Question

A bullet of mass 4.00 g is fired horizontally into a wooden block of mass 1.30 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.170. The bullet remains embedded in the block, which is observed to slide a distance 0.240 m along the surface before stopping. Part A What was the initial speed of the bullet

Answer 1

`291.67\ \text{m/s}`

Step-by-step explanation:

`m_1` = Mass of bullet = 4 g

`m_2` = Mass of block = 1.3 kg

`\mu` = Coefficient of friction = 0.17

`s` = Displacement of block = 0.24 m

`v_1` = Velocity of bullet

`v` = Velocity of combined mass

`g` = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

The energy balance of the system is given by

`(1)/(2)(m_1+m_2)v^2=\mu(m_1+m_2)gs\\\Rightarrow v=√(2\mu gs)`

As the momentum is conserved in the system we have

`m_1v_1=(m_1+m_2)v\\\Rightarrow m_1v_1=(m_1+m_2)√(2\mu gs)\\\Rightarrow v_1=((m_1+m_2)√(2\mu gs))/(m_1)\\\Rightarrow v_1=((4* 10^(-3)+1.3)* √(2* 0.17* 9.81* 0.24))/(4* 10^(-3))\\\Rightarrow v_1=291.67\ \text{m/s}`

The initial speed of the bullet is
`291.67\ \text{m/s}`.