# effect of vertical collision on kinetic friction and subsequent change in horizontal velocity

Suppose somehow a block of mass $m$ is moving on ground, and the coefficient of kinetic friction between the block and the block is $\mu_k$. If I drop a tennis ball(of same mass) on it from a height, say $h$ and it bounces back, coefficient of restitution being $e$ for the collision, I need to work out what effect would this have on its velocity, which was say at the time of collision equal to $v(t_0)=v$. The speed may or may not be changing due to everything else, but the change the collision would cause is what I want to find.

My view is that, at the time of collision, when the ball and the block are in contact, at that time ($t_0$), the normal reaction to the block will become $2mg$ instead of $mg$, which would change the kinetic friction at that time.

We say there is no change in momentum as impulse is due to opposite and equal magnitude forces. We say nothing about time. But here that time of contact is needed so that change in friction somehow changes the velocity.

Otherwise there is no extra change and so horizontal velocity shouldn't be affected by the collision. Does the velocity change due to collision?

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The kinetic energy the ball has when it hits the block is $mgh$. We have the relation:

$$\frac{mv^2}{2}=mgh$$

This can be rewritten:

$$m^2v^2=2m^2gh$$

Which means the downward momentum of the ball is:

$$p=mv=m\sqrt{2gh}$$

The block will excert a force on the ball to cancel this momemntum and give it a momentum of $e\cdot p$ in the opposite direction:

$$F\Delta t=m\sqrt{2gh}(1+e)$$

This will require a corresponding increase in the normal force on the block and thereby in the friction force:

$$\Delta F_f=\mu\Delta N=\mu F$$

The change in moementum of the block will be:

$$\Delta p_B=\Delta F_f\Delta t=\mu F\Delta t=\mu m\sqrt{2gh}(1+e)$$

The change (decrease) in speed of the block will thus be $\mu\sqrt{2gh}(1+e)$.

I have not included the extra force due to the weight of the ball or due to giving it a horizontal velocity component or spin, but these are likely to be small.

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1st: you must take care that friction is a impulsive force and momentum never remains conserved when $f$ changes.
2nd:the normal reaction will become $mg+I/\Delta t$ where $I$ is the impulse imparted on the ball ie:$mv(1+e);v=\sqrt{2gh}\ and\ \Delta t .$is the small time of impact.So, $$N=m(g+v(1+e)/\Delta t)$$
@AshishGaurav thanks. I just had skipped out $\Delta t$ –  Mr.ØØ7 Mar 29 at 14:54