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JEE Main 2020 (6 Sep, Shift 2) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2020 (6 Sep, Shift 2)

Q27JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Inertia, First & Second Laws

A particle moving in the xy plane experiences a velocity dependent force

F⃗=k(vyi^+vxj^)\vec{F} = k\left(v_y\hat{i} + v_x\hat{j}\right), where vx and vy are the

x and y components of its velocityv⃗\vec{v}. Ifa⃗\vec{a}is the

acceleration of the particle, then

which of the following statements is true for the particle?

Show answer and solution

Answer: (b)

  1. a⃗=km(vyi^+vxj^)\vec{a} = \frac{k}{m}(v_y\hat{i} + v_x\hat{j})
  2. (v⃗×a⃗)z=vxay−vyax=km(vx2−vy2)(\vec{v} \times \vec{a})_z = v_xa_y - v_ya_x = \frac{k}{m}(v_x^2 - v_y^2)
  3. ddt(vx2−vy2)=2vxax−2vyay=2km(vxvy−vyvx)=0\frac{d}{dt}(v_x^2 - v_y^2) = 2v_xa_x - 2v_ya_y = \frac{2k}{m}(v_xv_y - v_yv_x) = 0, so v⃗×a⃗\vec{v} \times \vec{a} is constant.
  4. (a), (c) fail: v⃗⋅a⃗=2kmvxvy≠0\vec{v} \cdot \vec{a} = \frac{2k}{m}v_xv_y \neq 0 and changes; (d) fails: a magnetic force is always ⊥v⃗\perp \vec{v}.
  5. So the answer is (b).
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