An open carriage in a goods train is moving with a uniform...

An open carriage in a goods train is moving with a uniform velocity of $10 \mathrm{~ms}^{-1}$. If the rain adds water with zero velocity at the rate of $5 \mathrm{kgs}^{-1}$, then the additional force applied by the engine to maintain the same velocity of the train is (a) $0.5 \mathrm{~N}$ (b) $2.0 \mathrm{~N}$ (c) $50 \mathrm{~N}$ (d) $25 \mathrm{~N}$

An open carriage in a goods train is moving with a uniform velocity of $10 \mathrm{~ms}^{-1}$. If the rain adds water with zero velocity at the rate of $5 \mathrm{kgs}^{-1}$, then the additional force applied by the engine to maintain the same velocity of the train is
(a) $0.5 \mathrm{~N}$
(b) $2.0 \mathrm{~N}$
(c) $50 \mathrm{~N}$
(d) $25 \mathrm{~N}$

Key Concepts

Newton's Second Law of Motion

This law states that the net force acting on an object is equal to the rate of change of its momentum. In cases where mass varies with time, such as when additional mass is added to a moving system, the force required to maintain a constant velocity must account for the momentum imparted to the newly added mass.

Momentum Conservation

Momentum, defined as the product of mass and velocity, must be conserved in the absence of external forces. When mass is added to a moving object at a different initial velocity, the system must undergo a change in momentum. This principle helps in determining the additional force required to bring incoming mass to the velocity of the moving object.

Variable Mass Systems

In systems where the mass changes over time, standard applications of Newton's second law require modification to include the effects of mass addition or loss. The analysis involves calculating the force needed to accelerate the added mass from its initial state to the velocity of the system, thereby ensuring the overall momentum of the system adjusts in accordance with the desired motion.

Transcript

00:01 In this question there is a train which travels on a track.

00:10 Now we know that the track exerts a force on the train, force of friction, which is equal to 2 times 10 to the power of 5 newton.

00:26 And this is this force to the left here.

00:30 We also know that the time it takes for the train to stop is 50 seconds.

00:40 Now we need to calculate in the first part of the question, the average force that would be needed to stop the train in 25 seconds.

00:51 So half this time.

00:53 Now to do this, we first calculate the change in momentum, the impulse.

01:00 We know that the change in momentum is defined as the force times the time interval delta t.

01:11 So we can calculate it by simply timing the force, 2 times 10 to the power of 5 by the time interval 15.

01:25 And this is equal to 1 times 10 to the power of 7 kilograms.

01:33 Meters per second and we know that this stays the same so once we have the change in momentum we can use it to then calculate what the force would be if the time interval was now halved so we have 25 seconds now and not 50 anymore we know that the change in momentum is one times 10 to the power of 7 kilograms meters per second.

02:10 So we can rearrange this equation to calculate the force.

02:17 And the force is given by changing momentum divided by time interval.

02:26 So we put the numbers in, we have 1 times 10 to the power of 7 divided by the new time interval, which is 25...

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