Question 281: Laws of Motion

Question

A particle moves in the X - Y plane under the influence of a force such that its linear momentum is $$\overrightarrow p \left( t \right) = A\left[ {\widehat i\cos (kt) - \widehat j\sin (kt)} \right]$$, where A and k are constants. The angle between the force and the momentum is

Accepted Answer

Given that,
$$\overrightarrow{\mathrm{P}}(t)=\mathrm{A}[\hat{i} \cos (k t)-\hat{j} \sin (k t)]$$
where, $$\mathrm{A}$$ and $$k=$$ constant
$$\begin{aligned}	ext { Force }\overrightarrow{\mathrm{F}}) & =\frac{d \overrightarrow{\mathrm{P}}}{d t}\left(	ext { from Newton's } 2^{	ext {nd }}ight. 	ext { law) } \\
\Rightarrow \quad \overrightarrow{\mathrm{F}} & =\frac{d}{d t}[\mathrm{~A}(\hat{i} \cos (k t)-\hat{j} \sin (k t))] \\& =\mathrm{A} k[\hat{i} \cos (k t)-\hat{j} \sin (k t)]\end{aligned}$$
By applying dot product between $$\vec{F} \& \vec{P}$$ we have
$$\begin{aligned}& \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{P}}=0 \\
& \Rightarrow|\vec{F}||\vec{P}| \cos 	heta=0 \\
& \Rightarrow \quad \cos 	heta=0 \\
& 	ext {Hence,} 	heta=90^{\circ}\end{aligned}$$

Answer

$$0^\circ $$

Answer

$$30^\circ $$

Answer

$$45^\circ $$

Answer

$$90^\circ $$

Laws of Motion

A particle moves in the X - Y plane under the influence of a force such that its linear momentum is $\overrightarrow p \left( t \right) = A\left[ {\widehat i\cos (kt) - \widehat j\sin (kt)} \right]$, where A and k are constants. The angle between the force and the momentum is

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