JEE Mains 02-Sep-2020 Shift 2 Solved Paper

By momentum conservation along y:
m1u1‌sin‌θ1=m2u2‌sin‌θ2
i.e. mu1‌sin‌θ1=10mu2‌sin‌θ2
⇒ [u1‌sin‌θ1=10u2‌sin‌θ2] ......(i)
kf m1=

1

2

kim1 i.e.

1

2

mu12=12×

1

2

mu2
i.e. [u1=

u

√2

] ........(ii)
Also collision is elastic :ki=kf

1

2

mu2=

1

2

mu12+

1

2

.10m.u22

1

2

mu2=

1

2

×

1

2

mu2 +

1

2

×10m.u22

1

4

mu2=

1

2

×10×mu22
[u2=

u

√20

] ......(iii)
Putting(ii) & (iii) in (i)

u

√2

‌sin‌θ1=10.

u

√20

‌sin‌θ2
[sin‌θ1=√10‌sin‌θ2]→ Hence n=10