A point mass is shot vertically up from ground level with a velocity of 4 m/s at time, t = 0. It loses 20% of its impact velocity after each collision with the ground. Assuming that the acceleration due to gravity is 10 m/s² and that air resistance is negligible, the mass stops bouncing and comes to complete rest on the ground after a total time (in seconds) of

The correct answer is 4.

Let us analyze the motion of the point mass step-by-step to find the total time before it comes to a complete rest.

Step 1: Time taken for the first flight (before the first collision)
The mass is projected vertically upwards from the ground at time $t=0$ with an initial velocity $u_0=4\text{m/s}$.
Under the influence of gravity $g=10{\text{m/s}}^2$ and neglecting air resistance, the motion is symmetric. The time taken to reach the maximum height is:
$t_{\text{up}}=\frac{u_0}{g}$
The time taken to return to the ground is equal to the time taken to rise. Thus, the total time for the first flight $T_1$ is:
$T_1=2·\frac{u_0}{g}=2·\frac{4}{10}=0.8\text{seconds}$

Step 2: Impact velocity and coefficient of restitution
Since air resistance is negligible, the impact velocity on the ground just before the first collision is equal in magnitude to the projection velocity, i.e., $v_{\text{impact}}=4\text{m/s}$.
It is given that the mass loses 20% of its impact velocity after each collision. This means the velocity immediately after collision is 80% of the impact velocity:
$u_1=0.8·u_0$
In general, the velocity after the n-th collision is:
$u_n=e·u_{n-1}=e^n·u_0$
where $e=0.8$ is the coefficient of restitution.

Step 3: Calculating subsequent flight times
The time interval between the 1st and 2nd collision is the time of flight for the second bounce, $T_2$:
$T_2=2·\frac{u_1}{g}=2·\frac{e·u_0}{g}=e·T_1$
Similarly, the time of flight for the n-th bounce is:
$T_n=e^{n-1}·T_1$

Step 4: Summing the infinite series for total time
The mass continues to bounce infinitely many times, with the bounce duration decreasing in a geometric progression. The total time $T_{\text{total}}$ until it comes to a complete rest is the sum of all these time intervals:
$T_{\text{total}}=T_1+T_2+T_3+...$
$T_{\text{total}}=T_1+e{T}_1+e^2{T}_1+...$
This is an infinite geometric series with the first term $a=T_1=0.8$ and the common ratio $r=e=0.8$.
Since $\left|r\right|<1$, the sum of the infinite series is:
$T_{\text{total}}=\frac{T_1}{1-e}$
Substituting the values:
$T_{\text{total}}=\frac{0.8}{1-0.8}=\frac{0.8}{0.2}=4\text{seconds}$

Thus, the point mass stops bouncing and comes to a complete rest after a total time of 4 seconds.