A particle of mass m comes down on a smooth inclined plane from a point B at a height of h from rest. The
magnitude of change in momentum of the particle between the position A and C (assuming the angle of inclination
of the plane as 8 with respect to the horizontal) is:
Attachments:
Answers
Answered by
0
Answer:
2m
2gh
sin(
2
θ
Explanation:
Let velocity of the particle at the base of the inclined plane be v
Work-energy theorem : W
g
=ΔK.E
mgh=
2
1
mv
2
−0
⟹v=
2gh
Momentum of the particle at point C,
P
c
=mv
i
^
Momentum of the particle at point A,
P
A
=mvcosθ
i
^
−mvsinθ
j
^
Change in momentum between A and C, Δ
P
=
P
c
−
P
A
Δ
P
=mv(1−cosθ)
i
^
−mvsinθ
j
^
∣Δ
P
∣=
m
2
v
2
(1−cosθ)
2
+m
2
v
2
sin
2
θ
∣Δ
P
∣=mv
1+cos
2
θ−2cosθ+sin
2
θ
∣Δ
P
∣=mv
2(1−cosθ)
(∵1−cosθ=2sin
2
2
θ
)
⟹ ∣Δ
P
∣=2mvsin(
2
θ
) =2m
2gh
sin(
2
θ
)
Similar questions