_{1}^2 +\frac{1}{2}m_{2}u_{2}^2 ] - \frac{1}{2}(m_{1}+m_{2})v^2\\\Delta E = [\frac{1}{2}\times1150\times5}^2 +\frac{1}{2}\times750 \times 25^2 ] - \frac{1}{2}(1150+750})10.3^2\\\Delta E = 147,964.5 J$

oeerivona oeerivona · Answer 2 · 2021-11-22T13:24:16-05:00

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

$v = \dfrac{1150 \times (-5)\cdot \mathbf {i} + 750 \times 25\cdot \mathbf {j} }{1150 + 750} = -\dfrac{115}{38} \cdot \mathbf {i} +\dfrac{375}{38} \cdot \mathbf {j}$

$v = \underline{ -\dfrac{115}{38} \cdot \mathbf {i} +\dfrac{375}{38} \cdot \mathbf {j}}$

Magnitude of the velocity is therefore;

$|v| = \sqrt{\left (-\dfrac{115}{38} \right)^2 +\left (\dfrac{375}{38} \right)^2 } = \dfrac{5 \cdot \sqrt{6154} }{38}$

$\mathrm{The \ final \ velocity \ of \ the \ two \ cars}, v = \dfrac{5 \cdot \sqrt{6154} }{38} \ m/s \approx \underline{ 10.32 \ m/s}$

(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

$K.E._2 = \dfrac{1}{2} \times (1150 + 750) \times \left(\dfrac{5 \cdot \sqrt{6154} }{38} \right)^2 = 101,217.11 \ m^2$

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