# Question Video: Combining Equations Using Rearrangement and Substitution Physics • 9th Grade

The momentum of an object is given by the formula momentum = mass Γ velocity, expressed in symbols as π = ππ£. The velocity of an object is given by the formula velocity = displacement/time, expressed in symbols as π£ = π /π‘. Which of the following is a correct formula? [A] ππ£ = ππ  [B] ππ£ = ππ‘ [C] ππ  = ππ‘ [D] ππ‘ = π£π  [E] ππ‘ = ππ

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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 π equals ππ£. The velocity of an object is given by the formula velocity equals displacement over time, expressed in symbols as π£ equals π  over π‘. Which of the following is a correct formula? A) ππ£ equals ππ , B) ππ£ equals ππ‘, C) ππ  equals ππ‘, D) ππ‘ equals π£π , or E) ππ‘ equals ππ .

To answer this question, weβll need to look at each answer option in turn and decide whether itβ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 weβll be looking for expressions with ππ£ on the left-hand side, with ππ  on the left-hand side, and with ππ‘ on the left-hand side.

Itβs important to notice that each of the answer options contains symbols from the equations π equals ππ£ and π£ equals π  over π‘. This means weβll need to combine these two equations together in order to find the expressions that weβre looking for. So, first, letβs find an expression where the left-hand side is π times π£.

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

Because we want to find out whether one of these expressions is correct, then weβre looking for a formula that not only has ππ£ on the left-hand side, but also contains the variables π  or π‘ 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, weβll express any π£s in terms of π  and π‘. So, we can just replace this π£ with π  over π‘. Notice that weβve used parentheses here just to keep things clear.

Because π£ is equal to π  over π‘, that means π£ squared is equal to π  over π‘ squared. So, we need to make sure that we square the entire fraction that weβve substituted in. When squaring any fraction, we square the numerator, and we square the denominator. So, here, that gives us π times π  squared over π‘ squared. Okay, so weβve now found an expression with ππ£ on the left and the desired variables on the right. But if we look at A and B, weβll see that those expressions only contain π  or π‘ 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 π  or just π‘. But weβd quickly find that we canβt do this without also introducing extra factors of π£, which arenβt present in our answer options. So, we know that neither A nor B are the correct answer.

Next, letβs find a formula with ππ  on the left-hand side. And again, weβ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 π equals ππ£ into our expression. So, ππ  becomes ππ£π . Notice that this gives us an expression with π 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 and π‘.

Once again, substituting equation two into our expression can help us. Substituting π  over π‘ in place of π£ gives us π times π  over π‘ times π , which we can simplify to give us ππ  squared over π‘. And we can stop there because weβve managed to express ππ  in terms of π and π‘. But itβ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, ππ‘ equals π£π  and ππ‘ equals ππ . To determine which of these is correct, weβll need to come up with an expression that has ππ‘ on the left-hand side. So, once again, weβ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π£ in place of π, which changes ππ‘ into ππ£π‘.

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

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