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A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be:

1. 2mv
2. mv√2
3. mv/√2
4. zero

Correct answer: 2mv

Solution

The horizontal component of momentum remains unchanged during the motion, but the vertical component reverses direction. The initial vertical momentum is mv/√2 upward, and the final vertical momentum is mv/√2 downward. The change in vertical momentum is 2mv/√2 = mv√2. Adding the horizontal component (unchanged) results in a total change of 2mv.

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