## highschoolmom2010 one year ago If a stationary cart is struck head-on by a cart with twice mass of the stationary one and a velocity of 5 m/s, what will be the new velocity of the stationary cart if the collision is inelastic? Show all calculations leading to an answer.

1. anonymous

in general a collision problem can be solved considering the conservation of the total momentum: $$\Delta P_{system}=0$$ if this is an "extreme inelastic collision", the colliding objects stick together after the collision, so we can express the momentum of the system before the collision as: $P_{before}=m_1v_1$ where $$m_1$$ is the moving cart, the stationary cart does not appear because $$v_2=0$$ the momentum after the collision is: $P_{after}=(m_1+m_2)v_f$ because the carts now form one mass together. You can now equal $$P_{before}=P_{after}$$ and solve to find the final velocity $$v_f$$

2. highschoolmom2010

i still dont understand this

3. jim_thompson5910

$\Large P_{\text{before}}=P_{\text{after}}$ $\Large m_1*v_1=(m_1+m_2)*v_f$ $\Large (m_1+m_2)*v_f = m_1*v_1$ $\Large v_f = \frac{m_1*v_1}{m_1+m_2}$ $\Large v_f = \frac{m_1*5}{m_1+m_2}$ The only thing wrong here is that we don't know the mass of either cart. So we can't find a numeric value of the final velocity $$\Large v_f$$

4. anonymous

to make an example, note that if we assume $$m_1=m_2$$, both cars have equal mass, we get a value for $$v_f$$: $$v_f=\frac{v_1}{2}=2.5m/s$$

5. highschoolmom2010

@jim_thompson5910 that is what i didnt understand because we dont know the mass of either cart all is said was the moving cart was 2 times that of the stationary one.

6. anonymous

@highschoolmom2010 if the moving cart has twice the mass than the stationary one, that means: $$m_1=2m_2$$, so we can use that in the expression for the final velocity and get: $$v_f=\frac{m_1v_1}{m_1+m_2}=\frac{2m_2v_1}{3m_2}=\frac{2}{3}v_1=\frac{10}{3}m/s$$

7. anonymous

@highschoolmom2010 This type of question shows us that we don't necessarily need to know values of every variable in our equation IF INSTEAD we have a way to compare one variable to the other. In this case saying $m_1 = 2*m_2$ allows us to replace two unknown variables with just one unknown variable. $v_f = (m_1*v_1)/(m_1+m_2) = (2*m_2*v_1)/(2*m_2+m_2) = (2*m_2*v_1)/(3*m_2)$ This seems counterintuitive at first but if we remember that there are lots of little algebraic math tricks that allow us to cancel out a variable like $m_2$ then the problem gets a lot simpler. $v_f = (2*v_1)/3$ Remember this for future problems in physics class there are many times this sort of variable replacement makes our lives easier!

8. anonymous

@akoziol4u good comment!