MHB Calculating Kinetic Energy After Collision of Balls A & B

[Shah 72]

Ball A of mass 2kg, is moving in a straight line at 5 m/s. Ball B of mass 4kg is moving in the same straight line at 2 m/s. Ball B is traveling directly towards Ball A. The balls hit each other and after the impact each ball has reversed its direction of travel. The kinetic energy lost in the impact is 12.5J

a) show that the speed of ball A after the impact is 3/10 m/s
b) Find the speed of ball B after the impact.
I don't understand how to calculate this.
Pls help

 

[topsquark]

Ball A of mass 2kg, is moving in a straight line at 5 m/s. Ball B of mass 4kg is moving in the same straight line at 2 m/s. Ball B is traveling directly towards Ball A. The balls hit each other and after the impact each ball has reversed its direction of travel. The kinetic energy lost in the impact is 12.5J

a) show that the speed of ball A after the impact is 3/10 m/s
b) Find the speed of ball B after the impact.
I don't understand how to calculate this.
Pls help

Momentum is virtually always conserved in impacts and you can generally take that to be an assumption.

Say Ball A is initially moving to the right and B is initially moving to the left. I'm going to define a +x axis to the right. So [math]v_{0A} = 5 \text{ m/s}[/math] and [math]v_{0B} = --2 \text{ m/s}[/math]. I'm going to label [math]v_A[/math] as negative in the equations since it is moving to the left after the collision and [math]v_B[/math] as positive. (Always write down your coordinate system!) The momentum equation, along with conservation of energy says
[math]\begin{cases} m_A v_{0A} + m_B v_{0B} = m_A (-v_A) + m_B v_B \\ \dfrac{1}{2} m_A v_{0A}^2 + \dfrac{1}{2} m_B v_{0B}^2 + 12.5 \text{ J} = \dfrac{1}{2} m_A (-v_A)^2 + m_B v_B^2 \end{cases}[/math]

or
[math]\begin{cases}2 \cdot 5 + 4 \cdot (-2) = -2 v_A + 4 v_B \\ 5^2 + 2 (-2)^2 + 12.5 = v_A^2 + 2 v_B^2 \end{cases}[/math]

or
[math]\begin{cases}2 = -2 v_A + 4 v_B \\ 45.5 = v_A^2 + 2 v_B^2 \end{cases}[/math]

Let's see if you can do it from here. My strategy would be to solve the momentum equation for [math]v_A[/math] (to avoid any unneccessary fractions) and plug that into the energy equation. (As a check I get [math]v_A = 5.154 \text{ m/s}[/math] and [math]v_B = 3.077 \text{ m/s}[/math].)

-Dan

Addendum: For the record what I did above abuses the notation a bit. What I really should have done was to ignore what A and B did after the collision and simply set things up as if they were positive. Then at the end I'd choose the solution that had [math]v_A[/math] negative and [math]v_B[/math] positive. But I'm betting that's not how your instructor would have set it up.

 

[DaalChawal]

Apply Momentum conservation and KE conservation.
How to delete a post? @admin

 

[skeeter]

I get A’s post collision speed as 10/3 … :unsure:

 

[Shah 72]

Momentum is virtually always conserved in impacts and you can generally take that to be an assumption.

Say Ball A is initially moving to the right and B is initially moving to the left. I'm going to define a +x axis to the right. So [math]v_{0A} = 5 \text{ m/s}[/math] and [math]v_{0B} = --2 \text{ m/s}[/math]. I'm going to label [math]v_A[/math] as negative in the equations since it is moving to the left after the collision and [math]v_B[/math] as positive. (Always write down your coordinate system!) The momentum equation, along with conservation of energy says
[math]\begin{cases} m_A v_{0A} + m_B v_{0B} = m_A (-v_A) + m_B v_B \\ \dfrac{1}{2} m_A v_{0A}^2 + \dfrac{1}{2} m_B v_{0B}^2 + 12.5 \text{ J} = \dfrac{1}{2} m_A (-v_A)^2 + m_B v_B^2 \end{cases}[/math]

or
[math]\begin{cases}2 \cdot 5 + 4 \cdot (-2) = -2 v_A + 4 v_B \\ 5^2 + 2 (-2)^2 + 12.5 = v_A^2 + 2 v_B^2 \end{cases}[/math]

or
[math]\begin{cases}2 = -2 v_A + 4 v_B \\ 45.5 = v_A^2 + 2 v_B^2 \end{cases}[/math]

Let's see if you can do it from here. My strategy would be to solve the momentum equation for [math]v_A[/math] (to avoid any unneccessary fractions) and plug that into the energy equation. (As a check I get [math]v_A = 5.154 \text{ m/s}[/math] and [math]v_B = 3.077 \text{ m/s}[/math].)

-Dan

Addendum: For the record what I did above abuses the notation a bit. What I really should have done was to ignore what A and B did after the collision and simply set things up as if they were positive. Then at the end I'd choose the solution that had [math]v_A[/math] negative and [math]v_B[/math] positive. But I'm betting that's not how your instructor would have set it up.

Thanks!

 

[Shah 72]

I get A’s post collision speed as 10/3 … :unsure:

The ans in the textbook also is 10/3. Can you pls tell me how did you get the ans

 

[Shah 72]

I get A’s post collision speed as 10/3 … :unsure:

Thank you I got the ans!

 

[topsquark]

I get A’s post collision speed as 10/3 … :unsure:

Where did I make the goof?

-Dan

 

[Shah 72]

Where did I make the goof?

-Dan

The loss is taken as positive so you add it to the impact

 

[topsquark]

The loss is taken as positive so you add it to the impact

On the wrong side of the equation. Oops!

Thanks for letting me know!

-Dan