A particle travelling horizontally with speed u collides and joins with a particle of equal mass hanging from a light inextensible string of length 2l. If the string rotates through an angle of 60degrees before first coming to rest, show that u^2=8gl
Answers
Answer:
Speed of the particle = u
Let the mass of the particle be m
When it collides with another mass m, the linear momentum will be conserved
If the velocity of the combined mass after collision is v then
When string rotates 60° before, coming to rest, the vertical height covered by the string with combined mass attached
upto height h, the whole kinetic energy will have converted into potential energy
Thus,
(Proved)
Hope this helps.
Explanation:
Speed of the particle = u
Let the mass of the particle be m
When it collides with another mass m, the linear momentum will be conserved
If the velocity of the combined mass after collision is v then
mu=(m+m)vmu=(m+m)v
\implies v=\frac{u}{2}⟹v=
2
u
When string rotates 60° before, coming to rest, the vertical height covered by the string with combined mass attached
h=2l-2l\cos60^\circh=2l−2lcos60
∘
\implies h=2l-\frac{2l}{2}⟹h=2l−
2
2l
\implies h=l⟹h=l
upto height h, the whole kinetic energy will have converted into potential energy
Thus,
\frac{1}{2}\times 2m\times v^2=2mgh
2
1
×2m×v
2
=2mgh
\implies v^2=2gh⟹v
2
=2gh
\implies \frac{u^2}{4}=2gl⟹
4
u
2
=2gl
\implies u^2=8gl⟹u
2
=8gl (Proved)
Hope this helps.