AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Let the mass of the man is m and speed is v1, thus the mass of the boy is

m

2

and let his initial velocity is v2
According to given condition,
(KE)man=

1

2

(KE)boy

1

2

mv12=

1

2

(

1

2

m

2

v22)

v12

v22

=

1

4

v1=

v2

2

After the increase in velocity of man by 1 m/s, the new velocity of the man is v1′=(

v2

2

+1)
According to given condition,
(KE)man′=(KE)boy

1

2

mv1′2=

1

2

(

m

2

)v22
v1′2=

v22

2

v1′=

v2

√2

Substituting the values, we get
(

v2

2

+1)=

v2

√2

√2v2+2√2=2v2
v2=

2

(√2−1)

v2=2(√2+1)