AP EAPCET 07-Jul-2022 Shift 2 Paper

For an elastic collision of 2 masses m1 and m2 with initial velocities u1 and u2, final velocities of masses after collision are
v1=(

m1−m2

m1+m2

)u1+

2m2u2

(m1+m2)

and
v2=(

m2−m1

m1+m2

)u2+

2m1u2

m1+m2

These are obtained as standard results by applying conservation of energy and momentum equations. Now, in given case at first bullet is embedded in mass M1. Let speed of bullet + Block system be v1 after collision.

Then by conservation of momentum we have.
mv=(m+M1)v1
or
v1=(

mv

m+M1

)
Note Here we are not using standard result as first collision is perfectly inelastic.
Now, bullet +M1 strikes M2 and it is given in problem that this collision is elastic.

Now, by formula we have,
v2=(

(m+M1)−M2

m+M1+M2

)v1+0
and v3=(

M2−(m+M1)

M2+m+M1

×0)+(

2(m+M1)v1

m+M1+M2

)
⇒v3=

2(m+M1)

m+M1+M2

⋅

mv

(m+M1)

=

2mv

m+M1+M2

Note You can remember these standard and useful results for finding final velocities in case of elastic collisions.