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Question

A projectile of mass m is thrown with velocity v making an angle 60 with the horizontal. Neglecting air resistance, the change in momentum from the departure A to its arrival at B, along the vertical direction is :
184032.png

A
2mv
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B
3mv
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C
mv
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D
mv3
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Solution

The correct option is A 3mv
As the figure drawn above shows that at points A and B the vertical component of velocity is vsin60 but their direction are opposite.
Hence , change in momentum is given by:
Δp=mvsin60(mvsin60)=2mvsin60
=2mv32=3mv

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