Question

A ball of mass 0.1 kg starting from rest false a height of 4 metre and then collides with the ground after the collision the ball bounces up to a height of 2 metre the collision with the ground text place overtime 4 into 10 to the power minus 3 second determine the momentum of the ball immediately before the collision and immediately after the collision and the average force exerted by the ground on the ball take G is equal to 10​

Answer 1

Question:-

A ball of mass $\displaystyle\sf{0.1\ kg}$, starting from rest, falls a height of $\displaystyle\sf{4\ m}$ and then collides with the ground. After collision, the ball bounces up to a height of $\displaystyle\sf{2\ m.}$ The collision with the ground takes place over a time $\displaystyle\sf{4\times10^{-3}\ s.}$ Determine the momentum of the ball immediately before the collision and immediately after the collision and the average force exerted by the ground on the ball. Take $\displaystyle\sf{g=10\ m\ s^{-2}.}$

Answer:-

Momentum immediately before collision = $\displaystyle\large\sf{\underline{\underline{0.894\ kg\ m\ s^{-1}.}}}$

Momentum immediately after collision = $\displaystyle\large\sf{\underline{\underline{0.632\ kg\ m\ s^{-1}.}}}$

Average force exerted by the ground on the ball = $\displaystyle\sf{\underline{\underline{381.5\ N.}}}$

Solution:-

Consider the incident immediately before collision.

By third kinematic equation,

$\displaystyle\longrightarrow\sf{v_1=\sqrt{2gh_1}}$

since $\displaystyle\sf{u_1=0\quad;\quad a_1=g\quad;\quad s=h_1}$

Then, the momentum of the ball immediately before collision,

$\displaystyle\longrightarrow\sf{mv_1=m\sqrt{2gh_1}}$

Here, $\displaystyle\sf{m=0.1\quad;\quad g=10\quad;\quad h_1=4}$

Then,

$\displaystyle\longrightarrow\sf{mv_1=0.1\times\sqrt{2\times10\times4}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{mv_1=0.894\ kg\ m\ s^{-1}}}}$

Now, consider the incident immediately after collision and the rebounding of the ball.

In this case, $\displaystyle\sf{a_2=g\quad;\quad v_2=0\quad;\quad s=-h_2=-2}$

since the downward motion is considered as positive.

Again by third kinematic equation,

$\displaystyle\longrightarrow\sf{(u_2)^2=(v_2)^2+2gh_2}$

$\displaystyle\longrightarrow\sf{u_2=-\sqrt{2gh_2}}$

Negative sign is because the ball usually moves upward after collision.

Then, the momentum of the ball immediately after collision,

$\displaystyle\longrightarrow\sf{mu_2=-m\sqrt{2gh_2}}$

$\displaystyle\longrightarrow\sf{mu_2=-0.1\times\sqrt{2\times10\times2}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{mu_2=-0.632\ kg\ m\ s^{-1}}}}$

The average force exerted by the ground on the ball is given by Second Law, i.e.,

$\displaystyle\longrightarrow\sf{F=\dfrac{dp}{dt}}$

$\displaystyle\longrightarrow\sf{F=\dfrac{mu_2-mv_1}{dt}}$

since the impact is occurred for a very short time $\displaystyle\sf{dt=4\times10^{-3}\ s.}$

Then,

$\displaystyle\longrightarrow\sf{F=\dfrac{-0.632-0.894}{4\times10^{-3}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{F=-381.5\ N}}}$

Negative sign shows that the force is acting upwards.