Question

11, Glider A, with a mass of 0.355 kg, moves along a frictionless air

track with a velocity of 0.095 m/s, as in Figure 9-7. It collides
with glider B, with a mass of 0.710 kg and a speed of 0.045 m/s
in the same direction, After the collision, glider A continues in
the same direction at 0.035 m/s. What is the speed of glider B
MA

Answer 1

The collision confirms linear momentum conservation, with glider B's post-collision velocity calculated as approximately 0.075 m/s, validating the principle within the system.

The problem involves the conservation of linear momentum, which states that the total momentum of an isolated system remains constant before and after a collision. Mathematically, this is expressed as:

m_A * v_A + m_B * v_B = m_A * v_A' + m_B * v_B'

Given the masses and velocities of gliders A and B before and after the collision, we can substitute these values into the equation.

(0.355 kg * 0.095 m/s) + (0.710 kg * 0.045 m/s) = (0.355 kg * 0.035 m/s) + (0.710 kg * v_B')

Now, solving for v_B', we get:

0.033725 kg*m/s + 0.03195 kg*m/s = 0.012425 kg*m/s + 0.710 kg * v_B'

Combine like terms:

0.065675 kg*m/s = 0.710 kg * v_B'

Finally, solve for v_B':

v_B' = 0.065675 kg*m/s / 0.710 kg ≈ 0.075 m/s

Therefore, the velocity of glider B after the collision is approximately 0.075 m/s. This result confirms the conservation of linear momentum in the system.