Question Video: Finding the Change in Momentum of a Body Given its Mass and Acceleration as a Function of Time | Nagwa Question Video: Finding the Change in Momentum of a Body Given its Mass and Acceleration as a Function of Time | Nagwa

Question Video: Finding the Change in Momentum of a Body Given its Mass and Acceleration as a Function of Time Mathematics • Third Year of Secondary School

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A body of mass 5 kg moves along a straight line. At time 𝑑 seconds, its acceleration is given by π‘Ž = (βˆ’6𝑑 βˆ’ 8) m/sΒ². Find the change in its momentum in the time interval 6 ≀ 𝑑 ≀ 9.

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Video Transcript

A body of mass 5 kilograms moves along a straight line. At time 𝑑 seconds, its acceleration is given by π‘Ž is equal to negative six 𝑑 minus eight meters per second squared. Find the change in its momentum in the time interval 𝑑 is greater than or equal to six and less than or equal to nine.

In this question, we’re asked to find the change in momentum. And we know that momentum 𝐻 is equal to the mass π‘š multiplied by the velocity 𝑣. We are told in the question that the body has a mass of 5 kilograms. However, we are not given any information about its velocity. We are told, however, that the acceleration of the body is equal to negative six 𝑑 minus eight meters per second squared. We know that we can calculate an expression for a body’s velocity by integrating the expression for acceleration with respect to 𝑑. In this question, 𝑣 is equal to the integral of negative six 𝑑 minus eight d𝑑.

We are interested in the momentum and therefore the velocity in the time period from six to nine seconds. We can therefore add these values as our lower and upper limits of our integral. Recalling that the integral of π‘Ž multiplied by π‘₯ to the power of 𝑛 with respect to π‘₯ is equal to π‘Ž multiplied by π‘₯ to the power of 𝑛 plus one all divided by 𝑛 plus one, we can integrate term by term. Integrating negative six 𝑑 gives us negative six 𝑑 squared over two, which simplifies to negative three 𝑑 squared. Integrating the constant negative eight gives us negative 8𝑑.

Our next step is to substitute our upper and lower limits. When 𝑑 is equal to nine, 𝑣 is equal to negative 315. And when 𝑑 equals six, 𝑣 is equal to negative 156. The velocity is therefore equal to negative 315 minus negative 156, which is equal to negative 159. Our units are meters per second. And we now have a value for the mass and velocity that we can use to calculate the change in momentum. 𝐻 is equal to five multiplied by negative 159. This is equal to negative 795. The change in momentum of the body in the time interval 𝑑 is greater than or equal to six and less than or equal to nine is negative 795 kilogram-meters per second.

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