91Ó°ÊÓ

Given that a 0.150-kg glider is moving to the right with a speed of 0.80m/s on a frictionless, horizontal air track. The glider has a head-on collision with a 0.300-kg glider that is moving to the left with a speed of 2.20 m/s.

Mass of first glider${\mathrm{m}}_1=0.15\mathrm{kg}$

Mass of second glider${\mathrm{m}}_2=0.3\mathrm{kg}$

Initial speed of first glider${\mathrm{u}}_1=0.8\mathrm{m}/\mathrm{s}$

Initial speed of second glider${\mathrm{u}}_2=2.2\mathrm{m}/\mathrm{s}$

Conservation of momentum states the if no external force is acting, then total momentum of the system is conserved.

After collision, let ${\mathrm{v}}_1\mathrm{and}{\mathrm{v}}_2$be speeds of first and second glider respectively.

During collision, momentum is conserved.

So, Initial momentum of gliders = final momentum of gliders

$$\begin{gathered} {\mathrm{m}}_1{\mathrm{u}}_1+{\mathrm{m}}_2{\mathrm{u}}_2={\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2 \\ \left(0.15\mathrm{kg}\right)\left(0.8\mathrm{m}/\mathrm{s}\right)+\left(0.3\mathrm{kg}\right)\left(2.2\mathrm{m}/\mathrm{s}\right)=0.15\mathrm{kg}\left({\mathrm{v}}_1\right)+0.3\mathrm{kg}\left({\mathrm{v}}_2\right) \\ -3.6={\mathrm{v}}_1+2{\mathrm{v}}_2 \end{gathered}$$

Since it is an elastic collision, therefore, from relative velocity equation

$$\begin{gathered} {\mathrm{v}}_2-{\mathrm{v}}_1=-\left({\mathrm{u}}_2-{\mathrm{u}}_1\right) \\ =-\left(-2.2\mathrm{m}/\mathrm{s}-0.8\mathrm{m}/\mathrm{s}\right) \\ 3={\mathrm{v}}_2-{\mathrm{v}}_1 \end{gathered}$$

Adding the two equations gives

$$\begin{gathered} {\mathrm{v}}_2=-0.2\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_1=3.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, first glider is moving to the left with speed 3.2 m/s and second glider is moving to the left with speed of 0.2 m/s.