TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Given,
Mass of ball P and Q be mP and mQ
where, mp=m and mQ=2m
Initial velocity of P and Q is uP,uQ i.e.
up=v and uQ=0
and final velocity of P and Q is vP,vQ.
Coefficient of restitution, e=1∕3
As we know that
e=‌

vP−vQ

uQ−uP

∴‌‌‌

1

3

=‌

vP−vQ

0−v

⇒‌

1

3

=‌

vP−vQ

−v

⇒‌‌‌‌vP−vQ=−‌

v

3

. . . (i)
‌ By using law of conservation of momentum, ‌
‌‌mPuP+mQuQ=mPvP+mQvQ
⇒‌‌mv+2m.0=mvP+2mvQ
⇒‌‌v=vP+2vQ. . . (ii)
‌ Now, subtracting Eq. (i) from Eq. (ii), as ‌
‌‌+v=+vP+2vQ
−v∕3=‌+vP−vQ
+‌−
4v∕3=3vQ⇒vQ=4v∕9
Substituting in Eq. (ii), we get
vP‌‌=v−2vQ
‌‌=v−2⋅‌

4v

9

=‌

9v−8v

9

=v∕9
Now, ‌

vP

vQ

=‌

v∕9

4v∕9

=‌

1

4

or ‌

vQ

vP

=4