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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.

User Gravstar
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2 Answers

4 votes

Answer:

(a) 1360 Ns

(b) 10 m/s

Step-by-step explanation:

(a)

Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns

From Newton's third law,

The expression for impulse is given as,

I = F×t............................ Equation 1

Where I = Impulse, F = Force, t = Time.

Given: F = 3400 N, t = 0.4 s

Substitute into equation 1

I = 3400×0.4

I = 1360 Ns

(b)

From newton third law of motion,

Impulse = Change in momentum.

I = m(v-u)................. Equation 2

Where I = impulse, m = mass of the car plus the driver, v = final velocity of the bumper car, u = initial velocity of the car.

make v the subject of the equation

v = (I/m)+u................ Equation 3

Given: I = 1360 Ns, m = 200 kg, u = 3.2 m/s

Substitute into equation 3

v = (1360/200)+3.2

v = 6.8+3.2

v = 10 m/s

User Quirky Purple
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3 votes

Answer:

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

Step-by-step 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

User Nils Magne Lunde
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