Question Video: Finding the Impulse and the Average Force Exerted by a Barrier at Its Collision with a Moving Body Mathematics

Video Transcript

A railway carriage of mass 23 tons was moving at 14 meters per second before it crashed into a barrier. Given that it took four seconds for the carriage to come to rest, determine the magnitude of the impulse 𝐼 and the average force 𝐹 to the nearest kilogram-weight. Consider the acceleration due to gravity to be 9.8 meters per second squared.

So we’re thinking about a railway carriage that’s initially moving at 14 meters per second. We’re told that this carriage crashes into a barrier, and this crash causes the carriage to come to rest over the course of four seconds. We’re being asked to calculate the magnitude of the impulse 𝐼 and the average force 𝐹 that act on the carriage during this time. In order to answer this question, we can start by recalling that the impulse 𝐼 experienced by an object is equal to that object’s mass 𝑚 multiplied by its change in velocity Δ𝑉.

In this question, we’re told that the mass of our carriage is 23 tons. And we can calculate the change in velocity Δ𝑉 by subtracting the initial velocity of the carriage from its final velocity. We know that the initial velocity of the carriage is 14 meters per second. And since it came to a rest when it crashed into the barrier, we know that its final velocity is zero meters per second. Now, at this point, it’s important to remember that velocity is a vector quantity. So generally speaking, we should decide which direction is the positive direction and which direction is the negative direction.

For this question, let’s say that vectors acting to the right are positive and vectors acting to the left are negative. Using this convention, we would say that the initial velocity of the carriage is positive 14 meters per second, since the velocity vector points in the positive direction. This means that when we calculate the change in velocity Δ𝑉, the final velocity minus the initial velocity is zero minus 14, meaning Δ𝑉 is equal to negative 14 meters per second. So now, to find the impulse that acts on our carriage, we simply need to multiply this value of Δ𝑉 by the mass of the carriage.

Now, we’re told in the question that the mass of the carriage is 23 tons. So let’s convert this into the standard units for mass kilograms. To do this, we can recall that one ton is equal to 1000 kilograms. So we can see that to convert from some number of tons to the equivalent number of kilograms, we would multiply that number by 1000. This means that 23 tons is equal to 23000 kilograms. So 𝑚 times Δ𝑉 is equal to 23000 kilograms multiplied by negative 14 meters per second, which gives us negative 322000 kilogram meters per second. This is the impulse that acted on the carriage during the collision. And we can also think of this as being the change in the carriage’s momentum.

Now the fact that this value is negative just means that the impulse vector that acted on the carriage was pointing in the negative direction. We can actually ignore this negative sign for two reasons. Firstly, the negative direction is something that we made up rather than something that was actually defined in the question. And secondly, the question asks us to find the magnitude of the impulse. Either way, we just take the positive version of this value. We can also write this a bit more concisely by using standard notation. 322000 is equal to 3.22 times 10 to the power of five. So this is our final answer to the first part of the question. The magnitude of the impulse that acts on the carriage is 3.22 times 10 to the power of 5 kilogram meters per second. We can also think of this as the change in the carriage’s momentum.

Okay, so the second thing we need to do is find the average force that acts on the carriage. To do this, we can recall that an alternative way of calculating the impulse that acts on an object is to multiply the average force 𝐹 acting on that object by the amount of time Δ𝑡 for which that force act. And since we’re looking for the average force that acts on the object, we want to rearrange this equation to make 𝐹 the subject. To do this, we just need to divide both sides of the equation by Δ𝑡 giving us 𝐹 equals 𝐼 over Δ𝑡. And we’ve calculated that the impulse that’s acting on the carriage is 3.22 times 10 to the power of 5 kilogram meters per second.

Δ𝑡 is the amount of time for which the force acts on the object. And in this case, we’re told that the collision took four seconds, which means that Δ𝑡 is four seconds. Dividing 3.22 times 10 to the power of five by four gives us 80500. Now, since we expressed impulse in the standard units of kilogram meters per second and we expressed time Δ𝑡 in the standard units of seconds, this means that the force we calculated is expressed in the standard units for force, newtons.

However, the question asked us to express our answer to the nearest kilogram-weight. Now to convert from newtons to kilogram-weight, we can recall that one kilogram-weight is the weight of a one-kilogram mass on the surface of Earth. Now, since weight is equal to mass times gravitational acceleration 𝑔, we know that the weight of a one-kilogram mass is one times 𝑔. In the question, we’re told to use a value for 𝑔 of 9.8 meters per second squared. So the weight of the one-kilogram object is one times 9.8. And since weight is a force, we would express this quantity as 9.8 newtons. In other words, one kilogram-weight is equal to 9.8 newtons.

So in order to convert our average force of 80500 newtons into kilograms-weight, we just need to divide this value by 9.8, which gives us a value of 8214.29, and so on, kilograms-weight. And rounding this to the nearest kilogram-weight, we have 8214. So we’ve now answered all parts of the question. If a railway carriage of mass 23 tons moving at 14 meters per second comes to rest over a period of four seconds, the magnitude of the impulse acting on that carriage is 3.22 times 10 to the power of five kilogram meters per second. And the average force acting on the carriage is 8214 kilograms-weight.