A 900- kg car collides with a 1500- kg car that was initially at rest at the ori

Question

A 900- kg car collides with a 1500- 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 40 o with respect to the positive x axis. The heavier car moves at 23 km/h at -48 o 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)?

A 900- kg car collides with a 1500- 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 40 o with respect to the positive x axis. The heavier car moves at 23 km/h at -48 o 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

900ucos(x)=900*20cos(40)+1500*23cos(48) (momentum conservation in x)

So

u*cos(x)=40.971----(i)

900usin(x)=-900*20sin(40)+1500*23sin(48)

So

usin(x)=15.631466----(ii)

So

x=tan^-1(15.631466/40.971)=20.88 degrees anticlockwise from x-axis

and

u=43.85km/h

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.