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

[looi76]

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?

 

[dx]

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?