JEE Main 18 March 2021 Shift 2 Solved Paper

The given situation is shown belowInitial velocity of the projectile,
u=u‌cos‌45∘‌

∧

i

+u‌sin‌45∘‌

∧

j

Final velocity of the projectile,
v=v‌cos‌45∘‌

∧

i

−v‌sin‌45∘‌

∧

j

At point B,v‌cos‌45∘=u‌cos‌45∘
and v‌sin‌45∘=u‌sin‌45∘
v=u‌cos‌45∘‌

∧

i

−u‌sin‌45∘‌

∧

j

The magnitude of change in the momentum between the points A and B
|∆p|=|m(v−u)|
⇒‌‌∆p=2mu‌sin‌45∘
⇒‌‌∆p=2×0.005×5√2×‌

1

√2

⇒‌‌∆p=5×10−2kg−m/s
Hence, the value of the x is 5 .