Solve Dynamics Q#5: 0.6kg Ball's Final Velocity

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SUMMARY

The discussion centers on calculating the final velocity of a 0.60 kg ball rolling North at 4.0 m/s after experiencing a force of 0.18 N East for 10 seconds. The correct approach involves recognizing that the acceleration due to the force is perpendicular to the initial velocity. The acceleration is calculated as 0.30 m/s², leading to a final velocity of 12 m/s. However, the Northward component remains unchanged, and the Eastward component must be calculated separately to determine the resultant velocity vector.

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Homework Statement


A ball of mass 0.60kg rolls North at 4.0ms^{-1}. It meets a slope which causes a force of 0.18N East. This force lasts for 10s. Calculate the final velocity of the ball. Neglect any friction effects.

Homework Equations


F = m.a
v = u+at

The Attempt at a Solution


m = 0.60kg \ , \ v = 4.0ms^{-1} \ , \ F = 0.18N \ , \ t = 10s

F = m.a
a = \frac{F}{m} = \frac{0.18}{0.60} = 0.30ms^{-2}

v = u + at
v = 4.0 + 0.3 \times 10
v = 12 ms^{-1}

Where did I go wrong?
 
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The acceleration is not in the same direction as the initial velocity. Infact, the acceleration is perpendicular to the initial velocity, so the velocity northward doesn't change. What is the eastward component of its velocity at the end?
 

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