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A 4.10 g bullet moving at 837 m/s strikes a 820 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 467 m/s. (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

Respuesta :

Answer:

(a) 1.85 m/s

(b) 4.1 m/s

Explanation:

Data

  • bullet mass, Mb = 4.10 g
  • initial bullet velocity, Vbi = 837 m/s
  • wooden block mass, Mw = 820 g
  • initial wooden block  velocity, Vwi = 0 m/s
  • final bullet velocity, Vbf = 467 m/s

(a) From the conservation of momentum:

Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf

Mb*(Vbi - Vbf)/Mw = Vwf

4.1*(837 - 467)/820 = Vwf

Vwf = 1.85 m/s

(b) The speed of the center of mass speed is calculated as follows:

V = Mb/(Mb + Mw) * Vbi

V = 4.1/(4.1 + 820) * 837

V = 4.1 m/s

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