Question

A bullet of mass m is fired into a block of mass M that is at
rest. The block, with the bullet embedded, slides distance d across
a horizontal surface. The coefficient of kinetic friction is
?_{k}.

What is the speed of a 9.0 g bullet that, when fired into a 8.0 kg stationary wood block, causes the block to slide 5.6 cm across a wood table? Assume that ?k=0.20.

Answer #1

**Given **

**mass of bullet m = 9 g**

**mass of block M = 8 kg**

**distancemoved (m+M) is s = 5.6 cm**

**coefficient of kinetic friction is mue_k =
0.20**

**due to friction the bullet and block system velocity
goes on decrease from t=0s**

**so the work done by the friction force
is **

**Ff = mue_k *(m+M)g cos 0 = 0.2*(0.009+8)*9.8cos 0 N =
15.69764 N**

**the work done by Ff is = Ff*s = 15.69764 *0.056 J =
0.87906784 J**

**this is equal to teh change in kientic
energy **

**0.5*(m+M)*V^2 = 0.87906784 J**

**0.5*(0.009+8)*V^2 = 0.87906784**

**V = 0.468529 m/s**

**this is final speed **

**from conservation of moemntum **

**m*u = (m+M)V**

**u = (m+M)V /m**

**u = (0.009+8)*0.468529 / (0.009) m/s**

**u = 416.939 m/s**

**so the speed of the bullet is u = 413.939
m/s**

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