The correct option is A 0.212 m,28.54 J
Here,
Mass of bullet (m)=0.012 kg
Mass of block (M)=0.4 kg
Speed of the bullet (u)=70 ms−1
Let v be the velocity of the [block+bullet] system after collision
Now, using law of conservation of linear momentum, we can write that, mu=(M+m)v
or v=(muM+m)
=(0.012×700.4+0.012)=0.840.412
=2.04 ms−1
Thus, the speed of bullet + block system after collision is 2.04 ms−1.
Let h be the height up to which the block rises after striking.
Applying law of conservation of energy,
Increase in P.E. of the Bullet+Block system = Decrease in K.E. of the Bullet + Block system
Hence, we can write that,
(M+m)gh=12(M+m)v2
or h=v22g=(2.04)22×9.8
=0.212 m
Heat produced = Loss in K.E of Bullet + Block system.
=12mu2−12(M+m)v2
=12(0.012)×(70)2−12×0.412×(2.04)2
=28.54 J
Thus, option (a) is the correct answer.