Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3400 N on the car for 0.400 s.(a) What impulse (in kg · m/s) is imparted by this force?(b) Find the final velocity (in m/s) of the bumper car if its initial velocity was 3.20 m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor.

a) 1360kgm/s b) 5.14m/s

Explanation:

a) According to Newton's second law of motion,

Force = mass × acceleration

F = ma

since acceleration = change in velocity (v-u)/time(t)

F = m(v-u)/t

Ft = m(v-u) = Impulse.

Impulse is defined as change in momentum of a body. It can be expressed as;

Impulse = Ft = m(v-u)

Given force exerted on the car = 3400N

time taken = 0.400seconds

Impulse = 3400×0.4

Impulse = 1360kgm/s

b) To find the final velocity/(v) of the car, we will use the equation of motion below;

v = u+at where;

v is the final velocity = ?

u is the initial velocity = 3.2m/s

a is the acceleration

t is the time taken = 0.4second

To get the acceleration, we will use the formula F = ma where;

m is the mass

F is the force

a is the acceleration

a = F/m = 3400/200

a = 17m/s²

Substituting the values to get the final velocity 'v' we have;

v² = 3.2²+17(0.4)

v² = 10.24+6.8

v² = 17.04

v =√17.04

v = 4.13m/s

The final velocity of the bumper car is 4.14m/s

pe = weightxheight = 2000x1000=2,000,000joules

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge