A particle of mass m is projected with velocity v making of an angle of 45 degree with the horizontal. when the particle lands on the level ground, then what is the magnitude of the change in its momentum?
Answers
Answered by
571
Initial momentum=m(vcos45i+vsin45j)
Final momentum=m(vcos45i-vsin45j)
change in momentum=-2mvsin45j
Magnitude of it is 2mvsin45 or √2mv.
Final momentum=m(vcos45i-vsin45j)
change in momentum=-2mvsin45j
Magnitude of it is 2mvsin45 or √2mv.
Answered by
254
The change in momentum is .
Given:
Mass = m
Velocity = v
Angle = 45°
Solution:
The formula for momentum is given below:
p = mv
From the question, momentum change when the particle reaches ground level is asked.
Thus, the horizontal momentum remains unchanged.
Initial momentum = mv sin45
Final momentum = -(mv sin45)
.
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