(a) 18.3 m/s
According to the law of conservation of momentum, the total initial momentum of the system must be equal to the total final momentum, so we have
[tex]p_i = p_f\\m u = (m+M)v[/tex]
where
m = 0.0240 kg is the mass of the bullet
u = 400 m/s is the initial speed of the bullet
M = 0.5 kg is the mass of the block
v is the final speed of the block+bullet together
Solving for v, we find the velocity after the collision
[tex]v=\frac{mu}{m+M}=\frac{(0.0240 kg)(400 m/s)}{0.0240 kg+0.5 kg}=18.3 m/s[/tex]
(b) 17.0 m/s
The frictional force acting on the bullet-block system is
[tex]F_f = -\mu (m+M)g[/tex]
where
[tex]\mu = 0.30[/tex] is the coefficient of kinetic friction
The acceleration due to the frictional force, therefore, will be equal to the frictional force divided by the total mass:
[tex]a=\frac{F_f}{m+M}=\-mu g = -(0.30)(9.8 m/s^2)=-2.94 m/s^2[/tex]
The system travels for a distance of
d = 8.0 m
So we can find the final velocity using the equation:
[tex]v_f^2 = v^2 + 2ad[/tex]
where
v = 18.3 m/s is the initial velocity, found at point a). Substituting,
[tex]v_f = \sqrt{(18.3 m/s)^2+2(-2.94 m/s^2)(8.0 m)}=17.0 m/s[/tex]
(c) 2.1 m
We can use again the law of conservation of momentum:
[tex](m+M) v = (m+M+M')v'[/tex]
where
v = 17.0 m/s is the initial velocity of the initial bullet+block system
M = 2.00 kg is the mass of the second block
v' is the final velocity of the system
Solving for v',
[tex]v=\frac{(m+M)v}{m+M+M'}=\frac{(0.0240 kg+0.5 kg)(17.0 m/s)}{0.0240 kg+0.5 kg+2.00 kg}=3.5 m/s[/tex]
The acceleration of the system on the rough surface is still
[tex]a=-2.94 m/s^2[/tex]
So we can find the distance covered by using again the formula used before, and requiring that the final velocity should be zero (v''=0):
[tex]v''^2 - v'^2 = 2ad\\d=\frac{v''^2-v'^2}{2a}=\frac{0-(3.5 m/s)^2}{2(-2.94 m/s^2)}=2.1 m[/tex]
