We can solve this problem using the principle of conservation of momentum. Before the bullet strikes the bag, the total momentum of the system is zero because the bag is at rest. After the bullet strikes the bag and exits, the momentum of the system is still zero because the bullet and bag are moving together.
Let's define the positive direction as upward. Then, the initial momentum of the bullet is:
p_i = m_bullet * v_bullet = 0.02 kg * 650 m/s = 13 kg m/s
After the collision, the bullet and bag move together as a single system. Let's assume that the final velocity of the bullet-bag system is v_f, and the mass of the bullet and bag combined is:
m_total = m_bullet + m_bag = 0.02 kg + 5 kg = 5.02 kg
Using the conservation of momentum, we can equate the initial and final momenta:
p_i = p_f
m_bullet * v_bullet = m_total * v_f
Solving for v_f, we get:
v_f = (m_bullet * v_bullet) / m_total
v_f = (0.02 kg * 650 m/s) / 5.02 kg
v_f = 2.6 m/s
The final velocity of the bullet-bag system after the collision is 2.6 m/s. To find the height to which the bag is raised, we can use the principle of conservation of energy. The initial energy of the system is all gravitational potential energy stored in the bag, and the final energy of the system is the sum of the gravitational potential energy of the raised bag and the kinetic energy of the bullet-bag system:
m_bag * g * h = (1/2) * m_total * v_f^2
where g is the acceleration due to gravity (9.81 m/s^2), and h is the height the bag is raised.
Solving for h, we get:
h = [(1/2) * m_total * v_f^2] / (m_bag * g)
h = [(1/2) * 5.02 kg * (2.6 m/s)^2] / (5 kg * 9.81 m/s^2)
h = 0.337 m
Therefore, the bag is raised to a vertical height of about 0.337 meters when hit by the bullet.