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One car has two and a half times 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? first car

User Zahlii
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1 Answer

5 votes

Answer:

velocity of first car = 7.91 m

Velocity of second car = 17.72 m/s

Step-by-step explanation:

Let the mass of second car B is m and the kinetic energy of this car is K.

First car A:

mass, mA = two and half of mass of second car = 2.5 m

Kinetic energy, KA = K/2

Let the speed is vA.

Second car B:

mass, mB = m

kinetic energy, KB = K

Let the speed is vB.

Kinetic energy of car B

K = 1/2 m vB² ....(1)

Kinetic energy of car A

K/2 = 1/2 x 2.5 m x vA² .... (2)

Put the value of K from equation (1), we get

1/2 m vB² x 1/2 = 1/2 x 2.5 m x vA²

vB² = 5 vA²

vB = 2.24 vA ....(3)

Now the speed of both the car is increased by 9 m/s so that thir kinetic energies are equal.

So,

1/2 x 2.5m x (vA +9)² =1/2 x m x (vB +9)²

2.5 (vA+9)² = (vB +9)²

1.58 (vA + 9) = (vB + 9)

Put teh value of vB from equation (3)

1.58 vA + 14.22 = 2.24 vA + 9

vA = 7.91 m/s

So, vB = 2.24 x vA = 2.24 x 7.9 = 17.72 m/s

User Arjun Raj
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