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 :
- mv 2
- zero
- 2mv
- mv/ 2
Answer
(1) mv 2
Solution
The horizontal component of velocity (v 45 ) does not change during the flight.
The vertical component changes from v 45 upwards to v 45 downwards.
| p| = m(v 45 ) - (-m v 45 )
| p| = 2mv 45 = 2mv 1 2
| p| = mv 2
Answer: (1)