A block of a mass of 10kg rests on a horizontal table. The coefficient of friction between the block and the table is 0.05. When hit by a bullet of mass 50g moving with speed v, that gets embedded in it, the block moves and comes to rest after moving a distance of 2m on the table. If a freely falling object were to acquire speed v/10 after being dropped from height H, then neglecting energy losses and taking g=10ms⁻², the value of H is close to: 0.04km, 0.05km, 0.02km, 0.03km

From conservation of linear momentum:
mbulletv = (mblock + mbullet)v′
KE of block+bullet system is given by:
E = (mbullet + mblock)v′²/2 = (mbulletv)²/2(mblock + mbullet)
The system stops at distance s on the on the table.
KE loss due to friction is equal to work done against friction
∴ μ(mblock + mbullet)gs = (mbulletv)²/2(mblock + mbullet)
If the freely falling object acquires speed v/10 in falling a distance H,
(v/10)² = 2gH ⇒ v² = 200gH ⇒ 2μgs(mblock + mbullet)²/m²bullet = 200gH
⇒ 0.05 × 2 × 10 × (10.05)²/0.05² = 200H
⇒ H = 40.4m = 0.04km