Question

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

Answer 1

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

$mu=(m+m)v$

$\implies v=\frac{u}{2}$

When string rotates 60° before, coming to rest, the vertical height covered by the string with combined mass attached

$h=2l-2l\cos60^\circ$

$\implies h=2l-\frac{2l}{2}$

$\implies h=l$

upto height h, the whole kinetic energy will have converted into potential energy

Thus,

$\frac{1}{2}\times 2m\times v^2=2mgh$

$\implies v^2=2gh$

$\implies \frac{u^2}{4}=2gl$

$\implies u^2=8gl$                  (Proved)

Hope this helps.

Answer 2

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.