AP EAPCET 05-Jul-2022 Shift 2 Paper

The given situation is shown below
Let us draw the free body diagram for the small body of mass m. Let block breaks off the plane at t=t0, i.e at t=t0, normal reaction (R) will become zero.
‌ i.e. ‌‌R=0
⇒mg−Fsin‌θ=0
⇒mg−at0sin‌θ=0
⇒t0=‌

mg

asin‌θ

......(i)
From the figure,
max‌=Fx
m‌

dv

dt

‌=F‌cos‌θ=at‌cos‌θ
⇒‌‌mdv‌=at‌cos‌θ‌dt
Integrating both sides, we have
‌‌‌m‌

v

∫

0

dvx=a‌cos‌θ‌

t0

∫

0

t‌dt
⇒‌‌mv=‌

at02‌cos‌θ

2

⇒‌‌v=‌

at02‌cos‌θ

2m

....(ii)
From Eqs. (i) and (ii), we get
Putting m=1‌kg
‌a=1ms−2
‌θ=45∘
‌v=‌

(1102)‌cos‌45∘

2×1×sin‌245∘

=50√2ms−1
‌