TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Mass of the ball, m=2kg Patential energy, U=(12x+16y)J
∴‌ Force applied on it, ‌F=−‌

dU

dx

∧

i

−‌

dU

dy

∧

j

⇒‌‌F=−12

∧

i

−16

∧

j

Initially (at t=0), the ball is at origin (0,0) and is moving with velocity vi=(15

∧

i

+20

∧

j

)‌m∕s
Now, F=ma=m‌

vf−vi

t

( where vf is the velocity of the ball after t=2s ) or ‌‌‌

Ft

m

=vf−vi
Substituting the given values in above equation, we get
∴‌‌(−12

∧

i

−16

∧

j

)×‌

t

m

‌‌=vf−vi
⇒‌‌(−12

∧

i

−16

∧

j

)×‌

2

2

‌‌=vf−(15

∧

i

+20‌

∧

j

)
⇒‌‌vf‌‌=15

∧

i

−12

∧

i

+20

∧

j

−16

∧

j

‌‌=(3

∧

i

+4

∧

j

)‌m∕s
‌vf∣‌‌=√32+42=5m∕s