How much momentum will an object of mass 10 kg is transfer to the floor if it fall from a height of 5m (g=10m s-2)
natsu1044q:
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Answered by
1
Assuming It is an Inelastic collision.
Momentum = mv -0= 10*(2*10*5)^1/2= 100 kg-m/s
Momentum = mv -0= 10*(2*10*5)^1/2= 100 kg-m/s
Answered by
1
Hi friend,
We know that ,
S=1/2 at²
t²= 2S/a
t²= 1
t= 1 sec.
:- v= 10m/s
Momentum (p)= MV
P= 10×10
P=100 Kg m/s.
Hope this helped you a little!!!
We know that ,
S=1/2 at²
t²= 2S/a
t²= 1
t= 1 sec.
:- v= 10m/s
Momentum (p)= MV
P= 10×10
P=100 Kg m/s.
Hope this helped you a little!!!
That is the formula
No I mean Like the text in the answers.
This is also a form of solution to this answer
No how do you write it. How did you write the t^2.
S=ut+1/2 at²
By bring all the others to left side t² will be get
s-ut-1/2=t^2
sorry
No probs.
Uh.. Thanks But I am not asking about a theory related doubt. It was a technical. Never Mind.
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