Question

If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the 1200.-kg pick-up truck behind him continues traveling at 25.0 m/s, with what velocity will the two move if they lock bumpers after a rear-end collision?

Answer 1

Answer: 20.2 m/s

Explanation:

From the question above, we have the following data;

M1 = 800kg

M2 = 1200kg

V1 = 13m/s

V2 = 25m/s

U (common velocity) =?

M1V1 + M2V2 = (M1 + M2). U

(800*13) + (1200*25) = (800+1200) * U

10400 + 30000 = 2000u

40400 = 2000u

U = 40400 / 2000

U = 20.2 m/s

Answer 2

Answer:

20.2 m/s

Explanation:

Using the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').................... Equation 1

Where m = mass of the sport car, m' = mass of the pick-up truck, u = Initial velocity of the sport car, u' = Initial velocity of the pick-up truck V = Common velocity.

make V the subject of the equation

V = (mu+m'u')/(m+m')............... Equation 2

Given: m = 800 kg, m' = 1200 kg, u = 13 m/s, u' = 25 m/s

Substitute into equation 2

V =[(800×13)+(1200×25)]/(800+1200)

V = (10400+30000)/2000

V = 40400/2000

V = 20.2 m/s

Hence both will move with a velocity of 20.2 m/s