Question Video: Combining Equations Using Rearrangement and Substitution | Nagwa Question Video: Combining Equations Using Rearrangement and Substitution | Nagwa

Question Video: Combining Equations Using Rearrangement and Substitution Physics

The momentum of an object is given by the formula momentum = mass Γ— velocity, expressed in symbols as 𝑝 = π‘šπ‘£. The velocity of an object is given by the formula velocity = displacement/time, expressed in symbols as 𝑣 = 𝑠/𝑑. Which of the following is a correct formula? [A] 𝑝𝑣 = π‘šπ‘  [B] 𝑝𝑣 = π‘šπ‘‘ [C] 𝑝𝑠 = π‘šπ‘‘ [D] 𝑝𝑑 = 𝑣𝑠 [E] 𝑝𝑑 = π‘šπ‘ 

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Video Transcript

The momentum of an object is given by the formula momentum equals mass times velocity, expressed in symbols as 𝑝 equals π‘šπ‘£. The velocity of an object is given by the formula velocity equals displacement over time, expressed in symbols as 𝑣 equals 𝑠 over 𝑑. Which of the following is a correct formula? A) 𝑝𝑣 equals π‘šπ‘ , B) 𝑝𝑣 equals π‘šπ‘‘, C) 𝑝𝑠 equals π‘šπ‘‘, D) 𝑝𝑑 equals 𝑣𝑠, or E) 𝑝𝑑 equals π‘šπ‘ .

To answer this question, we’ll need to look at each answer option in turn and decide whether it’s correct or incorrect. A good method for doing this is to start with the left-hand side from each answer option and try to match the right-hand side of the expression by substituting and rearranging the equations given in the question. This means we’ll be looking for expressions with 𝑝𝑣 on the left-hand side, with 𝑝𝑠 on the left-hand side, and with 𝑝𝑑 on the left-hand side.

It’s important to notice that each of the answer options contains symbols from the equations 𝑝 equals π‘šπ‘£ and 𝑣 equals 𝑠 over 𝑑. This means we’ll need to combine these two equations together in order to find the expressions that we’re looking for. So, first, let’s find an expression where the left-hand side is 𝑝 times 𝑣.

Let’s start just by writing down the left-hand side of the equation that we want. Then, let’s think about how we can use the two equations we’ve been given to come up with a right-hand side for our formula. Equation number one tells us that 𝑝 is equivalent to π‘š times 𝑣. This means that we can replace the 𝑝 in our expression with π‘šπ‘£. So, 𝑝 times 𝑣 becomes π‘šπ‘£ times 𝑣, which we can simplify to π‘šπ‘£ squared. This expression is correct. But we’re not done yet.

Because we want to find out whether one of these expressions is correct, then we’re looking for a formula that not only has 𝑝𝑣 on the left-hand side, but also contains the variables 𝑠 or 𝑑 on the right-hand side. The way we can achieve this is by substituting equation two into the right-hand side of our expression. That way, we’ll express any 𝑣s in terms of 𝑠 and 𝑑. So, we can just replace this 𝑣 with 𝑠 over 𝑑. Notice that we’ve used parentheses here just to keep things clear.

Because 𝑣 is equal to 𝑠 over 𝑑, that means 𝑣 squared is equal to 𝑠 over 𝑑 squared. So, we need to make sure that we square the entire fraction that we’ve substituted in. When squaring any fraction, we square the numerator, and we square the denominator. So, here, that gives us π‘š times 𝑠 squared over 𝑑 squared. Okay, so we’ve now found an expression with 𝑝𝑣 on the left and the desired variables on the right. But if we look at A and B, we’ll see that those expressions only contain 𝑠 or 𝑑 on the right-hand side and not both.

Now, we could attempt to keep substituting this equation into our expression in order to express the right-hand side in terms of just 𝑠 or just 𝑑. But we’d quickly find that we can’t do this without also introducing extra factors of 𝑣, which aren’t present in our answer options. So, we know that neither A nor B are the correct answer.

Next, let’s find a formula with 𝑝𝑠 on the left-hand side. And again, we’ll start just by writing down the left-hand side of the equation that we want. And just like before, we can start by substituting 𝑝 equals π‘šπ‘£ into our expression. So, 𝑝𝑠 becomes π‘šπ‘£π‘ . Notice that this gives us an expression with π‘š on the right-hand side, just like in our answer option. However, if we want to see if this option is correct, then we want to try to express the right-hand side in terms of π‘š and 𝑑.

Once again, substituting equation two into our expression can help us. Substituting 𝑠 over 𝑑 in place of 𝑣 gives us π‘š times 𝑠 over 𝑑 times 𝑠, which we can simplify to give us π‘šπ‘  squared over 𝑑. And we can stop there because we’ve managed to express 𝑝𝑠 in terms of π‘š and 𝑑. But it’s not the same as option C. Then, we know that option C is incorrect as well.

That leaves us with just two possible answers, 𝑝𝑑 equals 𝑣𝑠 and 𝑝𝑑 equals π‘šπ‘ . To determine which of these is correct, we’ll need to come up with an expression that has 𝑝𝑑 on the left-hand side. So, once again, we’ll start by just writing down the left-hand side of the equation that we want. And again, we can use equation one to substitute π‘šπ‘£ in place of 𝑝, which changes 𝑝𝑑 into π‘šπ‘£π‘‘.

Here, the possible answers both have 𝑠 on the right-hand side. So, we want to make a substitution that will give us 𝑠 on the right-hand side of our equation. And once again, we can do that by using equation two, which allows us to replace 𝑣 with 𝑠 over 𝑑. That gives us π‘š times 𝑠 over 𝑑 times 𝑑, which we can simplify to give us π‘šπ‘ π‘‘ divided by 𝑑.

Our final step here is to simplify this fraction. Because there is a factor of 𝑑 in the numerator and in the denominator, they effectively cancel each other out. Leaving us with the expression 𝑝𝑑 equals π‘šπ‘ , which is the same as Option E. So, we know that D is incorrect as well. And the correct answer is option E, 𝑝𝑑 equals π‘šπ‘ .

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