Two particles are originally placed at P and Q distant d apart- At t-0 they start moving such that velocity v- of P is aimed towards Q and velocity u- of Q is perpendicular to v- The two projectiles meet at time T-

The correct option is B vd/(v^2 u^2) Relative velocity of P and Q is (v u cosθ) . #160;The particles will meet when: ∫_0^π(v ucosθ ...

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Byju's Answer

Standard XII

Physics

Horizontal and Vertical Components of Projectile Motion

Two particles...

Question

Two particles are originally placed at $P$ and $Q$ distant $d$ apart. At $t = 0$ they start moving such that velocity $\to v$ of $P$ is aimed towards $Q$ and velocity $\to u$ of $Q$ is perpendicular to $\to v$ . The two projectiles meet at time $T =$

\frac{(v^{2} - u^{2}) d^{2}}{v^{3} d}

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\frac{(v + u) d}{v^{2}}

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\frac{v (v - u)}{d}

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\frac{v d}{(v^{2} - u^{2})}

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Solution

The correct option is B $\frac{v d}{(v^{2} - u^{2})}$
Relative velocity of $P$ and $Q$ is $(v - u cos θ)$ .

The particles will meet when:
$\int_{0}^{π} (v - u cos θ) d t = d$ and $\int_{0}^{π} v cos θ d t = d$
and $\int_{0}^{π} cos θ d t = \frac{u T}{v}$

$\Rightarrow v T - u \frac{u T}{v} = d$

$\Rightarrow (v^{2} - u^{2}) T = v d$
$\Rightarrow T = \frac{v d}{v^{2} - u^{2}}$

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Two particles are originally placed at P and Q distant d apart. At t=0 they start moving such that velocity →v of P is aimed towards Q and velocity →u of Q is perpendicular to →v. The two projectiles meet at time T=

The correct option is B vd(v2−u2)Relative velocity of P and Q is (v−ucosθ). The particles will meet when:∫π0(v−ucosθ)dt=d and ∫π0vcosθdt=dand ∫π0cosθdt=uTv ⇒vT−uuTv=d ⇒(v2−u2)T=vd⇒T=vdv2−u2

Two particles are originally placed at $P$ and $Q$ distant $d$ apart. At $t = 0$ they start moving such that velocity $\to v$ of $P$ is aimed towards $Q$ and velocity $\to u$ of $Q$ is perpendicular to $\to v$ . The two projectiles meet at time $T =$