AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

Radius of first steel ball,
r1=2cm=2×10−2m
Radius of second steel ball,
r2=4cm=4×10−2m
∴ Mass of first steel ball,
m1=V1×ρ=‌

4

3

πr13ρ‌‌[ρ= density ]
Similarly, ‌‌m2=‌

4

3

πr23ρ
∴‌‌‌

m1

m2

=‌

r13

r23

=‌

(2×10−2)3

(4×10−2)3

=‌

1

8

⇒‌‌m2=8m1 .....(i)
For elastic collision, velocity of first ball,
v1=(‌

m1−m2

m1+m2

)u1+(‌

2m2

m1+m2

)u2
Here, u1=0 and u2=81cms−1
∴‌‌v1=(‌

m1−m2

m1+m2

)×0+(‌

2m2

m1+m2

)×8
=0+‌

2×8m1

m1+8m1

×81 [from Eq. (i)]
=‌

16

9

×81=144cms−1