# Question Video: Calculating the Length of Time for Which a Force Is Applied Physics • 9th Grade

A force of 8 N is applied to an object and the object’s momentum increases by 2 kg ⋅ m/s. How long is the force applied for?

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### Video Transcript

A force of eight newtons is applied to an object and the object’s momentum increases by two kilogram meters per second. How long is the force applied for?

So in this question, we have some object and a force of magnitude eight newtons is applied to that object. We’ll label this force as 𝐹 so that we have 𝐹 is equal to eight newtons. We are then told that as a result of this force, the object’s momentum increases by two kilogram meters per second. We’ll label this change in momentum as Δ𝑝, where the Δ means that we’re measuring the change in the quantity 𝑝, the momentum of the object.

So we have that Δ𝑝 is equal to two kilogram meters per second, where this value is positive since we’re told that the momentum of the object increases. The question is asking us to work out how long this force is applied for. We’ll label this length of time as Δ𝑡. And it represents the amount of time that passes or the change in the value of time between when the force starts to act and when the force finishes acting.

To answer this question, we can recall that Newton’s second law tells us that the force on an object is equal to the rate of change of the object’s momentum. In other words, the force acting on an object is equal to the change in momentum of that object divided by the time taken for that change to occur. In our situation, we know the value of the force 𝐹 and we know the change in momentum Δ 𝑝. We are looking to find the value of Δ𝑡. So let’s take this equation and rearrange it to make Δ𝑡 the subject.

If we multiply both sides of the equation by Δ𝑡, then on the right-hand side, the Δ𝑡’s in the numerator and denominator cancel each other out. Then, if we divide both sides by 𝐹, we can cancel the 𝐹s in the numerator and denominator on the left-hand side. This gives us an equation that says that Δ𝑡, the length of time for which a force acts on an object, is equal to Δ𝑝, the change in momentum of that object, divided by 𝐹, the value of the force. Now all that we need to do is take our values of 𝐹 and Δ𝑝 and substitute them into this equation to calculate our value of Δ𝑡.

Substituting in these values gives us that Δ𝑡 is equal to our momentum change of two kilogram meters per second divided by the force of eight newtons. Let’s take a quick moment to check how the units work out. On the right-hand side of this equation, we have a quantity in the numerator with units of kilogram meters per second, while in the denominator we have a quantity with units of newtons.

We can recall that Newton’s second law can also be written as 𝐹 equals 𝑚𝑎. In other words, the force 𝐹 acting on an object is equal to the mass 𝑚 of that object multiplied by the acceleration 𝑎 experienced by that object. Now, a force is usually expressed in units of newtons, but since force is equal to mass times acceleration and mass has units of kilograms while acceleration has units of meters per second squared, then we can see that units of newtons must be equal to units of kilogram meters per second squared.

So if we look back at our equation for Δ𝑡, we can see that our force of eight newtons could also be written as eight kilogram meters per second squared. Then for the units on the right-hand side of the equation, we have kilogram meters per second divided by kilogram meters per second squared, which leaves us with units of seconds. This, of course, makes sense since on the left-hand side of the equation, we have a value of time. Evaluating this right-hand side gives a result of 0.25 seconds. And so our answer to the question of how long is the force applied for is 0.25 seconds.