A 2.00 kilogram block initially hangs at rest at the end of two 1.00 meter strings of negligible mass. A 0.003

A 2.00 kilogram block initially hangs at rest at the end of two 1.00 meter strings of negligible mass. A 0.00300 kilogram bullet, moving horizontally with a speed of 1000 meters per second, strikes the block and becomes embedded in it. After the collision, the bullet/block combination swings upward, but does not rotate.
a. Calculate the speed v of the bullet/block combination just after the collision.
b. Calculate the ratio of the initial kinetic energy of the bullet to the kinetic energy of the bullet/block combination immediately after the collision.
c. Calculate the maximum vertical height above the initial rest position reached by the bullet/block combination.

By principle of momentum
mu = (M+m)v

V = mu / (M+m) 0.003*1000/ 2.003 =1.5 m/s
--------------------------------------…
The ratio required is = mu^2 / (M+ m) V^2 = u / V = 1000/ 1.5 = 667
--------------------------------------…
h = v^2 /2g = 1.5*1.5 / 19.6 =0.11 m
===============================

a)Mass of bullet=m and of block =M
momentum of bullet =3.0 (kg.m/s)
Momentum of block is equal to 0>
so total momentum=3
so velocity=3/M+m
=3/2.003=1.4977=1.5

b)kinetic energy of the bullet = 1/2m.v^2=1/2*0.003*1000^2=1500J
kinetic energy of the bullet+block=1/2*2.003*1.5^2
=2.253375J
so ratio =1500/2.253375
=665.668
c)now kinetic energy will be equal to potential energy (considering no friction)
so 1/2m.v^2=m.g.h
2.253375=2.003*9.8*h
so h=0.1147metres.
MAN U took a lot of my time!