Conservation of Momentum

WhoCares357 · #1 $Old$ November 1st, 2009, 04:07 PM

Quote:

A railroad flatcar of weight 7840 N can roll without friction along a straight horizontal track. Initially, a man of weight 596 N is standing on the car, which is moving to the right with speed 25 m/s; see the figure. What is the change in velocity of the car if the man runs to the left so that his speed relative to the car is 8.0 m/s?

I set up the equation.
[math ]m_1*v_1_i+m_2*v_2_i=m_1*v_1_f+m_2*v_2_f[/math]
[math ]0+0=\frac{596}{9.8}*8+\frac{7840}{9.8}*v_2_f[/math]

I came out with the answer -.076 m/s. Can anyone confirm this?

skeeter · #2 $Old$ November 1st, 2009, 04:56 PM

Quote:

Originally Posted by WhoCares357 $View Post$

I set up the equation.
[math ]m_1*v_1_i+m_2*v_2_i=m_1*v_1_f+m_2*v_2_f[/math]
[math ]0+0=\frac{596}{9.8}*8+\frac{7840}{9.8}*v_2_f[/math]

I came out with the answer -.076 m/s. Can anyone confirm this?

$(M+m)v_0 = Mv_{1f} + mv_{2f}$

$\frac{(M+m)v_0 - mv_{2f}}{M} = v_{1f}$

$\frac{843.6(25) - 59.6(17)}{784} = 25.61$ m/s

$\Delta v = v_{1f} - v_0 = 25.61 - 25 = 0.61$ m/s