# Impulse and momentum steel balls

I am trying to figure out why they said that the right answer is 2?

The left figure depicts the path of two colliding steel balls A and B. Which of the arrows labeled from 1 to 5 represents teh direction of the impulse exerted on ball B by ball A?

Shouldn't it be 5? because the impulse exerted on B,$~~\vec{I}$, is the variation of the momentum $~~\vec{p}$,

$$\vec{I} = \vec{p_{f}}-\vec{p_{i}} = m (\vec{v_{f}}-{\vec{v_{i}}})$$

that means that the impulse has the same direction of the vector $~\vec{v_{f}}-{\vec{v_{i}}}$ but as we can see from the picture, $~\vec{v_{f}}-{\vec{v_{i}}}$ is corresponding to the arrow 5, so the impulse too has to have the same direction, isn't it ?

• Note the phrasing the direction of the impulse exerted on ball B by ball A. – a CVn Dec 28 '16 at 15:08
• It means the impulse of B , doesn'it ? – Hilbert Dec 28 '16 at 15:12
• It means "How does B's momentum change after the collision?". Your reasoning is right, your vector sum is wrong. – FrodCube Dec 28 '16 at 15:18
• Yes , actually , I was adding $\vec{v_{f}}$ and $\vec{v_{i}}$ instead of $\vec{v_{f}}$ and $- \vec{v_{i}}$ that's why I was getting arrow 5 instead of arrow 2 . Thanks . – Hilbert Dec 28 '16 at 15:26

The change in momentum of B is given by,

$I = p_f - p_i = m(v_f - v_i)$

Now $v_f$ and $v_i$ each have a vertical and a horizontal component.

The horizontal component of $v_f$ is equal to the horizontal component of $v_i$ (both pointing to the right).

Therefore, the difference in moment is due to only the difference in the vertical component.

The initial vertical component $v_{i_y}$ is equal and opposite to the final vertical component, $v_{f_y}$.

Therefore,

$I = m(v_{f_y} - v_{i_y}) = m(-v_{i_y} - v_{i_y}) = m(-2v_{i_y})$

and thus the impulse is in the opposite direction to the initial velocity.

An impulse changes the velocity components along its direction. Along the horizontal direction the velocity components do not change and hence there shouldn't be any impulse in this direction. On the other hand, the velocity components in the vertical direction change drastically (switches direction) and hence it can be assume that the majority of the impulse is in the vertical direction.

To distinguish between 1) and 2) look at the velocities before and after and figure out with what sense does the impulse need to act to give the object its new speed. In this case, ball (B) switches from moving upwards to moving downwards and hence the impulse must be acting downwards.

Mathematically, if an impulse of magnitude $J$ is acting along a direction $\hat{n}$ then the change is velocity vector is

\begin{align} \Delta \vec{v} & = \frac{1}{m} \hat{n} J \\ \vec{v}_{\rm after} & = \vec{v}_{\rm before} + \frac{1}{m} \hat{n} J \end{align}

The impulse exerted on the balls is in the direction of the line joining their centers.

• I don't really see how this helps answer the question. Could you expand more on it? – Kyle Kanos Feb 14 '18 at 14:05
• As you can see in the picture Centers are joined vertically. – Expert Mathematician Feb 14 '18 at 14:09
• Also, A imparts an impulse in downward direction on B. – Expert Mathematician Feb 14 '18 at 14:10

Contact forces exerted on surfaces have to be perpendicular to those surfaces. Because each ball is a sphere, the direction of the force has to be towards the center of ball B, because geometrically, the perpendicular line with respect to each point on a sphere points towards the center of the sphere. This means that the correct answer is answer 2.