Physics, asked by gis2422, 4 months ago

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:

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Answers

Answered by 1223448
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

​  

)

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