Question

A 500 gram bullet is fired at 550 m/s into a 3.5 kg wooden block at rest. The velocity of block afterward is 65 m/s. The bullet passes through the block and emerges with what velocity?

Answer 1

Answer:

this is a answer is 1050 this is correct. answee

Answer 2

Given:

$Mass of bullet, m_{b} = 500gm = 500\times 10^{-3} \\Velocity of the bullet, u_{b} = 550 m/s\\Mass of wooden, m_{w} = 3.5kg\\Velocity of the wooden, u_{w} = 0 m/s\\\\Velocity of the block afterward, v_{ba} = 65m/s$

Find: The bullet passes through the block and emerges with what velocity

Step by Step Solution:

So, the conservation of the momentum,

We have the expression of the conservation of the momentum,

$m_{b} u_{b} + m_{w} u_{w} = m_{b} v_{b} + m_{w} v_{w}$

$500\times 10^{-3}\times 550 + 3.5\times 0 = 500\times 10^{-3} \times v_{b} + 3.5\times 65\\\\275 = 500\times 10^{-3} \times v_{b} +227.5 \\\\275 - 227.5 = 500\times 10^{-3} \times v_{b}\\47.5 = 500\times 10^{-3} \times v_{b}\\\\\\v_{b} = \dfrac{47.5}{500\times 10^{-3}} \\\\v_{b} = 0.095\times 10^{3}\\\\v_{b} = 95m/s$

So, the bullet passes through the block and emerges at the velocity of 95m/s.