Question

4. Object A has a mass of 10 kg and is moving at 5 m/s. Object B has a mass of 8 kg and is moving at 2 m/s. They then collide, and object B continues moving with a velocity of 4 m/s after the collision.a. What is the total initial momentum of object A and B?b. What should be the total final momentum of object A and B?c. What is the final velocity of object A?

Answer 1

Answer:

a. 66 kg m/s

b. 66 kg m/s

c. 3.4 m/s

Explanation:

Part a.

The total initial momentum of objects A and B is calculated as

$p_i=m_Av_{Ai}+m_Bv_{Bi}$

Where m is the mass and v is the initial velocity for each object. Replacing the values, we get:

$\begin{gathered} m_A=10kg \\ v_{Ai}=5\text{ m/s} \\ m_B=8kg \\ v_{Bi}=2m/s_{} \\ p_i=(10kg)(5m/s)+(8kg)(2m/s) \\ p_i=50kg\text{ m/s + 16 }kg\text{ m/s} \\ p_i=66kg\text{ m/s} \end{gathered}$

Therefore, the total initial momentum of objects A and B is 66 kg m/s

Part b.

By the conservation of momentum, the total final momentum is equal to the total initial momentum, so

$\begin{gathered} p_f=p_i \\ p_f=66kg\text{ m/s} \end{gathered}$

Therefore, the total final momentum of objects A and B is 66 kg m/s.

Part c.

The final momentum is also equal to:

$p_f=m_Av_{fA}+m_Bv_{fB}$

Solving for the final velocity of object A, we get:

$\begin{gathered} p_f-m_Bv_{Bf}=m_Av_{fA} \\ v_{fA}=\frac{p_f-m_Bv_{Bf}}{m_A} \end{gathered}$

Then, we can replace the values to get:

$\begin{gathered} p_f=66kg\text{ m/s} \\ m_A=10kg \\ m_B=8kg_{} \\ v_{Bf}=4m/s \\ v_{fA}=\frac{66kg\text{ m/s - (8}kg)(4m/s)}{10kg} \\ v_{fA}=\frac{66kg\text{ m/s - 32 }kg\text{ m/s}}{10kg} \\ v_{fA}=\frac{34kg\text{ m/s}}{10kg} \\ v_{fA}=3.4m/s \end{gathered}$

Therefore, the final velocity of object A is 3.4 m/s