Two particle A and B having initial position 0-0 m and 485- 365m- Their initial velocity is 4-3- m-s and -4-3- m-s respectively- Both have same acceleration a-9- m-s 2- Then- two particles will collide after time

The correct option is B 1.2 s Tanθ = 36/5 / 48/5 = 36 / 48 = 3 / 4 V_A/B=4î+3ĵ ( 4î 3ĵ)=8î+6ĵ Now,magnitude nbsp;of relative nbsp;velocity of ap ...

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Standard XII

Physics

Applying NLM along 3 Axis

Two particle ...

Question

Two particle A and B having initial position $(0, 0)$ m and $(\frac{48}{5}, \frac{36}{5}) m$ . Their initial velocity is $(4^i + 3^j)$ m/s and $(- 4^i - 3^j)$ m/s respectively. Both have same acceleration $\to a = - 9^j$ m/s $^{2}$ . Then, two particles will collide after time

1.2 s

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12 s

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2.4 s

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They will not collide

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Solution

The correct option is B 1.2 s
$T a n θ = \frac{36 / 5}{48 / 5} = \frac{36}{48} = \frac{3}{4}$
$V_{A / B} = 4^i + 3^j - (- 4^i - 3^j) = 8^i + 6^j$
Now,magnitude of relative velocity of approach
$\Rightarrow v_{r e l} = \sqrt{8^{2} + 6^{2}} = \sqrt{100} = 10 m / s$
Now, $S_{r e l} = \sqrt{\frac{48^{2}}{5^{2}} + \frac{36^{2}}{5^{2}}} = 12 m$
Time $= \frac{12}{10} = 1.2 s e c o n d s$

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Two particle A and B having initial position (0,0) m and (485,365)m. Their initial velocity is (4^i+3^j) m/s and (−4^i−3^j) m/s respectively. Both have same acceleration →a=−9^j m/s2. Then, two particles will collide after time

The correct option is B 1.2 sTanθ=36/548/5=3648=34VA/B=4^i+3^j−(−4^i−3^j)=8^i+6^jNow,magnitude of relative velocity of approach⇒vrel=√82+62=√100=10m/sNow, Srel=√48252+36252=12mTime =1210=1.2seconds

Two particle A and B having initial position $(0, 0)$ m and $(\frac{48}{5}, \frac{36}{5}) m$ . Their initial velocity is $(4^i + 3^j)$ m/s and $(- 4^i - 3^j)$ m/s respectively. Both have same acceleration $\to a = - 9^j$ m/s $^{2}$ . Then, two particles will collide after time