AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

According to question, the given situation is shown in the following figure.Mass of block, M=2kg
Mass of bullet, m1=0.01kg
Speed of bullet, v=500ms−1
h=0.1 {m}
If v1 and v2 be the velocities of the bullet and blocks after collision, then by conservation of energy,
‌

1

2

Mv22=Mgh
v2=√2gh=√2×9.8×0.1=1.4ms−1
According to the law of conservation of momentum,
m1v=m1v1+Mv2
⇒‌‌v1=‌

m1v−Mv3

m1

=‌

0.01×500−2×1.4

0.01

=‌

5−2.8

0.01

=‌

2.2

0.01

=220ms−1