Question

phet simulation virtual lab: momentum and collisions

1. navigate to the simulation at https://tinyurl.com/fbhsuwa and click \intro.\
2. notice that the collision is elastic. (the slider bar labeled \elasticity\ is currently 100%)
3. click \more data\ so you can see details about each ball. in this lab assignment, the position isnt super important, but the mass, velocity, and momentum are very important.
4. dont change any of the values yet! fill out the table below. youll have to press \play\ to see the values after the collision. (2 pts)

	mass	velocity before the collision	momentum before the collision	velocity after the collision	momentum after the collision
ball 2 (pink)	1.5 kg	-0.5 m/s	-0.75 kg m/s
total momentum:

1. what happened to the velocity of ball 1 after they collided? what about ball 2? (answer for both of them!) (2 pts)
2. what happened to the momentum of ball 1 after they collided? what about ball 2? (answer for both of them!) (2 pts)
3. what happened to the total momentum after the collision? (1 pt)

Explanation:

Step1: Recall elastic - collision principles

In an elastic collision, both momentum and kinetic - energy are conserved. The equations for conservation of momentum and kinetic - energy for two - body collision are:
Conservation of momentum: $m_1u_1 + m_2u_2=m_1v_1 + m_2v_2$
Conservation of kinetic - energy: $\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$
where $m_1$ and $m_2$ are the masses of the two balls, $u_1$ and $u_2$ are the initial velocities, and $v_1$ and $v_2$ are the final velocities.
Given $m_1 = 0.5\ kg$, $u_1=1.00\ m/s$, $m_2 = 1.5\ kg$, $u_2=- 0.5\ m/s$.

Step2: Calculate total initial momentum

$P_{total - initial}=m_1u_1 + m_2u_2=(0.5\times1.00)+(1.5\times(-0.5))=0.5 - 0.75=-0.25\ kg\cdot m/s$

Step3: Solve for final velocities

From conservation of momentum: $0.5\times1.00+1.5\times(-0.5)=0.5v_1 + 1.5v_2$
$0.5 - 0.75 = 0.5v_1+1.5v_2$
$-0.25 = 0.5v_1+1.5v_2$
From conservation of kinetic - energy:
$\frac{1}{2}\times0.5\times(1.00)^2+\frac{1}{2}\times1.5\times(-0.5)^2=\frac{1}{2}\times0.5\times v_1^2+\frac{1}{2}\times1.5\times v_2^2$
$0.25 + 0.1875=0.25v_1^2 + 0.75v_2^2$
$0.4375=0.25v_1^2 + 0.75v_2^2$
Solving the system of equations $-0.25 = 0.5v_1+1.5v_2$ (or $v_1=-0.5 - 3v_2$) and $0.4375=0.25v_1^2 + 0.75v_2^2$:
Substitute $v_1=-0.5 - 3v_2$ into $0.4375=0.25v_1^2 + 0.75v_2^2$
$0.4375=0.25(-0.5 - 3v_2)^2+0.75v_2^2$
$0.4375=0.25(0.25 + 3v_2+9v_2^2)+0.75v_2^2$
$0.4375 = 0.0625+0.75v_2 + 2.25v_2^2+0.75v_2^2$
$0.4375=0.0625+0.75v_2 + 3v_2^2$
$3v_2^2+0.75v_2 - 0.375 = 0$
Using the quadratic formula $v_2=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ with $a = 3$, $b = 0.75$, $c=-0.375$
$v_2=\frac{-0.75\pm\sqrt{(0.75)^2-4\times3\times(-0.375)}}{2\times3}=\frac{-0.75\pm\sqrt{0.5625 + 4.5}}{6}=\frac{-0.75\pm\sqrt{5.0625}}{6}=\frac{-0.75\pm2.25}{6}$
We get two solutions for $v_2$. Taking the physical - relevant solution $v_2 = 0.25\ m/s$
Substitute $v_2 = 0.25\ m/s$ into $v_1=-0.5 - 3v_2$ gives $v_1=-1.25\ m/s$

Step4: Calculate final momenta

For Ball 1: $P_1=m_1v_1=0.5\times(-1.25)=-0.625\ kg\cdot m/s$
For Ball 2: $P_2=m_2v_2=1.5\times0.25 = 0.375\ kg\cdot m/s$
$P_{total - final}=P_1 + P_2=-0.625+0.375=-0.25\ kg\cdot m/s$

5.
The velocity of Ball 1 changed from $1.00\ m/s$ to $-1.25\ m/s$. Its direction reversed and its magnitude increased.
The velocity of Ball 2 changed from $-0.5\ m/s$ to $0.25\ m/s$. Its direction reversed and its magnitude decreased.
6.
The momentum of Ball 1 changed from $0.50\ kg\cdot m/s$ to $-0.625\ kg\cdot m/s$. Its direction reversed and its magnitude increased.
The momentum of Ball 2 changed from $-0.75\ kg\cdot m/s$ to $0.375\ kg\cdot m/s$. Its direction reversed and its magnitude decreased.
7.
The total momentum before the collision was $-0.25\ kg\cdot m/s$ and the total momentum after the collision was also $-0.25\ kg\cdot m/s$. So, the total momentum remained the same.

Answer:

1. Ball 1: Velocity changed from $1.00\ m/s$ to $-1.25\ m/s$. Ball 2: Velocity changed from $-0.5\ m/s$ to $0.25\ m/s$.
2. Ball 1: Momentum changed from $0.50\ kg\cdot m/s$ to $-0.625\ kg\cdot m/s$. Ball 2: Momentum changed from $-0.75\ kg\cdot m/s$ to $0.375\ kg\cdot m/s$.
3. The total momentum remained the same ($-0.25\ kg\cdot m/s$ before and after the collision).