# Video: Finding the Change in Momentum of a Body of a Variable Mass given Its Mass and Displacement as Functions of Time

A body with variable mass moves in a straight line. At time 𝑡 seconds, its mass is given by 𝑚 = 7(𝑡 + 4) kg, and its displacement from a fixed point on the line is given by 𝑠 = (1/4)(𝑡² − 2𝑡 + 10) m. Calculate the change in the body’s momentum between 𝑡 = 10 s and 𝑡 = 13 s.

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### Video Transcript

A body with variable mass moves in a straight line. At time 𝑡 seconds, its mass is given by 𝑚 equals seven times the quantity 𝑡 plus four kilograms. And its displacement from a fixed point on the line is given by 𝑠 equals one-quarter times the quantity 𝑡 squared minus two 𝑡 plus 10 meters. Calculate the change in the body’s momentum between 𝑡 equals 10 seconds and 𝑡 equals 13 seconds.

We’ll call the time values at either end of our interval of interest, 𝑡 equals 10 seconds and 𝑡 equals 13 seconds, 𝑡 sub 𝑖 and 𝑡 sub 𝑓, respectively. We want to solve for the change in the body’s momentum over this time interval. We’ll call that change Δ𝐻. To help us solve it, we’re given the mass of the object as a function of time as well as its displacement, 𝑠, also as a function of time.

Let’s start our solution by recalling the mathematical equation for momentum. The momentum of an object with mass is equal to the object’s mass times its velocity 𝑣. In our case, we want to solve for a change in momentum, Δ𝐻. Based on our expression for momentum, we can write this as the magnitude of the difference between the final mass times the final velocity and the initial mass times the initial velocity. We’ll need velocities in order to calculate Δ𝐻. But we’re given displacement 𝑠 in our problem statement. The two quantities are related through the expression: 𝑣 is equal to the time derivative of displacement 𝑠.

In order to eventually solve for 𝑣 sub 𝑖 and 𝑣 sub 𝑓, let’s first find out what 𝑣 as a function of 𝑡 is. We’ll do this by taking the time derivative of 𝑠. Plugging in for our displacement 𝑠, we calculate the time derivative to be one-quarter times the quantity two 𝑡 minus two meters per second. Factoring out a two from inside the parentheses, we now have our velocity as a function of time. We recall that we’re specifically interested in our particle’s velocity at 𝑡 equals 𝑡 sub 𝑖 and 𝑡 equals 𝑡 sub 𝑓. And likewise, we’re interested in our particle’s mass at those two times.

We can write 𝑚 sub 𝑖 times 𝑣 sub 𝑖 as seven times 10 plus four kilograms multiplied by one-half times 10 minus one meters per second. This is equal to 49 times 9 kilograms meters per second. So that’s 𝑚 sub 𝑖 times 𝑣 sub 𝑖.

Moving on to 𝑚 sub 𝑓 𝑣 sub 𝑓, that’s equal to seven times 13 plus four kilograms multiplied by one-half times 13 minus one meters per second. This is equal to 42 times 17 kilograms meters per second.

With this plugged in to our expression, we’re now ready to calculate Δ𝐻, the change in momentum. Entering these values on our calculator, we find it’s 273 kilograms meters per second. That’s the change in momentum of this object over the given time interval.