TS EAMCET 4 May 2018 Shift 2 Solved Paper

Consider the following figure.By conserving momentum at initial and final condition of ball we get,
m1u1+m2u2=m1v1+m1v2
Given that,
m1=m2=m,u1=v,u2=0, and v=0.15v
Therefore,
mv=m(0.15v)+mv2
v2=0.85v
The final kinetic energy is given by,

1

2

mv12+

1

2

mv22 =

1

2

m(0.15v)2+

1

2

m(0.85v)2
The change in kinetic energy is given by,
ΔKE=(KE)i−(KE)f
=

1

2

mv2−[

1

2

m(0.15v)2+

1

2

m(0.85v)2]
=12mv2(1−(0.15)2+(0.85)2)
The percentage change in kinetic energy is given by,
% change=

ΔKE

(KE)i

×100
=

1 2 mv2(1−(0.15)2+(0.85)2)

1 2 mv2

≈25%