Question

A 10 g bullet is fired into a 9.0 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. The coefficient of kinetic friction for wood sliding on wood is 0.20. What was the speed of the bullet?

Answer #1

Using force balance

Ff = Fnet

uk*m*g = m*a

a = uk*g

a = 0.2*9.81 = 1.962 m/sec^2

Now initial speed of the (block + bullet) will be

Vf^2 = Vi^2 + 2*a*d

Vf = 0

Vi = sqrt (2*a*d)

Vi = sqrt (2*1.962*5*10^-2)

Vi = 0.443 m/sec

Now using momentum conservation

Pi = Pf

Mb*Vb + mb*vb = (Mb + mb)*Vi

Mb = mass of block = 9 kg

mb = mass of bullet = 10 gm = 0.01 kg

Vb = initial speed of block = 0 m/sec

vb = initial speed of bullet = ?

vb = (Mb + mb)*Vi/mb

Vb = (9 + 0.01)*0.443/0.01 = 399.143 m/sec

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