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Chapter 7: Q70P (page 170)

A 4800-kg open railroad car coasts at a constant speed of 7.60 m/s on a level track. Snow begins to fall vertically and fills the car at a rate of 3.80 kg/min. Ignoring friction with the tracks, what is the car鈥檚 speed after 60.0 min? (See Section 7鈥�2.)

Short Answer

Expert verified

The speed of the car is \(v' = 7.25\,{\rm{m/s}}\).

Step by step solution

01

Understanding conservation of momentum

In the horizontal direction of the railroad car, the momentum will be conserved. Also, the final speed of the car is equal to the final speed of the snow.

02

Given data

Given data

The mass of the car is\(m = 4800\;{\rm{kg}}\).

The constant speed of the car is\(v = 7.60\;{\rm{m/s}}\).

The rate of fall is\(c = 3.80\;{\rm{kg/min}}\).

The time is\(t = 60\;{\rm{min}}\).

03

Calculate the speed of the car

The relation from conservation of momentum is given by the following:

\(\begin{array}{c}P = P'\\mv = \left( {m + m'} \right)v'\\v' = \frac{{mv}}{{\left( {m + m'} \right)}}\end{array}\)

Here,\(m'\)is the mass of the snow, and\(v'\)is the speed of the car.

Plugging the values in the above equation,

\(\begin{array}{l}v' = \left[ {\frac{{\left( {4800\;{\rm{kg}}} \right)\left( {7.60\;{\rm{m/s}}} \right)}}{{\left( {4800\;{\rm{kg}} + 3.80\;{\rm{kg/min}} \times {\rm{60}}\,{\rm{min}}} \right)}}} \right]\\v' = 7.25\,{\rm{m/s}}\end{array}\)

Thus, \(v' = 7.25\,{\rm{m/s}}\) is the required number of rebounds made by the ball.

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