Question

A running man has half the
the kinetic energy of that of a boy of half of his mass. The man speeds up by
1 m/s so as to have same kinetic energy as that of the boy. The
same kinetic energy as that of the boy. The original speed of the man will be​

Answer 1

Answer:

Man's ke

mvv/2

Explanation:

boy's ke

(m/2)*vv/2

Man speeds up by 1m/s

so his ke

=m(v+1)(v+1)/2

This= boy's ke

=(m/2)(vv/2)

so

(v+1)^2=v sq/2

or

2*(v+1)^2=v sq

so 2*(v sq +2v+1)- v sq=0

or v sq+ 4v+2=0

so v sq + 4v + 4=2

so (v+2)=√2

or v=(√2)-2

= original speed of ma...ANSWER

Answer 2

Answer:1+√2

Explanation:

Okay so let the man initially have a velocity V and the small guy v

Acc. To condition

1/2 M x V^2 = 1/2 [ 1/2 x M/2 x v^2]

=> M x V^2 = 1/2 x M /2 x v^2

[After all the merciless cancellation, it comes out as]

v = 2V .............. (I)

Now, A/Q

1/2 M (V+1)^2 = 1/2 x M/2 x v^2

=>(V+1)^2 = v^2/2

= 2V^2 [Squaring eq ( I )

=>V^2+ 2V+1= 2V^2

=> V^2- 2V+1=0

=>V= (2+-√8)/2

= 1+√2 [We ignore -ve answer cause dude... he's never moving in opp direction]