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diamong [38]
3 years ago
7

Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmove

s forward at 25 m/s and car A is at rest. What fraction of theinitial kinetic energy is lost in the collision?
My approach:
How exactly do you approach this question? At first, I thoughtusing the the formula ΣKEi = ΣKEf.Then, to find the fraction just divide the former from the latter.But is it that cut and dry?
Before collision:
viA= 35 m/s
viB = 0 m/s
vfA = 0 m/s
vfB =25 m/s
m1 = m2
Physics
1 answer:
kolezko [41]3 years ago
7 0

Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2

kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2

Since the masses are equals:

m=ma=mb

For the known velocities, the kinetics energies result:

ki=0.5*mVa_{i}^2

ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m

kf=0.5*mbVb_{f}^2

kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m

The lost energy in the collision is the difference between the initial and final kinectic energies:

kl=ki-kf

kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m

Finally the relation between the lost and the initial kinetic energy:

kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m

kl/ki = 24/49=0.49

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Answer:

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Explanation:

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r = radius of merry-go-round = \frac{d}{2} =  \frac{4}{2} = 2 m

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Using conservation of angular momentum

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