Question Video: Finding the Momentum of a Body at a Given Time given an Expression of Its Position with Time | Nagwa Question Video: Finding the Momentum of a Body at a Given Time given an Expression of Its Position with Time | Nagwa

Question Video: Finding the Momentum of a Body at a Given Time given an Expression of Its Position with Time Mathematics • Third Year of Secondary School

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A body of mass 7 kg is moving in a straight line. Its position vector at a time π‘ is given by the relation π(π‘) = (π‘Β² + 5)π + (π‘Β³ + π‘)π, where βπβ is measured in metres and π‘ in seconds. Determine the bodyβs momentum after 2 seconds.

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Video Transcript

A body of mass seven kilograms is moving in a straight line. Its position vector at a time π‘ is given by the relation π as a function of π‘ equals π‘ squared plus five quantity in the π direction plus π‘ cubed plus π‘ quantity in the π direction, where the magnitude of π is measured in metres and π‘ in seconds. Determine the bodyβs momentum after two seconds.

We can record the bodyβs mass of seven kilograms as π. The time value weβre interested in β two seconds β weβll call π‘ sub two. Weβre given the position of our object as a function of time β π as a function of π‘ β and based on that want to solve for its momentum, which weβll represent by the letter capital π».

To begin on our solution, letβs recall the mathematical relationship describing momentum. The momentum of an object with mass is equal to that objectβs mass times its velocity. We know that an objectβs velocity is defined as its change in position over time β ππ ππ‘.

Combining these relationships, we can write that the momentum we want to solve for is equal to the objectβs mass multiplied by its changing position over time. First, letβs solve for that time derivative of position. The time derivative of π as a function of π‘ is equal to π ππ‘ of π‘ squared plus five in the π direction and π‘ cubed plus π‘ in the π direction.

Using the chain rule to differentiate, this is equal to two π‘ in the π direction plus three π‘ squared plus one in the π direction. We can rewrite our expression for π» now to include ππ ππ‘. But we want to solve for π» at a particular value of π‘ when π‘ is equal to what weβve called π‘ sub two or two seconds. So we plug in π‘ sub two for our π‘ values on our equation.

We now have an expression for momentum given entirely in terms of constants or variables whose value we know π and π‘ sub two. Weβre ready to plug in and solve for π». When we do and simplify our expression for our velocity, we find that at this time value itβs equal to four π plus 13 π. Multiplying through by our mass, we find that π» is 28 π plus 91 π. This is the momentum of our mass at π‘ equals two seconds.

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