AIEEE 2002 Physics Solved Paper

We know the work energy theorem, W=∆K=FS
For first penetration, by applying work energy theorem we get,
‌

1

2

mv2−‌

1

2

m(‌

v

2

)2=F×3...(i)
For second penetration, by applying work energy theorem we get,
‌

1

2

m(‌

v

2

)2−0=F×S....(ii)
On dividing (ii) by (i)
‌

1∕4

3∕4

=S∕3
∴S=1‌cm