@Catch.me @Mashy @Abhisar
Please help me...
A particle of mass m travelling with a velocity u makes a head-on elastic cpllision with another particle of mass M which is at rest. If m
Do you know how to derive this equation \(\huge\sf \boxed{u_2=\frac{m-M}{m+M}u_1}\)
@MichelleShum ?
?
From the above equation we can see that the final velocity v for an object of mass m will be \(\huge\sf \boxed{v=\frac{m-M}{m+M}u}\) Now if m<M, value of v will be negative which means the direction of v will be opposite to u
ok...Thanks....
Do you know how to derive the below equation which i used ? \(\huge\sf \boxed{u_2=\frac{m-M}{m+M}u_1}\)
don't know...
Have u read ur books carefully ? `Conservation of linear momentum` and `conservation of kinetic energy` in an elastic collision ?
@Abhisar i know how to get it already...
That's Great !...in case if u don't u can use this
ok...
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