A 4 g bullet is fired horizontally with a speed of 300m/s into 0.8 kg block of wood at rest on a table. If the coefficient of friction between block and table is 0.3, how far will the block slide approximately?

momentum:
p₁ = m₁v₁
p₂ = m₂v₂

p₃ = (m₁+ m₂) v

p₁ + p₂ = p₃

v = p₁ + p₂ / (m₁ + m₂)

friction:
F = μ(m₁ + m₂)g = (m₁ + m₂)a ⇒ a = μg

equations of motion:
v = at = μgt ⇒ t = v/μg
x = 1/2 at² = 1/2 μg * v²/(μg)² = 1/2 v² / μg

v₁ = 300 m/s
v₂ = 0 m/s
m₁ = 0.004 kg
m₂ = 0.8 kg
μ = 0.3
g = 9.81 m/s²

x = displacement

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aristocles aristocles · Answer 2 · 2019-08-18T05:56:13+02:00

Answer:

d = 3784.5 m

Explanation:

First bullet goes into the block and fixed into it

so here the speed of the combined is given by momentum conservation

so it is given as

[tex]mv_o = (m + M) v[/tex]

here we know that

[tex]m = 4 g = 0.004 kg[/tex]

[tex]M = 0.8 kg[/tex]

[tex]v_o = 300 m/s[/tex]

now from above formula

[tex](0.004)(300) = (0.004 + 0.8) v[/tex]

[tex]v = 149.25 m/s[/tex]

now the coefficient of friction on the floor is given as

[tex]\mu = 0.3[/tex]

so the deceleration is given as

[tex]a = -\mu g[/tex]

[tex]a = -(0.3)9.81[/tex]

[tex]a = -2.943 m/s^2[/tex]

now from equation of kinematics we know that

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 149.25^2 = 2(-2.943)d[/tex]

[tex]d = 3784.5 m[/tex]