AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

Mass of the bullet =m
Speed of bullet =v
According to question, bullets gets embedded into the bag, then they will move with common velocity v1 (say).
This is the case of perfectly enelastic collision. ∴ Initial kinetic energy of bullet,
∴ Initial kinetic energy of bullet.
Ki=‌

1

2

mv2...(i)
By the conservation of linear momentum.
mv=(M+m)v1
⇒‌‌v1=‌

mv

M+m

........(ii)
∴ Final kinetic energy, Kf=‌

1

2

(M+m)v12
=‌

1

2

(M+m)‌

m2v2

(M+m)2

[from Eq. (ii)]
=‌

1

2

‌

m2v2

M+m

Loss in kinetic energy =Ki−Kf
=‌

1

2

mv2−‌

1

2

‌

m2v2

M+m

=‌

1

2

mv2(1−‌

m

M+m

)=‌

mMv2

2(M+m)