81.1k views
4 votes
The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9.0 m/s, they then have the same kinetic energy. What were the original speeds of the two cars? Express your answers using two significant figures separated by a comma?

1 Answer

6 votes

Answer:

The original speeds of the two cars were :


6.36(m)/(s),12.72(m)/(s)

Step-by-step explanation:

Let's start reading the question and making our equations in order to find the speeds.

The first equation is :


m_(1)=2m_(2) (I)

The kinectic energy can be calculated using the following equation :


K=((1)/(2)).m.v^(2) (II)

Where ''K'' is the kinetic energy

Where ''m'' is the mass and where ''v'' is the speed.

By reading the exercise we find that :


K_(1)=(K_(2))/(2) (III)

If we use (II) in (III) :


((1)/(2)).m_(1).v_(1)^(2)=((1)/(2)).m_(2).v_(2)^(2).((1)/(2))


m_(1).v_(1)^(2)=(m_(2).v_(2)^(2))/(2) (IV)

If we replace (I) in (IV) ⇒


2.m_(2).v_(1)^(2)=(m_(2).v_(2)^(2))/(2)


4.v_(1)^(2)=v_(2)^(2)


2.v_(1)=v_(2) (V)

'' When both cars increase their speed by
9.0(m)/(s), they then have the same kinetic energy ''

The last equation is :


((1)/(2)).m_(1).(v_(1)+9)^(2)=((1)/(2)).m_(2).(v_(2)+9)^(2) (VI)

If we use (I) in (VI) ⇒


2.m_(2).(v_(1)+9)^(2)=m_(2).(v_(2)+9)^(2)


2.(v_(1)+9)^(2)=(v_(2)+9)^(2)

If we use (V) in this last expression ⇒


2.(v_(1)+9)^(2)=(2.v_(1)+9)^(2)


2.(v_(1)^(2)+18v_(1)+81)=4v_(1)^(2)+36v_(1)+81


2v_(1)^(2)+36v_(1)+162=4v_(1)^(2)+36v_(1)+81


2v_(1)^(2)=81


v_(1)^(2)=40.5


v_(1)=√(40.5)=6.36

We find that the original speed
v_(1) is
6.36(m)/(s)

If we replace this value in the equation (V) ⇒


2.(6.36(m)/(s))=v_(2)


v_(2)=12.72(m)/(s)

User Ameya Pandilwar
by
7.7k points

No related questions found