AP EAPCET 25-Aug-2021 Shift 2 Paper

Given, mass of bullet, mb=30g=0.03‌kg
Velocity of bullet, vb=700ms−1
Mass of block, mB=4‌kg
Height upto which block rises, h=0.2m
Before collision, momentum of bullet =mb×vb
=0.03×700
=21‌kg−ms−1
Let v1,v2 be the velocities of bullet and block after the collision.
Using conservation of momentum,
⇒ 21=0.03×v1+4v2 ...(i)
Using conservation of energy for the block, Change in KE = Work done

1

2

mBv22=mBgh
⇒ v22=2gh
⇒ v2=√2×9.8×0.2
=1.979ms−1
Putting the value of v2 in Eq. (i), we get
21=0.03×v1+4×1.979
⇒ 21=0.03×v1+7.919
⇒ 21−7.919=0.03×v1
⇒ v1=

13084

0.03

ms−1
=436.13ms−1=433ms−1