enter image description here When we throw a ball to the ground then in which direction does the impulse acts on the ball?

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    $\begingroup$ We can define change in momentum as impulse so we can look for the change in the momentum vector to determine direction of impulse applied by the floor in the ball $\endgroup$ Jun 16, 2019 at 18:03
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    $\begingroup$ Show effort and thought before asking a question. Make the question specific. Don't just post your homework problem $\endgroup$
    – Andrew
    Jun 16, 2019 at 22:53
  • $\begingroup$ Its just a specific concept of a problem $\endgroup$
    – Sanjay
    Jun 17, 2019 at 3:28
  • $\begingroup$ No, it's really not. It's the entire problem testing your understanding of a concept by applying it to a physical scenario. At least specify what it is exactly that you are confused about. Don't simply post the whole problem and just expect us to pull a rabbit out of our hat that solves the work you need to put effort into. You really should accept responsibility and look to improve in the future considering 3/8 of your questions have been deemed poor quality. That's an above average bad ratio $\endgroup$
    – Andrew
    Jun 17, 2019 at 9:35

1 Answer 1


Think of $\vec{v}$ as a vector split up into two components: the vertical and the horizontal component.

Whenever two objects come in contact, the Normal Force acts. The Normal Force is "the component of a contact force that is perpendicular to the surface that an object contacts." So in the case of your diagram, the Normal force is acting directly vertical because that is perpendicular to the flat surface. Now impulse, $J$, in equation form is just

$$J = F\cdot\Delta t = \mathrm{m}\cdot\Delta v$$

which is just an extension of Newton's Second Law which states that

$$F = \frac{m\cdot\Delta v}{\Delta t} \implies F = ma$$

$F$, in this case, would be provided by the Normal Force exclusively because the angle of incidence and reflection are the same. If there were friction, the horizontal component of $\vec{v}$ would be reduced thus decreasing the angle of reflection and causing it to be less than the angle of incidence (assuming the ball had no initial spin). This also implies the collision between the ball and the surface is elastic which means no energy is lost in the collision so the vertical component of $\vec{v}$, i.e. $v_{y}$ before = $-{v_{y}}$ after.

If $F$ is entirely normal then this explains why $\Delta v$ points in the vertical direction ($\Delta t$ and $m$ are just scalars) - the horizontal component of $\vec{v}$ i.e. $v_{x}$ is unchanged and still points to the right with the same magnitude. So $J$ points vertically because $F$ is the normal force which points vertically.

Also, for future reference, show that you've at least tried to solve the problem and have already thought about it to the best of your ability. The StackExchange is not a place for people to do your homework. Don't simply post a problem and ask other people to solve it for you. I also notice that you never accept answers to your rather lazily asked questions. You need to accept people's answers so they get credit for spending time to answer your "questions"

  • $\begingroup$ How do you know that $F$ is normal? $\endgroup$
    – Elio Fabri
    Jun 16, 2019 at 13:40
  • $\begingroup$ Whenever two objects come into contact the component of the force that is perpendicular is ALWAYS the normal force. The magnitude of normal force is determined by the vertical component of v which determines the angle of incidence. We know the normal force is the only acting force because the angle of incidence and reflection are the same - that means no friction force (because friction would reduce the horizontal component of v and change the angle of reflection). So test, F is exclusively normal $\endgroup$
    – Andrew
    Jun 16, 2019 at 15:38
  • $\begingroup$ We also know this is an elastic collision because the angle of incidence/reflection is the same which means $v$ vertical before = - $v$ vertical after $\endgroup$
    – Andrew
    Jun 16, 2019 at 21:47

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