A particle A moves along the line y=30 m wit | vedclass

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(B) Let the time of collision be $t$. At the time of collision,both particles must have the same $x$ and $y$ coordinates.For particle $A$:$x_A = v t$,

Vedclass · Accepted Answer

$3$

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(B) Let the time of collision be $t$. At the time of collision,both particles must have the same $x$ and $y$ coordinates.For particle $A$:$x_A = v t$,$y_A = 30 \ m$.For particle $B$:$a_x = a \sin 60^{\circ} = 0.4 	imes \frac{\sqrt{3}}{2} = 0.2\sqrt{3} \ m/s^2$.$a_y = a \cos 60^{\circ} = 0.4 	imes \frac{1}{2} = 0.2 \ m/s^2$.Since $B$ starts from the origin with zero initial velocity:$x_B = \frac{1}{2} a_x t^2 = \frac{1}{2} (0.2\sqrt{3}) t^2 = 0.1\sqrt{3} t^2$.$y_B = \frac{1}{2} a_y t^2 = \frac{1}{2} (0.2) t^2 = 0.1 t^2$.For collision,$y_A = y_B$:$30 = 0.1 t^2 \Rightarrow t^2 = 300 \Rightarrow t = 10\sqrt{3} \ s$.For collision,$x_A = x_B$:$v t = 0.1\sqrt{3} t^2 \Rightarrow v = 0.1\sqrt{3} t$.Substituting $t = 10\sqrt{3}$:$v = 0.1\sqrt{3} 	imes 10\sqrt{3} = 1 	imes 3 = 3 \ m/s$.Thus,the value of $|v|$ is $3 \ m/s$.

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