A 200 g projectile is fired with a velocity of 900 m/s toward the center of a 15 kg wooden block which rests on a rough surface.

If the projectile penetrates and emerges from the back at a velocity of 300 m/s, determine the velocity of the block just after the projectile emerges.

How long does the block slide on the rough surface after the projectile emerges before it comes to rest again?

The coefficient of kinetic friction between the block and surface is 0.2.

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex] where [tex]v_1[/tex] and [tex]v_2[/tex] are the final velocities of the fired projectile and wooden block respectively, [tex]u_1[/tex] and [tex]u_2[/tex] are initial velocities of the fired projectile and wooden block respectively, [tex]m_1[/tex] and [tex]m_2[/tex] are masses of the fired projectile and wooden block respectively

Substituting 200 g=0.2 Kg for [tex]m_1[/tex], 15 Kg for [tex]m_2[/tex], 900 m/s for [tex]u_1[/tex], o m/s for [tex]u_2[/tex] since it's at rest, 300 m/s for [tex]v_1[/tex] then

[tex]0.2 Kg\times 900 m/s + (15 Kg \times 0 m/s)=0.2 Kg \times 300 m/s + (15 Kg \times v_2)[/tex]

[tex]v_2=8 m/s[/tex]

This acts as the initial velocity

Since [tex]F=ma=\mu mg[/tex]

Therefore, [tex]a=\mu g[/tex]

Substituting [tex]\mu[/tex] with 0.2 and g with 9.81 then

[tex]a=0.2\times 9.81=1.962 m/s^{2}[/tex]

From kinematics equations

v=u-at

since v=0 then

at=u hence [tex]t=\frac {u}{a}[/tex]

Since we already obtained u as 8 m/s

[tex]t=\frac {8}{1.962}=4.077471967 s \approx 4.08 s[/tex]