name: ____________ date: ____________ block: ____________\nphet simulation virtual lab: momentum and…
name: ____________ date: ____________ block: ____________\nphet simulation virtual lab: momentum and collisions\n1. navigate to the simulation at https://tinyurl.com/fbhsuwa and click \intro.\\n2. notice that the collision is elastic. (the slider bar labeled \elasticity\ is currently 100%)\n3. 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.\n4. 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)\n| | mass | velocity before the collision | momentum before the collision | velocity after the collision | momentum after the collision |\n|--|--|--|--|--|--|\n| ball 1 (blue) | 0.5 kg | 1.00 m/s | 0.50 kg·m/s | | |\n| ball 2 (pink) | 1.5 kg | -0.5 m/s | -0.75 kg·m/s | | |\n| total momentum: | | | | | |\n5. what happened to the velocity of ball 1 after they collided? what about ball 2? (answer for both of them!) (2 pts)\n6. what happened to the momentum of ball 1 after they collided? what about ball 2? (answer for both of them!) (2 pts)\n7. what happened to the total momentum after the collision? (1 pt)
Answer
Explanation:
Step1: Recall conservation laws for elastic - collisions
In elastic collisions, both momentum and kinetic energy are conserved. The equations for conservation of momentum and kinetic energy are used to find the velocities after the collision. The conservation - of - momentum equation is $m_1u_1 + m_2u_2=m_1v_1 + m_2v_2$, and for 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, $u_1$ and $u_2$ are the initial velocities, and $v_1$ and $v_2$ are the final velocities.
Step2: Calculate the total initial momentum
The total initial momentum $P_{total - initial}=m_1u_1 + m_2u_2$. Given $m_1 = 0.5\ kg$, $u_1=1.00\ m/s$, $m_2 = 1.5\ kg$, and $u_2=- 0.5\ m/s$. So, $P_{total - initial}=(0.5\times1.00)+(1.5\times(-0.5))=0.5 - 0.75=-0.25\ kg\cdot m/s$.
Step3: Solve for final velocities
From the conservation of momentum $0.5\times1.00+1.5\times(-0.5)=0.5v_1 + 1.5v_2$, which simplifies to $0.5 - 0.75 = 0.5v_1+1.5v_2$ or $-0.25 = 0.5v_1+1.5v_2$. From the conservation of kinetic energy $\frac{1}{2}\times0.5\times1.00^2+\frac{1}{2}\times1.5\times(-0.5)^2=\frac{1}{2}\times0.5\times v_1^2+\frac{1}{2}\times1.5\times v_2^2$. After simplifying and solving the system of equations (the conservation - of - momentum and conservation - of - kinetic - energy equations simultaneously), we get $v_1=-0.5\ m/s$ and $v_2 = 0.5\ m/s$.
Step4: Calculate final momenta
The final momentum of ball 1, $P_{1 - final}=m_1v_1=0.5\times(-0.5)=-0.25\ kg\cdot m/s$. The final momentum of ball 2, $P_{2 - final}=m_2v_2=1.5\times0.5 = 0.75\ kg\cdot m/s$. The total final momentum $P_{total - final}=P_{1 - final}+P_{2 - final}=-0.25 + 0.75 = 0.5\ kg\cdot m/s$.
Answer for question 5:
The velocity of ball 1 changed from $1.00\ m/s$ to $-0.5\ m/s$ (it reversed direction). The velocity of ball 2 changed from $-0.5\ m/s$ to $0.5\ m/s$ (it also reversed direction).
Answer for question 6:
The momentum of ball 1 changed from $0.50\ kg\cdot m/s$ to $-0.25\ kg\cdot m/s$. The momentum of ball 2 changed from $-0.75\ kg\cdot m/s$ to $0.75\ kg\cdot m/s$.
Answer for question 7:
The total momentum before the collision is $-0.25\ kg\cdot m/s$ and the total momentum after the collision is $0.5\ kg\cdot m/s$. There is an error in the above calculation of total initial momentum. Recalculating the total initial momentum: $P_{i}=m_1u_1 + m_2u_2=(0.5\times1)+(1.5\times(-0.5))=0.5 - 0.75=-0.25\ kg\cdot m/s$. The total final momentum $P_{f}=m_1v_1 + m_2v_2=(0.5\times(-0.5))+(1.5\times0.5)=-0.25 + 0.75 = 0.5\ kg\cdot m/s$. In a correct elastic - collision calculation, the total momentum before and after the collision should be the same. Let's correct the calculation.
Using the correct formulas for elastic collision: The formula for final velocities in one - dimensional elastic collision are $v_1=\frac{(m_1 - m_2)u_1+2m_2u_2}{m_1 + m_2}$ and $v_2=\frac{(m_2 - m_1)u_2+2m_1u_1}{m_1 + m_2}$. Substituting $m_1 = 0.5\ kg$, $m_2 = 1.5\ kg$, $u_1 = 1\ m/s$, and $u_2=-0.5\ m/s$: $v_1=\frac{(0.5 - 1.5)\times1+2\times1.5\times(-0.5)}{0.5 + 1.5}=\frac{-1-1.5}{2}=\frac{-2.5}{2}=-1.25\ m/s$ $v_2=\frac{(1.5 - 0.5)\times(-0.5)+2\times0.5\times1}{0.5 + 1.5}=\frac{-0.5 + 1}{2}=\frac{0.5}{2}=0.25\ m/s$
The momentum of ball 1 after collision $P_{1 - final}=m_1v_1=0.5\times(-1.25)=-0.625\ kg\cdot m/s$ The momentum of ball 2 after collision $P_{2 - final}=m_2v_2=1.5\times0.25 = 0.375\ kg\cdot m/s$ The total initial momentum $P_{i}=m_1u_1+m_2u_2=(0.5\times1)+(1.5\times(-0.5))=0.5 - 0.75=-0.25\ kg\cdot m/s$ The total final momentum $P_{f}=P_{1 - final}+P_{2 - final}=-0.625+0.375=-0.25\ kg\cdot m/s$
Answer for question 5:
The velocity of ball 1 changed from $1.00\ m/s$ to $-1.25\ m/s$ (reversed direction and increased in magnitude). The velocity of ball 2 changed from $-0.5\ m/s$ to $0.25\ m/s$ (reversed direction).
Answer for question 6:
The momentum of ball 1 changed from $0.50\ kg\cdot m/s$ to $-0.625\ kg\cdot m/s$. The momentum of ball 2 changed from $-0.75\ kg\cdot m/s$ to $0.375\ kg\cdot m/s$.
Answer for question 7:
The total momentum before the collision is $-0.25\ kg\cdot m/s$ and the total momentum after the collision is also $-0.25\ kg\cdot m/s$. According to the law of conservation of momentum, the total momentum in an isolated system (such as this elastic - collision system) remains constant.
Answer:
- Ball 1: velocity changed from $1.00\ m/s$ to $-1.25\ m/s$. Ball 2: velocity changed from $-0.5\ m/s$ to $0.25\ m/s$.
- 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 $0.375\ kg\cdot m/s$.
- The total momentum remained the same at $-0.25\ kg\cdot m/s$ before and after the collision.