A 800- kg car collides with a 1400- kg car that was initially at rest at the ori

Question

A 800- kg car collides with a 1400- kg car that was initially at rest at the origin of an x-y coordinate system. After the collision, the lighter car moves at 10.0 km/h in a direction of 35 o with respect to the positive x axis. The heavier car moves at 13 km/h at -43 o with respect to the positive x axis.
What was the initial speed of the lighter car (in km/h)?

I've seen this worked out already on cramster, but I can't follow the equations. Please show work and use my numbers. THANKS

A 800- kg car collides with a 1400- kg car that was initially at rest at the origin of an x-y coordinate system. After the collision, the lighter car moves at 10.0 km/h in a direction of 35 degree with respect to the positive x axis. The heavier car moves at 13 km/h at -43 degree with respect to the positive x axis. What was the initial speed of the lighter car (in km/h)?

Explanation / Answer

P2y = m1v1fsin1 + m2v2fsin2 ---------------------- (4)

  According to Conserve x-component momentum:

    P1x = P2x

m1v1cos =m1v1fcos1 + m2v2fcos2

(800kg)v1cos = (800kg)(10km/h)cos35 + (1400kg)(13km/h)cos43

(800kg)v1cos = 19863.85 kg km/h

v1cos = 24.82 km/h

v1x = v1cos = 24.82 km/h

and  According to Conserve of y-component momentum:

P1y = P2y

m1v1sin =m1v1fsin1 + m2v2fsin2

(800kg)v1sin = (800kg)(10km/h)sin35 + (1400kg)(13km/h)sin(-43)

(800kg)v1sin = -7823.75 kg km/h

v1sin = -9.77 km/h

v1y = v1sin = -9.77 km/h

now the velocity,

v1 = sqrt(v1x^2 + v1y^2)

v1 = ((24.82 km/h)^2 + (-9.77 km/h)^2)

v1 = 26.67 km/h

the direction is

= tan-1(v1y/v1x)

= tan-1(9.77 km/h / 24.82 km/h)

= -21.48 degrees

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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