Answer: The correct answer is option (A).
Explanation:
Momentum of the car with 0.04 kg mass , which travelling with velocity of 2.00 m/s
[tex]P_1=mass\times velocity=m_1\times v_1=0.04 kg\times 2.00 m/s=8.00 kg m/s[/tex]
Then the maximum speed of the another car in order to not to break the eggs will be same as first car:
[tex]P_1=P_2[/tex]
[tex]0.08 kg m/s=m_2\times v_2=0.08 kg\times v_2[/tex]
[tex]v_2=1 m/s[/tex]
Speed slightly more than 1 m/s will increase the momentum of second car and the eggs will break. So, from the given options the minimum speed need by the second car will be 1.42m/s.
