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A 30-g bullet is fired with a horizontal velocity of 450 m/s andbecomes embedded in block B, which has a mass of 3 kg. After theimpact, block B slides on 30-kg carrier C until it impacts the endof the carrier. Knowing the impact between B and C is perfectlyplastic and the coefficient of kinetic friction between B and C is 0.2,determine (a) the velocity of the bullet and B after the first impact,(b) the final velocity of the carrier.

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whitneytr12

Answer:

a) [tex]v_{fb+B}=4.46\: m/s[/tex]

b) [tex]v_{fc}=1.21\: m/s[/tex]

Explanation:

a) Let's use the conservation of linear momentum.

[tex]m_{b}v_{iv}=m_{B}v_{f}+m_{b}v_{f}[/tex]

Where:

  • m(b) is mass of bullet (0.03 kg)
  • v(ib) is the initial velocity of the bullet (450 m/s)
  • m(B) is the mass of the block (3 kg)
  • v(f) is the final velocity of the bullet and the block  

[tex]m_{b}v_{iv}=m_{B}v_{f}+m_{b}v_{f}[/tex]

[tex]v_{f}=\frac{m_{b}v_{iv}}{m_{B}+m_{b}}[/tex]

[tex]v_{f}=\frac{0.03*450}{0.03+3}[/tex]

[tex]v_{fb+B}=4.46\: m/s[/tex]

b) As we have an external force between B and C we can not use the conservation of linear momentum here. We need to use the work definition.

The work here is due to the friction force.

[tex]W=\Delta K[/tex]

[tex]W=K_{f}-K_{i}[/tex]

[tex]-\mu m_{B}g=0.5(m_{b}+m_{B}+m_{C})v_{f}^{2}-0.5(m_{b}+m_{B})v_{i}^{2}[/tex]

[tex]-\mu m_{B}g=0.5(m_{b}+m_{B}+m_{C})v_{f}^{2}-0.5(m_{b}+m_{B})v_{i}^{2}[/tex]                  

[tex]-0.2*3*9.81=0.5(0.03+3+30)v_{f}^{2}-0.5(0.03+3)4.46^{2}[/tex]    

[tex]v_{f}^{2}=\frac{-0.2*3*9.81+0.5(0.03+3)4.46^{2}}{0.5(0.03+3+30)}[/tex]

[tex]v_{fC}=1.21\: m/s[/tex]

I hope it helps you!

 

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