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A 600-kg car makes a 52° turn. Its speed before the turn is 21.0 m/s and...

A 600-kg car makes a 52° turn. Its speed before the turn is 21.0 m/s and after the turn it is 17.0 m/s. What is the magnitude of the change in the car's momentum during the turn?

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Answer #1

vi = 21 i

Vf = -17*cos52 i + 17*sin52 j = -10.466 i +13.396 j

change in the car's momentum = Δ P = m*(vf-vi )   = 600* (-10.466 i +13.396 j - 21i )

Δ P = 600* ( -31.466 i +13.396 j )

magnitude of the change = √600^2*( 31.466^2+13.396^2)

  

= 600* √( 31.466^2+13.396^2) =20519.3155 kg*m/s answer

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