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A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in the car's momentum?

1 Answer

6 votes

Answer:

96046 Ns.

Step-by-step explanation:

We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.

40 km/h in the east

V₁ = 40 i

V₂ = 50j

momentum p₁ = mV₁

= 1500 X 40 i

= 60000 i

Momentum p₂ = mV₂

= 1500 X 50j

= 75000 j

Change in momentum

p₂ - p₁

75000j - 60000i

Magnitude of change

=
\sqrt{(750000)^2 +(60000)^2

= 96046 Ns.

User Dominik Neise
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