AMU Engineering 2020 solved paper Solved Paper

Let the speed of the bullet be v1. Let the common velocity of the bullet and the block, after the bullet is attached like a simple pendulum is v2.
By the principle of conservation of linear momentum
‌

10

1000

×140+‌

100

1000

×0=‌

100

1000

v1+‌

10

1000

v2
140=10v1+v2 . . . (i)
For the block, 0=v12−2g(‌

80

100

)
⇒‌‌v1=4m∕s
⇒‌‌140=10×4+v2‌‌‌ [from Eq. (i)] ‌
⇒‌‌v2=100m∕s
Final velocity of bullet =100m∕s