Question

Students were experimenting with objects in a collision. Ball A moves with a constant acceleration by sliding down a frictionless incline plane before colliding with Ball B. Students used motion detectors to measure the velocity of Ball A at various points. Ball A (mass of 1.00 kg) began from rest, at position 1, on a ramp at a height of 1.25 m. At position 2, ball A was moving at 5.00 m/s. Ball A continues to roll at a constant 5.00 m/s when it collides with Ball B (mass of 1.00 kg) at position 3. Ball A comes to a complete stop. Ball B moves at a constant velocity after the collision.FORMULAS: PE = m•g•h (g=9.8m/s²) KE = 1/2•m•v² momentum = m•v

Answer 1

Given data:

* The mass of ball A is m_1 = 1 kg.

* The mass of ball B is m_2 = 1 kg.

* The initial velocity of ball A is u_1 = 5 m/s.

* The final velocity of ball A is v_1 = 0 m/s.

* The initial velocity of the ball B is u_2 = 0 m/s.

Solution:

According to the law of conservation of momentum, the net momentum of the system before the collision is equal to the net momentum of the system after the collision.

Thus,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

Substituting the known values,

`\begin{gathered} 1*5+1*0=1*0+p_2 \\ 5+0=0+p_2 \\ p_2=5\text{ kgm/s} \end{gathered}`

where p_2 is the momentum of the ball B,

Thus, the momentum of ball B after the collision is 5 kgm/s.

Hence, the third option is the correct answer.