Chapter 7, Problem 10P

To determine

The speed of the car after $90\min$

Answer to Problem 10P

  $7.94\frac{\text{m}}{\text{s}}$

Explanation of Solution

Given:

Mass of the railroad car $=M=3800\text{kg}$

Initial total mass of the system of car and snow $=m_i=M=3800\text{kg}$

Initial velocity of the railroad car before snow fills in it $=v_i=8.60\frac{\text{m}}{\text{s}}$

Final velocity of the combination of railroad car and snow in it $=v_f$

Time for which snow falls $=t=90\text{min}$

Rate at which snow falls in the car

  $=R=3.50\frac{\text{kg}}{\text{min}}$

Mass of snow in the car $=m$

Final total mass of the system of car and snow $=m_f=M+m=3800+m$

Formula Used:

Momentum of an object is given as

  $\begin{array}{l}p=mv\\ \text{where }p=\text{momentum}\\ m=\text{mass}\\ v=\text{velocity}\end{array}$

According conservation of momentum principle, we have

  $\begin{array}{l}\text{Final momentum = initial momentum}\\ p_f=p_i\\ \text{where }{p}_f=\text{Final momentum of system }\\ p_i=\text{Initial momentum of system}\end{array}$

Calculation:

Mass of the snow in the car can be given as

  $\begin{array}{l}m=Rt\\ m=(3.50)(90)\\ m=315\text{kg}\end{array}$

Using conservation of momentum

  $\begin{array}{l}{p}_f=p_i\\ m_f{v}_f=m_i{v}_i\\ (3800+m)v_f=(3800)(8.60)\\ (3800+315)v_f=(3800)(8.60)\\ v_f=\frac{32680}{4115}\\ v_f=7.94\frac{\text{m}}{\text{s}}\end{array}$

Conclusion:

Therefore, speed of car comes out to be

  $7.94\frac{\text{m}}{\text{s}}$