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Laws of Motion question

2001 · Q155
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Laws of Motion question

2001 · Q155

NEETPhysicsLaws of MotionMCQ+4 / −1
A 1 kg stationary bomb is exploted in three parts having mass 1 : 1 : 3 respectively. Parts having same mass move in perpendicular direction with velocity 30 m/s, then the velocity of bigger part will be
  1. A
    10210\sqrt 2102​ m/sec
  2. B
    102{{10} \over {\sqrt 2 }}2​10​ m/sec
  3. C
    15215\sqrt 2152​ m/sec
  4. D
    152{{15} \over {\sqrt 2 }}2​15​ m/sec
View written solutionFree

Correct answer: A

Apply conservation of linear momentum. Total momentum before explosion = total momentum after explosion

0=m5v1i^+m5v2j^+3m5v→30 = {m \over 5}{v_1}\widehat i + {m \over 5}{v_2}\widehat j + {{3m} \over 5}{\overrightarrow v _3}0=5m​v1​i+5m​v2​j​+53m​v3​

3m5v→3=−m5[v1i^+v2j^]{{3m} \over 5}{\overrightarrow v _3} = - {m \over 5}\left[ {{v_1}\widehat i + {v_2}\widehat j} \right]53m​v3​=−5m​[v1​i+v2​j​]

v→3=−v13i^−v23j^{\overrightarrow v _3} = {{ - {v_1}} \over 3}\widehat i - {{{v_2}} \over 3}\widehat jv3​=3−v1​​i−3v2​​j​
     ∴\therefore∴ v1=v2=30{v_1} = {v_2} = 30v1​=v2​=30 m/sec

v→3=−10i^−10j^;v3=102{\overrightarrow v _3} = - 10\widehat i - 10\widehat j;{v_3} = 10\sqrt 2 v3​=−10i−10j​;v3​=102​ m/sec

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