Question

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.

Answer 1

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 \(v_f\)

Answer 2

i still dont understand this

Answer 3

\[\Large P_{\text{before}}=P_{\text{after}}\]
\[\Large m_1*v_1=(m_1+m_2)*v_f\]
\[\Large (m_1+m_2)*v_f = m_1*v_1\]
\[\Large v_f = \frac{m_1*v_1}{m_1+m_2}\]
\[\Large v_f = \frac{m_1*5}{m_1+m_2}\]
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 \(\Large v_f\)

Answer 4

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

Answer 5

@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.

Answer 6

@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:
\(v_f=\frac{m_1v_1}{m_1+m_2}=\frac{2m_2v_1}{3m_2}=\frac{2}{3}v_1=\frac{10}{3}m/s\)

Answer 7

@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!

Answer 8

@akoziol4u good comment!