Question

A 1200 kg car moving north at 50 km/h collides with a 1600-kg car moving east at 30 km/h. The cars stick together, and the wreckage moves off at an initial speed of
a. 27 km/h
b. 38 km/h
c. 39 km/h
d. 58 km/h
*follow up question
• The direction in which the wreckage of he preceding collision moves off is
a. 39° E of N
b. 45° E of N
c. 51° E of N
d. 54° E of N

Answer 1

consider the initial momentum and energy.

Answer 2

this is a perfectly inelastic collision. Use the formula:

\[m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}\]

Answer 3

Get \[vf_{x}\]
\[m_{1}v_{1ix}+ m_{2}v_{2ix}=(m_{1}+m_{2})Vf_{x}\]

Get \[vf_{z}\]
\[m_{1}v_{1iz}+ m_{2}v_{2iz}=(m_{1}+m_{2})Vf_{z}\]

Then \[vf=\sqrt{Vfx^2+Vfz^2}\]
\[\theta=\tan^{-1} [\frac{Vfz}{Vfx}]\]

Hope you understand :)