Two hard steel balls collide in a perfectly elastic collision. The first ball has a mass of 1kg and is traveling at 5m/s east. The second ball has a mass of 2kg and is initially traveling at 7 m/s west along the same path so they collide head on. After the collision the second ball is traveling at 1m/s east. What is the velocity of the 1st ball after the collision?

**How do I solve and show the work for this? :/ Thanks!!

Question

Two hard steel balls collide in a perfectly elastic collision. The first ball has a mass of 1kg and is traveling at 5m/s east. The second ball has a mass of 2kg and is initially traveling at 7 m/s west along the same path so they collide head on. After the collision the second ball is traveling at 1m/s east. What is the velocity of the 1st ball after the collision?

**How do I solve and show the work for this? :/ Thanks!!

michele_laino · Answer

here we have to apply the law of conservation of total momentum
and the law of conservation of kinetic energy, since the collision is elastic

michele_laino · Answer

the situation, before collision is:
|dw:1433610623171:dw|

anonymous · Answer

ok!

michele_laino · Answer

so, total momentum is:
$$\Large {m_1}{v_1} - {m_2}{v_2} = 1 \times 5 - 2 \times 7 = ...$$

anonymous · Answer

ok! so we get 5-14? = -9 ?

michele_laino · Answer

minus sign, means that the vector of the resultant momentum is oriented towards west

anonymous · Answer

ohh okay! so -9, meaning 9 m/s to the west is the velocity of the 1st ball after collision?

michele_laino · Answer

no, -9 is the magnitude and sign of the resultant momentum

anonymous · Answer

ohh okay! so it's like this?

5-14=-9 =resultant momentum ?

michele_laino · Answer

yes!
now total kinetic energy is:
$$\large KE = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 = \frac{1}{2} \times 1 \times 25 + \frac{1}{2} \times 2 \times 49 = ...$$

anonymous · Answer

ok! so we get this? 61.5 ?

michele_laino · Answer

ok! correct!

michele_laino · Answer

now let's consider the situation after the collision

anonymous · Answer

ohhh so 61.5 is just the KE right? so we solve for velocity now? :/

michele_laino · Answer

|dw:1433611229406:dw|

michele_laino · Answer

no, we have to apply the conservation of total momentum first

anonymous · Answer

oh ok! how do we do that?

michele_laino · Answer

I call with u the speed of the first ball after collision

anonymous · Answer

ok!

michele_laino · Answer

so we can write:
conservation of total momentum:
$$\Large  \begin{gathered}
  {m_1}u + {m_2}v{'_2} =  - 9 \hfill \
  1 \times u + 7 \times 1 =  - 9 \hfill \ 
\end{gathered} $$

michele_laino · Answer

what is u?

anonymous · Answer

-16?

michele_laino · Answer

in my reasoning i have considered positive velocity if the velocity is towards east, and negative velocity if that velocity is towards west

anonymous · Answer

oh! so u=-16?

anonymous · Answer

16 m/s to the west?

michele_laino · Answer

correct!
that means the first ball, after the collision is goinf towards west

michele_laino · Answer

going*

anonymous · Answer

ahh so that is our solution? :O

michele_laino · Answer

yes!

anonymous · Answer

yay!! thank you!:D

michele_laino · Answer

:)