15. If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the
1200.-kg pick-up truck behind him continues traveling at 25.0 m/s, with what
velocity will the two move if they lock bumpers after a rear-end collision?

Answer :

skyluke89

The final combined velocity after the collision is 20.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two car and of the truck must be conserved before and after the collision.

This means that we can write the following equation:

[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v[/tex]  

where:  

[tex]m_1 = 800 kg[/tex] is the mass of the sport car

[tex]u_1 = 13.0 m/s[/tex] is the initial velocity of the car (taking its direction as positive  direction)

[tex]m_2 = 1200 kg[/tex] is the mass of the truck

[tex]u_2 = 25.0 m/s[/tex] is the initial velocity of the truck

[tex]v[/tex] is the final combined velocity of the car and the truck, after the collision

Re-arranging the equation and substituting the values, we find the velocity after the collision:

[tex]v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(800)(13)+(1200)(25)}{800+1200}=20.2 m/s[/tex]

And the positive sign indicates their final direction is the same as the initial direction of the two vehicles.

Learn more about momentum here:

brainly.com/question/7973509  

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