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Chapter 5: Problem 30

A trolley containing water has total mass \(4000 \mathrm{~kg}\) and is moving at a speed of \(40 \mathrm{~m} / \mathrm{s}\). Now water starts coming out of a hole at the bottom of the trolley at the rate of \(8 \mathrm{~kg} / \mathrm{s}\). Speed of trolley after \(50 \mathrm{~s}\) is (A) \(44.44 \mathrm{~m} / \mathrm{s}\) (B) \(40 \mathrm{~m} / \mathrm{s}\) (C) \(44 \mathrm{~m} / \mathrm{s}\) (D) \(54.44 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The final speed of the trolley after 50 seconds is \(44.44 \mathrm{~m}/\mathrm{s}\). So, the correct answer is (A) \(44.44 \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Find the initial momentum of the system

First, we need to find the initial momentum of the system, which is equal to the mass of the trolley times its initial speed. The total mass of the trolley with the water is given as 4000 kg and the speed is given as 40 m/s. The initial momentum (P_initial) can be calculated using the formula: \( P_{initial} = m_{initial} \cdot v_{initial} \)
02

Calculate the mass ejected and the remaining mass of the trolley

The water is coming out of the hole at the rate of 8 kg/s. After 50 seconds, the amount of water ejected can be calculated as: Water_Ejected = rate of water leaving 脳 time Water_Ejected = 8 kg/s 脳 50 s = 400 kg Now, we can find the remaining mass of the trolley by subtracting the mass of ejected water from the initial total mass: Remaining_Mass = Initial total mass - Ejected mass Remaining_Mass = 4000 kg - 400 kg = 3600 kg
03

Apply the conservation of linear momentum

We know that the linear momentum is conserved in the system, so the final momentum will be equal to the initial momentum: \( P_{final} = P_{initial} \) As the final momentum is the product of the remaining mass and final speed (v_final), we can write the equation as: \( m_{final} \cdot v_{final} = m_{initial} \cdot v_{initial} \)
04

Solve for the final speed

Now, we can plug in the values to solve for the final speed of the trolley: \(3600 \mathrm{~kg} \cdot v_{final} = 4000 \mathrm{~kg} \cdot 40 \mathrm{~m/s} \) \(v_{final} = \frac{4000 \mathrm{~kg} \cdot 40 \mathrm{~m/s}}{3600 \mathrm{~kg}} \) \(v_{final} = 44.44 \mathrm{~m/s} \) The final speed of the trolley after 50 seconds is 44.44 m/s. Therefore, the correct answer is (A) \(44.44 \mathrm{~m} / \mathrm{s}\).

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