Question

A bullet with a mass m = 12.5 g and speed v = 97.4 m/s is fired into a wooden block with M = 113 g which is initially at rest on a horizontal surface. The bullet is embedded into the block. The block-bullet combination slides across the surface for a distance d before stopping due to friction between the block and surface. The coefficients of friction are µs = 0.753 and µk = 0.659.(a) What is the speed (m/s) of the block-bullet combination immediately after the collision?(b) What is the distance d (m)?

Answer 1

Step-by-step explanation:

`m_1=12.5 gm`

initial speed
`v_1=97.4 m/s`

Mass of block
`m_2=113 gm`

Initial momentum
`P_i=m_1\cdot v_1`

Final Momentum
`P_f=(m_1+m_2)\cdot v_2`

Conserving momentum

`P_i=P_f`

`m_1* v_1=(m_1+m_2)v_2`

`v_2=(m_1)/(m_1+m_2)* v_1`

`v_2=(12.5)/(12.5+113)* 97.4`

`v_2=9.701 m/s`

(b)Frictional Force on combined mass

`F_(r)=\mu _kN`

`F_(r)=0.659* (m_1+m_2)\cdot g`

`F_r=0.659* (0.0125+0.113)\cdot 9.8`

`F_r=0.81 N`

acceleration a=\frac{F_r}{m_1+m_2}[/tex]

`a=(0.81)/(0.1255)`

`a=6.45 m/s^2` (deceleration)

using
`v^2-u^2=2 as`

`0-(9.701)^2=2\cdot (-6.45)\cdot s`

`s=(94.109)/(12.9)`

`s=7.295 m`