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In this question $i$ and $j$ are horizontal unit vectors. A particle $P$ of mass $2\,\text{kg}$ moves under the action of two forces, $(p i + q j)\,\text{N}$ and $(2q i + p j)\,\text{N}$ , where $p$ and $q$ are constants. Given that the acceleration of $P$ is $(i - j)\,\text{ms}^{-2}$ (a) find the value of $p$ and the value of $q$ .

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. In this question

i

and

j

are horizontal unit vectors. A particle

P

of mass

2\,\text{kg}

moves under the action of two forces,

(p i + q j)\,\text{N}

and

(2q i + p j)\,\text{N}

, where

p

and

q

are constants. Given that the acceleration of

P

is

(i - j)\,\text{ms}^{-2}

(a) find the value of

p

and the value of

q

.

Addition of Forces: First, let's add the two forces together to get the total force acting on the particle $P$ . $(p+iq)\mathbf{N} + (2q+ip)\mathbf{N} = (p+2q)\mathbf{i} + (q+p)\mathbf{j}$
Newton's Second Law: Now, according to Newton's second law, $\text{Force} = \text{mass} \times \text{acceleration}$ . So, the total force should equal the mass of the particle times its acceleration. Let's write that down: $(p+2q)\mathbf{i} + (q+p)\mathbf{j} = 2\,\text{kg} \times (\mathbf{i}-\mathbf{j})\,\text{m/s}^{-2}$
Coefficient Equations: We can now equate the coefficients of $i$ and $j$ from both sides of the equation. $\newline$ For $i$ : $p + 2q = 2 \times 1$ $\newline$ For $j$ : $q + p = 2 \times (-1)$
Solving for p: Let's solve the first equation for p. $\newline$ $p + 2q = 2$ $\newline$ $p = 2 - 2q$
Plugging in Values: Now, let's plug the value of $p$ into the second equation. $q + (2 - 2q) = -2$ $q + 2 - 2q = -2$

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