Here, Mass of the bullet m1=0.012 kg
Mass of the block m2=0.4 kg
Initial velocity of the bullet, u1=70ms–1
Initial velocity of the block, u2=0
Since on striking the wooden block, the bullet comes to rest w.r.t. the block of wood, the collision is inelastic in nature.
Let v be the velocity acquired by the combination.
Applying principle of conservation of linear momentum
(m1+m2)v =m1u1+m2u2 =m1u1
v=

m1u1

m1+m2

=

0.012×70

0.012+0.4

=

0.84

0.412

=2.04ms−1
Let the block rise to a height h.

1

2

(m1+m2)v2=(m1+m2)gh
h=

v2

2g

=

(2.04)2

2×9.8

=21.2cm
The loss in kinetic energy of the system will appear as heat.
∴ Heat produced = initial K.E. of bullet – final K.E. of combination
=

1

2

m1u12−

1

2

(m1+m2)v2
=

1

2

×0.012×(70)2−

1

2

(0.412)(2.04)2
=29.4–0.86=28.54J