# In which direction does the impulse acts on ball when we throw it to ground? [closed]

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

## closed as off-topic by M. Enns, Thomas Fritsch, stafusa, Jon Custer, GiorgioPJun 21 at 4:23

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• 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 – Aditya Garg Jun 16 at 18:03
• Show effort and thought before asking a question. Make the question specific. Don't just post your homework problem – Andrew Jun 16 at 22:53
• Its just a specific concept of a problem – Sanjay Jun 17 at 3:28
• 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 – Andrew Jun 17 at 9:35

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.

• How do you know that $F$ is normal? – Elio Fabri Jun 16 at 13:40
• 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 – Andrew Jun 16 at 21:47