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A ball of mass $2 \mathrm{~kg}$ is moving in $x y$-plane with a potential energy given as $U=(12 x+16 y) \mathrm{J}, x$ and $y$ being in metre. Assume the initial position of the ball at $t=0$ is at origin $(0,0)$ and it is moving with a velocity of $(15 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. Then, identify the correct statement.
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Verified Answer
The correct answer is:
The speed of the ball at $t=2 \mathrm{~s}$ is $5 \mathrm{~m} / \mathrm{s}$.
Mass of the ball, $m=2 \mathrm{~kg}$
Patential energy, $U=(12 x+16 y) \mathrm{J}$
$\therefore$ Force applied on it, $\mathbf{F}=-\frac{d U}{d x} \hat{\mathbf{i}}-\frac{d U}{d y} \hat{\mathbf{j}}$
$$
\Rightarrow \quad \mathbf{F}=-12 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}
$$
Initially $(a t t=0)$, the ball is at origin $(0,0)$ and is moving with velocity $\mathbf{v}_i=(15 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$
Now, $\mathbf{F}=m \mathbf{a}=m \frac{\mathbf{v}_f-\mathbf{v}_i}{t}$ (where $\mathbf{v}_f$ is the velocity of the ball after $t=2 \mathrm{~s}$ ) or
$$
\frac{\mathbf{F} t}{m}=\mathbf{v}_f-\mathbf{v}_i
$$
Substituting the given values in above equation, we get
$$
\begin{aligned}
& \therefore \quad(-12 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}) \times \frac{t}{m}=\mathbf{v}_f-\mathbf{v}_i \\
& \Rightarrow \quad(-12 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}) \times \frac{2}{2}=\mathbf{v}_f-(15 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}) \\
& \Rightarrow \quad \mathbf{v}_f=15 \hat{\mathbf{i}}-12 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}-16 \hat{\mathbf{j}} \\
& =(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} \\
& \therefore \quad\left|\mathbf{v}_f\right|=\sqrt{3^2+4^2}=5 \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$$
Patential energy, $U=(12 x+16 y) \mathrm{J}$
$\therefore$ Force applied on it, $\mathbf{F}=-\frac{d U}{d x} \hat{\mathbf{i}}-\frac{d U}{d y} \hat{\mathbf{j}}$
$$
\Rightarrow \quad \mathbf{F}=-12 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}
$$
Initially $(a t t=0)$, the ball is at origin $(0,0)$ and is moving with velocity $\mathbf{v}_i=(15 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$
Now, $\mathbf{F}=m \mathbf{a}=m \frac{\mathbf{v}_f-\mathbf{v}_i}{t}$ (where $\mathbf{v}_f$ is the velocity of the ball after $t=2 \mathrm{~s}$ ) or
$$
\frac{\mathbf{F} t}{m}=\mathbf{v}_f-\mathbf{v}_i
$$
Substituting the given values in above equation, we get
$$
\begin{aligned}
& \therefore \quad(-12 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}) \times \frac{t}{m}=\mathbf{v}_f-\mathbf{v}_i \\
& \Rightarrow \quad(-12 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}) \times \frac{2}{2}=\mathbf{v}_f-(15 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}) \\
& \Rightarrow \quad \mathbf{v}_f=15 \hat{\mathbf{i}}-12 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}-16 \hat{\mathbf{j}} \\
& =(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} \\
& \therefore \quad\left|\mathbf{v}_f\right|=\sqrt{3^2+4^2}=5 \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$$
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