Question

a bullet of mass 10g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900g . what is the velocity acquired by the block.​

Answer 1

Explanation:

linear momentum remain conserved.

hence

mv = MV

10/1000 × 400 = (10+900)/1000 × V

4 = 10V

V = (4/10)m/s

V = 0.4 m/s

Answer 2

Answer: The velocity acquired by the wooden block is 4.39 m/s.

Explanation:

Mass of the bullet (m1) = 10 g = 0.01 kg

Mass of the wooden block (m2) = 900 g = 0.9 kg

Initial velocity of the bullet (u1) = 400 m/s

Initial velocity of the wooden block (u2) = 0 m/s

The bullet gets embedded into the block.

∴ Mass of the wooden block with the bullet (m) =  0.01 + 0.9 kg

                                                                                   = 0.91 kg

Final velocity of the wooden block with the bullet (v) = ?

According to the Law of Conservation of Momentum,

or, m1u1 + m2u2 = mv

or, 0.01 X 400 + 0.9 X 0 = 0.91 X v

or, 4 = 0.91 X v

or, v = 4 / 0.91

or, v = 4 X 100 / 91

or, v = 4.39

∴ The velocity acquired by the wooden block (v) = 4.39 m/s