# Do objects really move in the direction of the force applied?

According to the second law of motion:

Direction of the force is in the direction of change in momentum.

Now, I came before a problem. Though an easy one, I was perplexed at a point.

A cricket ball of mass $m$ moving with velocity $u$ is deflected by an angle $90^{\circ}$ with same magnitude of velocity after being struck by the batsman. What is the direction of the force applied on the ball?

Really an easy one: change of momentum $$|\vec{\Delta p}| = \sqrt{2}m|\vec{u}|.$$ Now, the direction of the force is given by $\theta$ such that $$tan{\theta} = \dfrac{| \vec{F}.\Delta t|}{| \vec{p_{initial}}|}$$. After doing calculations, I got $\theta = 135^{\circ}$ with the original momentum of the ball.

Here I was surprised: The force was applied at $135^{\circ}$, but the ball moved at $90^{\circ}$ with the initial momentum. Why? I thought the body moves in a direction the force is applied. Here,it is contradicting. The ball moves in one direction and the force is applied in another direction. Momentum changes in one direction, but the body moves in another direction.

How is it possible? Doesn't a body always move in the direction of the appliedforce?

• "Doesn't a body always move in the direction of the applied force?". Consider, for example, uniform circular motion where they object never moves in the direction of the force. – Alfred Centauri Dec 24 '14 at 12:38
• When you throw a ball, it keeps moving forward, even though the force from gravity is straight down. Force changes your velocity, it doesn't set it. – dfan Dec 24 '14 at 14:31

According to Newton's second law, $\overrightarrow F=m\overrightarrow a$ which means an object accelerates in the direction of the applied force. Acceleration means change in velocity is in the direction of the applied force. The final velocity will not necessarily be in the direction of the applied force, except when the initial velocity is zero. • You can integrate the total force as a function of time, $\vec{F}(t)$, to find the change in momentum--for example, from time $t_0$ to some later time $t_1$, the change in momentum $\Delta \vec{p}$ is $\int_{t_0}^{t_1} \vec{F}(t) \, dt$, so if the initial momentum at $t_0$ was $\vec{p}_0$, the new momentum at $t_1$ is $\vec{p}_0 + \int_{t_0}^{t_1} \vec{F}(t) \, dt$, and the total vector you get from that sum tells you the direction the object was moving at any desired time $t_1$. – Hypnosifl Dec 24 '14 at 3:48
• And note that if the force is constant the change in momentum $\Delta \vec{p}$ between $t_0$ and $t_1$ simplifies to $\vec{F} \times (t_1 - t_0)$, so in this case $\Delta \vec{p}$ is guaranteed to be parallel to $\vec{F}$, but if the initial momentum at $t_0$ was $\vec{p}_0$ then the momentum at $t_1$ is $\vec{p}_0 + \Delta \vec{p}$, which need not be parallel to the force even though $\Delta \vec{p}$ is. – Hypnosifl Dec 24 '14 at 3:58