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Work Energy and Power question

2011 · Q104
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Work Energy and Power question

2011 · Q104

NEETPhysicsWork Energy and PowerMCQ+4 / −1
A mass m moving horizontally (along the x-axis) with velocity vvv collides and sticks to a mass of 3m moving vertically upwards (along the y-axis) with velocity 2vvv. The final velocity of the combination is
  1. A
    32vi^+14vj^{3 \over 2}v\widehat i + {1 \over 4}v\widehat j23​vi+41​vj​
  2. B
    14vi^+32vj^{1 \over 4}v\widehat i + {3 \over 2}v\widehat j41​vi+23​vj​
  3. C
    13vi^+23vj^{1 \over 3}v\widehat i + {2 \over 3}v\widehat j31​vi+32​vj​
  4. D
    23vi^+13vj^{2 \over 3}v\widehat i + {1 \over 3}v\widehat j32​vi+31​vj​
View written solutionFree

Correct answer: B

AIPMT 2011 Mains Physics - Work, Energy and Power Question 36 English Explanation

According to conservation of momentum, we get
mvi^+(3m)2vj^=(m+3m)v′→mv\widehat i + \left( {3m} \right)2v\widehat j = \left( {m + 3m} \right)\overrightarrow {v'} mvi+(3m)2vj​=(m+3m)v′

where v′→\overrightarrow {v'} v′ is the final velocity after collision
v′→=14vi^+64vj^=14vi^+32vj^\overrightarrow {v'} = {1 \over 4}v\widehat i + {6 \over 4}v\widehat j = {1 \over 4}v\widehat i + {3 \over 2}v\widehat jv′=41​vi+46​vj​=41​vi+23​vj​

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