A particle moving in the xy plane experiences | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given force $\overrightarrow{F} = m\overrightarrow{a} = k(v_y \hat{i} + v_x \hat{j})$.Thus,$a_x = \frac{dv_x}{dt} = \frac{k}{m} v_y$ and $a_y = \f

Vedclass · Accepted Answer

The quantity $\overrightarrow{v} 	imes \overrightarrow{a}$ is constant in time.

Vedclass · Answer

(C) Given force $\overrightarrow{F} = m\overrightarrow{a} = k(v_y \hat{i} + v_x \hat{j})$.Thus,$a_x = \frac{dv_x}{dt} = \frac{k}{m} v_y$ and $a_y = \frac{dv_y}{dt} = \frac{k}{m} v_x$.Dividing the two equations: $\frac{dv_y}{dv_x} = \frac{v_x}{v_y} \implies v_y dv_y = v_x dv_x$.Integrating both sides: $\int v_y dv_y = \int v_x dv_x \implies v_y^2 = v_x^2 + C \implies v_y^2 - v_x^2 = 	ext{constant}$.Now,calculate $\overrightarrow{v} 	imes \overrightarrow{a}$:$\overrightarrow{v} 	imes \overrightarrow{a} = (v_x \hat{i} + v_y \hat{j}) 	imes \frac{k}{m}(v_y \hat{i} + v_x \hat{j})$$= \frac{k}{m} [v_x^2 (\hat{i} 	imes \hat{j}) + v_y^2 (\hat{j} 	imes \hat{i})]$$= \frac{k}{m} [v_x^2 \hat{k} - v_y^2 \hat{k}] = \frac{k}{m} (v_x^2 - v_y^2) \hat{k}$.Since $v_y^2 - v_x^2$ is constant,$\overrightarrow{v} 	imes \overrightarrow{a}$ is constant in time.

Similar Questions

An athlete completes one round of a circular track of radius $R$ in $40 \, s$. What will be his displacement at the end of $2 \, min \, 20 \, s$?

$A$ body of mass $1 \, kg$ starts moving from rest at $t = 0$ in a circular path of radius $8 \, m$. Its kinetic energy varies as a function of time as: $K.E. = 2t^2 \, J$,where $t$ is in seconds. Then:

The angle of projection of a particle is measured from the vertical axis as $\phi$ and the maximum height reached by the particle is $h_m$. Here $h_m$ as a function of $\phi$ can be represented as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module