A(n) 1600-kg car going at 6.32 m/s in the positive x direction collides with a 3

Question

A(n) 1600-kg car going at 6.32 m/s in the positive x direction collides with a 3100-kg truck at rest. The collision is totally inelastic and takes place over an interval of 0.203 s. Assume that no brakes are applied during the collision and the car strikes the rear of the truck. Neglect the friction between the vehicles and the ground. A) What is the average x component of the acceleration of the car during the collision (use appropriate units)? B) What is the average x component of the acceleration of the truck during the collision (use appropriate units)?

Explanation / Answer

Since collision is inelastic, so momentum is conserved.

mc uc + mt ut = (mc + mt) v

V = (mc uc + mt ut) /(mc + mt)

V = (1600 x 6.32)/(1600 + 3100) = 2.15 m/s

A)

X component of acceleration of car,

ac = (v - uc) /t = (2.15 - 6.32)/0.203 = - 20.53 m/s^2

So magnitude of acceleration of car is 20.53 m/s^2

B)

X component of acceleration of truck

at = (2.15 - 0)/0.203 = 10.59 m/s^2

Sk magnitude of acceleration of truck is 10.59 m/s^2

Comment in case any doubt please rate my answer ...

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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