Answer
a) $a_1=-10.71m/s^2$
b) $a_2=6m/s^2$
c) $t=0.078s$
d) $v_f=0.468m/s$
e) $(m_1+m_2)v_f=0.182Kg.m/s$
Work Step by Step
(a) We know that
$m_1a_1=-0.15N$
$\implies a_1=\frac{-1.5N}{0.14Kg}$
$\implies a_1=-10.71m/s^2$
(b) The acceleration of car 2 can be determined as
$m_2a_2=F$
$\implies a_2=\frac{F}{m_2}$
We plug in the known values to obtain:
$a_2=\frac{1.5N}{0.25Kg}$
$a_2=6m/s^2$
(c) We know the speed of cart 1 after collision is given as
$v_1=v_{\circ}+a_1t$
$\implies v_1=1.3+(-10.7)t$....eq(1)
and the speed of cart 2 after collision is given as
$v_2=v_{\circ}+a_2t$
$v_2=0+6t$....eq(2)
As the speeds of both cars are equal
$\implies 1.3+(-10.7)t=6t$
$\implies t=\frac{1.3}{16.7}$
$t=0.078s$
(d) We know that
$v_f=a_2t$
We plug in the known values to obtain:
$v_f=(6m/s^2)(0.078s)$
$v_f=0.468m/s$
(e) As $m_1v_{\circ}=(0.14kg)(1.3m/s)$
$\implies m_1v_{\circ}=0.182Kg.m/s$
Now $(m_1+m_2)v_f=(0.14Kg+0.25Kg)(0.4678m/s)$
$(m_1+m_2)v_f=0.182Kg.m/s$
Thus, we proved that $m_1v_{\circ}=(m_1+m_2)v_f$
