Question

To test the resiliency of its bumper during low-speed collisions, a 2 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 4.00 106 N/m and compresses 3.18 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall?

Answer 1

Vi = 2 m/s

Step-by-step explanation:

First we find the force applied to the car by wall to stop it. We use Hooke's Law:

F = kx

where,

F = Force = ?

k = spring constant = 4 x 10⁶ N/m

x = compression = 3.18 cm = 0.0318 m

Therefore,

F = (4 x 10⁶ N/m)(0.0318 m)

F = 127200 N

but, from Newton's Second Law:

F = ma

a = F/m

where,

m = mass of car = 2010 kg

a = deceleration = ?

Therefore,

a = 127200 N/2010 kg

a = 63.28 m/s²

a = - 63.28 m/s²

negative sign due to deceleration.

Now, we use 3rd equation of motion:

2as = Vf² - Vi²

where,

s = distance traveled = 3.18 cm = 0.0318 m

Vf = Final Speed = 0 m/s

Vi = Initial Speed = ?

Therefore,

2(- 63.28 m/s²)(0.0318 m) = (0 m/s)² - Vi²

Vi = √4.02 m²/s²

Vi = 2 m/s