Question

A bullet of mass m=26g is fired into a wooden block of mass M=4.7kg. The block is attached to a string of length 1.5m. The bullet is embedded in the block, causing the block to then swing. If the block reaches a maximum height of h=0.31m, what was the initial speed of the bullet?

Answer 1

Answer:

u = 449 m/s

Explanation:

Given,

Mass of the bullet, m = 26 g

Mass of the wooden block,M = 4.7 Kg

height of the block,h = 0.31 m

initial speed of the block, u = ?

Using conservation of energy

$(M+ m)gh = \dfrac{1}{2}(M+m)v^2$

$gh = \dfrac{1}{2}v^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.81\times 0.31}$

v = 2.47 m/s

Now, using conservation of momentum to calculate the speed of the bullet.

m u + M u' = (M+m)v

m u  = (M+m)v

0.026 x u  = (4.7+0.026) x 2.47

u = 449 m/s

Hence, the speed of the bullet is equal to 449 m/s.