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?

 

[Moetasim]

Your solution appears to be correct. The final velocity of the ball is indeed 12 ms^-1. If you are unsure, you can double check your calculations and make sure you are using the correct units for each variable. Additionally, you can try to solve the problem using a different method to see if you get the same answer.