TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

According to given situation,
Momentum conservation gives,
p1′+p2′=p1.....(i)
KE conservation gives,
‌

p′2

m1

+‌

p′22

m2

=‌

p12

m1

‌‌[∴k=‌

p2

2m

]....(ii)
Substituting the value of p{ from Eq. (i) into Eq. (ii), we have
‌

p1′2

m1

+‌

p2′2

m2

=‌

1

m1

(p1′+p2′)2
As particles fly off at 90° after collision,
∴‌‌p1⋅p2=0
p1⋅p2‌cos‌θ=0⇒p1p2=0
⇒‌‌‌

p′2

m2

=‌

p2′2

m1

or ‌

m2

m1

=1