Conservation of Momentum EX 4

(a) The momentum of the truck, $p$, is:

$p=mv=(10^{4})(30)=3\times 10^{5}{\tfrac {kgm}{s}}$

The momentum of the car, $p'$, is equal to the momentum of the truck, $p$:

$p'=p=3\times 10^{5}=m'v'=(1.2\times 10^{3})v'$

Solve for $v'$:

$v'={\frac {3\times 10^{5}}{1.2\times 10^{3}}}=2.5\times 10^{2}m/s$

(b) The kinetic energy of the truck, $K$, is:

$K={\frac {1}{2}}mv^{2}={\frac {1}{2}}(10^{4})(30)^{2}=4.5\times 10^{6}J$

The kinetic energy of the car, $K'$, is equal to the kinetic energy of the truck, $K$:

$K'=K=4.5\times 10^{6}={\frac {1}{2}}m'{v'}^{2}={\frac {1}{2}}(1.2\times 10^{3}){v'}^{2}$

Solve for $v'$:

$v'={\sqrt {2\times 4.5\times 10^{6} \over 1.2\times 10^{3}}}=86.6m/s$