Question

A 4.00 gg bullet is fired horizontally into a 1.10 kgkg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.200. The bullet remains embedded in the block, which is observed to slide 0.230 mm along the surface before stopping.

par a)

What was the initial speed of the bullet?

Express your answer in meters per second.

Answer 1

Solution) m = 4 g = 4×10^(-3) kg

M = 1.10 kg

Uk = 0.2

S = 0.230 m

(a) initial speed , v = ?

Work done , W = f.S

Here f is friction force

f = (Uk)(mg)

W = (Uk)(mg)(S)

W = Kinetic Energy (KE)

KE = (1/2)(m)(V^2)

(Uk)(mg)(S) = (1/2)(m)(V^2)

V^2 = 2(Uk)(g)(S)

V = ((2)(Uk)(g)(S))^(1/2)

V = (2×0.2×9.8×0.230)^(1/2)

V = 0.9495 m/s

Applying conservation of momentum

mv = (m + M)V

v = ((m + M)(V))/(m)

v = ((4×10^(-3) + 1.10)(0.9495))/(4×10^(-3))

v = 262.1 m/s