Car 1 headed north and car 2 headed west collide. They stick together and leave skid marks on the pavement, which show that car 1 was deflected 30°(so car 2 was deflected 60°). What can you conclude about the cars before the collision?

The given data can be listed below as,

• Momentum of car 1, $p_1$
• Momentum of car 2, $p_2$
• Initial momentum, ${\overrightarrow{p}}_i$
• Final momentum, ${\overrightarrow{p}}_f$
• Car 1’s deflection,30°
• Car 2’s deflection,60°

The space diagram is as follows:

The distance between centres perpendicular to the incoming velocity is calledthe “impact parameter” and is often denoted byb.

A head-on collision has an impact parameter of zero.The smaller the impact parameter, the more severe is the collision, and thelarger the deflection angle of the incoming particle (larger “scattering”), exceptfor a head-on collision, where if the masses are equal the incoming ball stopsdead and the target ball gets the entire momentum.

The net force on the system is zero, then the momentum of the system must be conserved:

${\overrightarrow{p}}_i={\overrightarrow{p}}_f$

Applying the momentum conservation inx direction:

$p_{2,i}=p_f\cos \left(60\right)$………. first equation

Applying the momentum conservation iny direction:

$p_{1,i}=p_f\sin \left(60\right)$………...second equation

By dividing the second equation by the first equation:

$\frac{p_{1,i}}{p_{2,i}}=\frac{p_f\sin \left(60\right)}{p_f\cos \left(60\right)}=\tan \left(60\right)$

tan (60) = √3

So,

$p_{1,i}=\sqrt{3}{p}_{2,i}$.

Hence, car 1’s momentum was $\sqrt{3}$times greater than the momentum of car 2.