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Chapter 3: Problem 342

A wagon weighing \(1000 \mathrm{~kg}\) is moving with a velocity $50 \mathrm{~km} \mathrm{~h}^{-1}\( on smooth horizontal rails. A mass of \)250 \mathrm{~kg}$ is dropped into it. The velocity with which it moves now is (A) \(2.5 \mathrm{~km} \mathrm{~h}^{-1}\) (B) \(20 \mathrm{~km} \mathrm{~h}^{-1}\) (C) \(40 \mathrm{~km} \mathrm{~h}^{-1}\) (D) \(50 \mathrm{~km} \mathrm{~h}^{-1}\)

Short Answer

Expert verified
The correct answer is (C) \(40\mathrm{~km} \mathrm{~h}^{-1}\).

Step by step solution

01

Calculate the initial momentum of the wagon

To do this, we need to first convert the velocity of the wagon from \(\mathrm{km} \mathrm{~h}^{-1}\) to \(\mathrm{m} \mathrm{~s}^{-1}\). Using the conversion factor \(\frac{1000 \mathrm{~m}}{1 \mathrm{~km}}\) and \(\frac{3600 \mathrm{~s}}{1 \mathrm{~h}}\), we convert the velocity: Initial velocity of the wagon \(=\frac{50 \cdot \frac{1000}{1} \mathrm{~m} \mathrm{~km}^{-1} \cdot \frac{1}{3600} \mathrm{~h} \mathrm{~s}^{-1}}{\mathrm{~km} \mathrm{~h}^{-1}} = 13.89\mathrm{~m}\mathrm{~s}^{-1}\) Now we can find the initial momentum of the wagon: Initial momentum \(=\) mass \(\times\) initial velocity \(= 1000\mathrm{~kg} \times 13.89\mathrm{~m}\mathrm{~s}^{-1} = 13,890\mathrm{~kg}\mathrm{~m}\mathrm{~s}^{-1}\)
02

Calculate the final momentum of the system

When the mass is dropped into the wagon, the combined mass becomes the sum of the initial mass of the wagon and the dropped mass: Final mass \(=\) initial mass \(+\) dropped mass \(=1000\mathrm{~kg}+250\mathrm{~kg} = 1250\mathrm{~kg}\) Since momentum is conserved during this process, the final momentum of the system will be equal to the initial momentum of the wagon. Final momentum \(= 13,890\mathrm{~kg}\mathrm{~m}\mathrm{~s}^{-1}\)
03

Calculate the final velocity of the system

To find the final velocity of the system, we use the following formula: Final velocity \(=\frac{\text{Final Momentum}}{\text{Final Mass}}\) Substituting the values we found earlier: Final velocity \(=\frac{13,890\mathrm{~kg}\mathrm{~m}\mathrm{~s}^{-1}}{1250\mathrm{~kg}} = 11.11\mathrm{~m}\mathrm{~s}^{-1}\) Now convert the final velocity back to \(\mathrm{km} \mathrm{~h}^{-1}\): \(\frac{11.11 \cdot 3600}{1000}\mathrm{~km} \mathrm{~h}^{-1}=40\mathrm{~km} \mathrm{~h}^{-1}\) So the final velocity of the wagon after the mass is dropped is \(40\mathrm{~km} \mathrm{~h}^{-1}\). Therefore, the correct answer is (C) \(40\mathrm{~km} \mathrm{~h}^{-1}\).

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Most popular questions from this chapter

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