Question

In: Physics

The mass of a high speed train is 4.5×105kg, and it is traveling forward at a velocity of 8.3×101m/s.

The mass of a high speed train is 4.5×105kg, and it is traveling forward at a velocity of 8.3×101m/s. Given that momentum equals mass times velocity, determine the values of m and n when the momentum of the train (in kg⋅m/s) is written in scientific notation.​

 

Solutions

Expert Solution

Concepts and reason

The main concepts required to solve this problem are momentum, speed, and mass. Initially, write the equation for the momentum of an object that is moving with velocity. Use this equation and find the momentum of the train and using the scientific notation, finally find the values of \(\mathrm{m}\) and \(\mathrm{n}\).

Fundamentals

Momentum can be defined as the product of the mass and speed of the object. The equation for the momentum of the object is, \(P=m v\)

Here, \(\mathrm{m}\) is the mass and \(\mathrm{v}\) is the speed of the object.

 

Momentum is the product of the mass and speed of an object. Momentum can be applicable only for the object which has mass. Momentum can be denoted by P. The momentum is measured with the unit of \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\). The equation for the momentum for the given train which traveling is, \(P=m v\)

Here, \(\mathrm{m}\) is the mass of the train and \(\mathrm{v}\) is the speed of the train.

The momentum of an object depends on its mass and speed. If the object's speed increases, the momentum will increase, and if the speed of the object decreases, then the momentum will decrease.

 

The equation for the momentum of the train is, \(P=m v\)

Here, \(\mathrm{m}\) is the mass of the train and \(\mathrm{v}\) is the speed of the train. Substitute \(4.5 \times 10^{5} \mathrm{~kg}\) for \(\mathrm{m}\), and \(8.3 \times 10^{1} \mathrm{~m} / \mathrm{s}\) for \(\mathrm{v}\) in above equation.

$$ \begin{array}{l} P=\left(4.5 \times 10^{5} \mathrm{~kg}\right)\left(8.3 \times 10^{1} \mathrm{~m} / \mathrm{s}\right) \\ =37.35 \times 10^{6} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\left(\frac{10^{1} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{10 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}\right) \end{array} $$

\(=3.735 \times 10^{7} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\)

According to the scientific notation, here the value of \(\mathrm{m}\) is 3.735 and the value of \(\mathrm{n}\) is 7 in the final answer of the momentum.

The value of \(\mathrm{m}=3.735\) and the value of \(\mathrm{n}=7\).

Here, the values of \(\mathrm{m}\) and \(\mathrm{n}\) are determined in the above step depending on the final results of the momentum of the train.

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