Question

10 gm bullet is fired from a riffle horizontally into a
5 kg block of wood suspended by a string and
the bullet get embedded in the block. The impact
causes the block to swing to a height of 2.5 cm
its initial level. Calculate the velocity of bullet?​

Answer 1

Answer:

Considering the bullet and block as a system,

using mechanical energy conservation,

ki + ui = kf + uf

1/2 mu^2 + 0 = 0 + (M+m)gh

Where,

m = mass of bullet

M = mass of block

u = Initial velocity of bullet

Finally at the highest point bullet and block both come to rest, so v = 0

So,

1/2 mu^2 = (M + m)gh

1/2 × 0.01 kg × u^2 =(5 + 0.01) × 10 × 0.025

0.01/2 u^2 = 5.01 × 10 ×0.025

u^2 = (50.1 × 0.025 × 2)/0.01

u^2 = (100.2 × 0.025)/0.01

u^2 = 2.505/0.01 = 250.5 m/s

u = root 250.5 = 15.8 m/s

Answer 2

Answer:

15.8 m/s

Explanation:

Considering the bullet and block as a system,

using mechanical energy conservation,

ki + ui = kf + uf

1/2 mu^2 + 0 = 0 + (M+m)gh

Where,

m = mass of bullet

M = mass of block

u = Initial velocity of bullet

Finally at the highest point bullet and block both come to rest, so v = 0

So,

1/2 mu^2 = (M + m)gh

1/2 × 0.01 kg × u^2 =(5 + 0.01) × 10 × 0.025

0.01/2 u^2 = 5.01 × 10 ×0.025

u^2 = (50.1 × 0.025 × 2)/0.01

u^2 = (100.2 × 0.025)/0.01

u^2 = 2.505/0.01 = 250.5 m/s

u = root 250.5 = 15.8 m/s