JEE Main 16 March 2021 Shift 2 Solved Paper

Given
Mass of large block of wood, M=5.99kg
Mass of bullet, m=10g
Height at which their centre of mass rise, h=9.8cm
From the law of conservation of energy,
Energy of the system when bullet gets embedded = Energy of the system till it momentarily comes to rest.
⇒‌‌‌

1

2

(M+m)v12=(M+m)gh
where, v1= velocity of bullet + block system
⇒‌‌v1=√2gh.....(i)
According to law of conservation of momentum,
Momentum before collision = Momentum after collision.
⇒‌‌mv=(M+m)v1
[where, v= velocity of bullet before collision]
⇒‌‌mv=(M+m)√2gh [using Eq. (i)]
⇒‌‌v=(‌

M+m

m

)√2gh
⇒‌‌v=‌

(5.99+0.01)

10×10−3

×√2×9.8×9.8×10−2
⇒‌‌v=831.55ms−1