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A 59 kg physics student is riding her 220 kg Harley at 12 m/s when she has a head-on collision with a 2.1 kg pigeon flying the opposite direction at 44m/s. The bird is still on the motorcycle after the collision. How fast is the motorcycle going after the collision?

Respuesta :

fortuneonyemuwa

Answer:

Explanation:

Using conservation of momentum

[tex]m_1=59+220=279kg[/tex]

[tex]m_2=2.1kg[/tex]

[tex]U_1=12m/s[/tex]

[tex]U_2=-44m/s[/tex]

[tex]m_1U_1+m_2U_2=(m_1+m_2)V\\\\\frac{278\times 12-2.1\times 44}{279+2.1}=V\\\\V=11.58m/s[/tex]

pstnonsonjoku

The velocity of the motorcycle after collision is 11.6 m/s.

Conservation of linear momentum

Using the principle of conservation of momentum, the total momentum of the system remains constant. Hence, momentum after collision must be equal to momentum before collision.

Thus;

[(55 + 220) Kg * 12 m/s] +  2.1 kg * (-44m/s) = (55 + 220 + 2.1) v

3300 - 92.4 = 277.1v

v = 3300 - 92.4/277.1

v = 11.6 m/s

Learn more  about velocity: https://brainly.com/question/9743981

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