(c7p50) A 900- kg car collides with a 1200- kg car that was initially at rest at
ID: 1281561 • Letter: #
Question
(c7p50) A 900- kg car collides with a 1200- kg car that was initially at rest at the origin of an x-y coordinate system. After the collision, the lighter car moves at 20.0 km/h in a direction of 30 owith respect to the positive x axis. The heavier car moves at 23 km/h at -42 o with respect to the positive x axis.
What was the initial speed of the lighter car (in km/h)? Tries 0/7 What was the initial direction (as measured counterclockwise from the x-axis)? Car Collision in 2D (c7p50) A 900- kg car collides with a 1200- kg car that was initially at rest at the origin of an x-y coordinate system. After the collision, the lighter car moves at 20.0 km/h in a direction of 30 degree with respect to the positive x axis. The heavier car moves at 23 km/h at -42 degree with respect to the positive x axis. What was the initial speed of the lighter car (in km/h)? What was the initial direction (as measured counterclockwise from the x-axis)?
Explanation / Answer
Use conservation of momentum, so
mv=mv'+MV', so we need to use this for both x and y directions
for x we have (900kg)(v)cos0=(900kg)(10km/h)cos20+(1...
for y we have (900kg)(v)sin0=(900kg)(10km/h)sin20+(1...
with these two equations we can solve for either v or 0.
The first equation simplifies to vcos0=
The second equation simplifies to vsin0=
so we divide the two to get vsin0/vcos0= or tan0= and 0= degrees from the +x axis.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.