Answer
The speed of the second cart after impact is $~~1.86~m/s$
Work Step by Step
In part (a), we found that the mass of the second cart is $98.7~g$
We can use conservation of momentum to find the velocity of the second cart after impact:
$p_f=p_i$
$(340~g)(0.66~m/s)+(98.7~g)~v_{2f}=(340~g)(1.2~m/s)$
$(98.7~g)~v_{2f}=(340~g)(1.2~m/s)-(340~g)(0.66~m/s)$
$v_{2f}=\frac{(340~g)(1.2~m/s)-(340~g)(0.66~m/s)}{98.7~g}$
$v_{2f} = 1.86~m/s$
The speed of the second cart after impact is $~~1.86~m/s$
