(a) 1.49 m/s
The conservation of momentum states that the total initial momentum is equal to the total final momentum:
[tex]p_i = p_f\\m u_b + M u_B = (m+M)v[/tex]
where
m = 0.016 kg is the mass of the bullet
[tex]u_b = 280 m/s[/tex] is the initial velocity of the bullet
M = 3 kg is the mass of the block
[tex]u_B = 0[/tex] is the initial velocity of the block
v = ? is the final velocity of the block and the bullet
Solving the equation for v, we find
[tex]v=\frac{m u_b}{m+M}=\frac{(0.016 kg)(280 m/s)}{0.016 kg+3 kg}=1.49 m/s[/tex]
(b) Before: 627.2 J, after: 3.3 J
The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):
[tex]K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.016 kg)(280 m/s)^2=627.2 J[/tex]
The final kinetic energy is the kinetic energy of the bullet+block system after the collision:
[tex]K_f = \frac{1}{2}(m+M)v^2=\frac{1}{2}(0.016 kg+3 kg)(1.49 m/s)^2=3.3 J[/tex]
(c) The Energy Principle isn't valid for an inelastic collision.
In fact, during an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not: this means that part of the kinetic energy of the system is losted in the collision. The principle of conservation of energy, however, is still valid: in fact, the energy has not been simply lost, but it has been converted into other forms of energy (thermal energy).
(a) The final speed of the block after the collision is 1.485 m/s.
(b) The kinetic energy before the collision is 627.2 J and The total kinetic energy of the system after the collision is 3.33 J.
(c) The difference in the two kinetic energy is due to energy lost to frictional force during the collision.
The final speed of the block after the collision is determined by applying principle of conservation of linear momentum.
m₁u₁ + m₂u₂ = v(m₁+ m₂)
0.016(280) + 3(0) = v(0.016 + 3)
4.48 = 3.016v
v = 4.48/3.016
v = 1.485 m/s
The kinetic energy before the collision is calculated as follows;
K.E i = ¹/₂mv²
K.Ei = 0.5 x 0.016 x 280²
K.Ei = 627.2 J
The total kinetic energy of the system after the collision is calculated as follows;
K.Ef = ¹/₂(m1 + m2) v²
K.Ef = ¹/₂(0.016 + 3) 1.485²
K.Ef = 3.33 J
The difference in the two kinetic energy is due to energy lost to frictional force during the collision.
Learn more about inelastic collision here: https://brainly.com/question/7694106
