Two particles A and B are projected simultane | vedclass

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(D) For the particles to collide,their vertical components of velocity must be equal,and their horizontal relative displacement must be covered in the

Vedclass · Accepted Answer

$30^{\circ}, 5 \sqrt{3} \ m$

Vedclass · Answer

(D) For the particles to collide,their vertical components of velocity must be equal,and their horizontal relative displacement must be covered in the given time.$(a)$ Vertical motion:The vertical component of velocity for particle $A$ is $V_{Ay} = V_A \sin 	heta = 20 \sin 	heta$.The vertical component of velocity for particle $B$ is $V_{By} = 10 \ ms^{-1}$ (upwards).Since they collide,their vertical positions must be the same at time $t = 0.5 \ s$. Assuming they start from the same horizontal level,$V_{Ay} = V_{By}$ is required for them to meet at the same height.$20 \sin 	heta = 10$$\sin 	heta = \frac{10}{20} = \frac{1}{2}$$	heta = 30^{\circ}$.$(b)$ Horizontal motion:The horizontal component of velocity for particle $A$ is $V_{Ax} = V_A \cos 	heta = 20 \cos 30^{\circ} = 20 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3} \ ms^{-1}$.Particle $B$ has no horizontal velocity component $(V_{Bx} = 0)$.To collide,particle $A$ must cover the horizontal distance $x$ in time $t = 0.5 \ s$.$x = V_{Ax} 	imes t = (10\sqrt{3}) 	imes 0.5 = 5\sqrt{3} \ m$.

Two particles $A$ and $B$ are projected simultaneously in the directions shown in the figure with velocities $V_A = 20 \ ms^{-1}$ and $V_B = 10 \ ms^{-1}$ respectively. They collide in the air after $0.5 \ s$. Find $(a)$ the angle $\theta$ and $(b)$ the distance $x$.

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