A man (weighing915N) stands on a long railroad flatcar (weighing2415N) as it rolls at $\mathbf{18}\mathbf{.}\mathbf{2}\frac{\mathbf{m}}{\mathbf{s}}$in the positive direction of an xaxis, with negligible friction, then the man runs along the flatcar in the negative xdirection at$\mathbf{4}\mathbf{.}\mathbf{00}\frac{\mathbf{m}}{\mathbf{s}}$relative to the flatcar. What is the resulting increase in the speed of the flatcar?

i) Speed of the man is${\mathrm{v}}_{\mathrm{mf}}=4.0\frac{\mathrm{m}}{\mathrm{s}}$

ii) Speed of the flat car is ${\mathrm{v}}_c=18.2\frac{\mathrm{m}}{\mathrm{s}}$

iii) Weight of the man is w=915 kg

iv) Weight of the car is W=2415 kg

Using the law of conservation of momentum and writingthefinal speed of the car in terms of final speed of man and car, find the final speed of the flatcar and thentheresulting increase in the speed of the flatcar.

Consider the relation between the final and the initial momentum as:

$P_i=P_f$

Here, $P_i$,and $P_f$are initial and final momentum.

From conservation of momentum solve as:

$\left(\frac{W}{g}+\frac{w}{g}\right)v_c=\frac{W}{g}v+\frac{w}{g}\left(v-v_{mf}\right)$

Multiply above equation by ‘g’,

$(W+w)v_f=Wv+w(v-v_{m}f)$

Substittue the values and solve as:

$(2415\mathrm{kg}+915\mathrm{kg})18.2\frac{\mathrm{m}}{\mathrm{s}}=2415\mathrm{v}+915\left(\mathrm{v}-4\frac{\mathrm{m}}{\mathrm{s}}\right)$

$60606\frac{m}{s}=2415v+915v-3660\frac{m}{s}$

$60606\frac{m}{s}=3330v-3660\frac{m}{s}$

$3330v=64266\frac{m}{s}$

$v=19.299\frac{m}{s}$

So, increase in the speed of the flat car is,

$△v=v-v_c$

$△v=19.30-18.2$

$△v=1.1\hspace{0.33em}\frac{m}{s}$

Hence, theresulting increase in the speed of the flatcar is$1.1\frac{\mathrm{m}}{\mathrm{s}}$

Therefore, the increase in the speed of an object can be found using the law of conservation of momentum.