Question

1. 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 conservation of momentum

The total momentum before the collision is $p_{total - before}=0.50\ kg\cdot m/s+( - 0.75\ kg\cdot m/s)= - 0.25\ kg\cdot m/s$. According to the law of conservation of momentum, the total momentum after the collision $p_{total - after}$ is equal to the total momentum before the collision, so $p_{total - after}=- 0.25\ kg\cdot m/s$.

Step2: Assume elastic - like collision (without more info)

For a one - dimensional elastic collision between two objects with masses $m_1 = 0.5\ kg$ and $m_2 = 1.5\ kg$ and initial velocities $u_1 = 1.00\ m/s$ and $u_2=-0.5\ m/s$, we can use the following equations for elastic collisions:
The velocity of the first object after the collision $v_1=\frac{(m_1 - m_2)u_1+2m_2u_2}{m_1 + m_2}$ and the velocity of the second object after the collision $v_2=\frac{(m_2 - m_1)u_2+2m_1u_1}{m_1 + m_2}$.
Substitute $m_1 = 0.5\ kg$, $m_2 = 1.5\ kg$, $u_1 = 1.00\ m/s$ and $u_2=-0.5\ m/s$ into the formula for $v_1$:
\[

$$\begin{align*} v_1&=\frac{(0.5 - 1.5)\times1.00+2\times1.5\times(-0.5)}{0.5 + 1.5}\\ &=\frac{- 1.00-1.5}{2}\\ &=\frac{-2.5}{2}=-1.25\ m/s \end{align*}$$

\]
Substitute into the formula for $v_2$:
\[

$$\begin{align*} v_2&=\frac{(1.5 - 0.5)\times(-0.5)+2\times0.5\times1.00}{0.5 + 1.5}\\ &=\frac{-0.5 + 1.00}{2}\\ &=\frac{0.5}{2}=0.25\ m/s \end{align*}$$

\]
The momentum of Ball 1 after the collision $p_1=m_1v_1=0.5\times(-1.25)=-0.625\ kg\cdot m/s$.
The momentum of Ball 2 after the collision $p_2=m_2v_2=1.5\times0.25 = 0.375\ kg\cdot m/s$.

1. For Ball 1: The velocity changed from $1.00\ m/s$ to $-1.25\ m/s$, so its speed increased and its direction reversed. For Ball 2: The velocity changed from $-0.5\ m/s$ to $0.25\ m/s$, so its speed decreased and its direction reversed.
2. For Ball 1: The momentum changed from $0.50\ kg\cdot m/s$ to $-0.625\ kg\cdot m/s$, so its magnitude increased and its direction reversed. For Ball 2: The momentum changed from $-0.75\ kg\cdot m/s$ to $0.375\ kg\cdot m/s$, so its magnitude decreased and its direction reversed.
3. According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. The total momentum before the collision is $0.50\ kg\cdot m/s+( - 0.75\ kg\cdot m/s)=-0.25\ kg\cdot m/s$, and the total momentum after the collision is $-0.625\ kg\cdot m/s+0.375\ kg\cdot m/s=-0.25\ kg\cdot m/s$. So the total momentum remained the same.

Answer:

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	0.25 m/s	0.375 kg·m/s
Total Momentum:		- 0.25 kg·m/s		- 0.25 kg·m/s

1. Ball 1: Speed increased, direction reversed. Ball 2: Speed decreased, direction reversed.
2. Ball 1: Magnitude of momentum increased, direction reversed. Ball 2: Magnitude of momentum decreased, direction reversed.
3. The total momentum remained the same.