Question

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?

Answer 1

(a) 1.85 m/s

(b) 4.1 m/s

Step-by-step 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