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3 years ago

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A 1200-kg station wagon is moving along a straight highway at Another car, with mass 1800 kg and speed has its center of mass 40

.0 m ahead of the center of mass of the station wagon (Fig. E8.54). (a) Find the position of the center of mass of the system consisting

1 answer:

taurus [48]3 years ago

5 0

Answer:

Center of mass lies 24 m in front of center of mass of second wagon.

Explanation:

Suppose A 1200 kg station wagon is moving along a straight highway at 12.0 m/s. Another car with mass 1800 kg and speed 20.0 m/s.

Given that,

Mass of first wagon = 1200 kg

Mass of second wagon = 180 kg

Distance = 40 m

We need to calculate the position of the center of mass of the system

Using formula of center mass

$x_{cm}=\dfrac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$

$x_{cm}=\dfrac{1200\times0+1800\times40}{1200+1800}$

$x_{cm}=24\ m$

Hence, Center of mass lies 24 m in front of center of mass of second wagon.

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Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

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Answer:

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Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

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and after the collision we have that

$p_{xf} =9m\ v_{1x}$

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<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

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We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

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The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

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$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

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