Question

Q.

a) find impulse provided by ground to particle in first collision, b) find total impulse provided by ground till particle come to rest. c) Total time of flight.

Answer 1

a) $\frac{3}{2}m\sqrt{2gh}$, b) $m\sqrt{gh}(2+3\sqrt{2})$, c) $\sqrt{\frac{2h}{g}}(3+\sqrt{2})$

Answer 2

The problem describes a particle of mass 'm' released from a height 'h'. It then undergoes a series of collisions with the ground. From the provided image, the particle rebounds to a height of $h/4$ after the first collision and to $h/8$ after the second collision. This implies a varying coefficient of restitution.

Let $h_0 = h$ be the initial height. The height after the first bounce is $h_1 = h/4$. The height after the second bounce is $h_2 = h/8$.

The coefficient of restitution 'e' for a collision is given by $e = \sqrt{h_{rebound}/h_{impact}}$. For the first collision, $e_1 = \sqrt{h_1/h_0} = \sqrt{(h/4)/h} = \sqrt{1/4} = 1/2$. For the second collision, $e_2 = \sqrt{h_2/h_1} = \sqrt{(h/8)/(h/4)} = \sqrt{1/2} = 1/\sqrt{2}$. Assuming this pattern continues, the coefficient of restitution for subsequent collisions ($n \ge 2$) will be $e_n = 1/\sqrt{2}$.

a) Impulse provided by ground to particle in first collision: The speed of the particle just before the first collision is $v_{1,down} = \sqrt{2gh_0} = \sqrt{2gh}$. The speed of the particle just after the first collision is $v_{1,up} = e_1 v_{1,down} = (1/2)\sqrt{2gh} = \sqrt{gh/2}$. Impulse is the change in momentum. Taking the upward direction as positive: $J_1 = m(v_{1,up} - (-v_{1,down})) = m(v_{1,up} + v_{1,down})$ $J_1 = m(\sqrt{gh/2} + \sqrt{2gh}) = m(\frac{\sqrt{2gh}}{2} + \sqrt{2gh}) = m\sqrt{2gh}(1/2 + 1) = \frac{3}{2}m\sqrt{2gh}$.

b) Total impulse provided by ground till particle comes to rest: The total impulse is the sum of impulses from all collisions. $J_{total} = \sum_{n=1}^{\infty} J_n$. For the n-th collision, $J_n = m(1+e_n)v_{n,down}$, where $v_{n,down} = \sqrt{2gh_{n-1}}$.

$J_1 = \frac{3}{2}m\sqrt{2gh}$.

For $n \ge 2$: $h_n = h_1 \cdot (e_2^2)^{n-1}$ is not correct as $e_2$ is for $h_2/h_1$. The height sequence is $h_0=h$, $h_1=h/4$, $h_2=h/8$, $h_3=h/16$, and so on. This means $h_{n-1} = h/4 \cdot (1/2)^{n-2}$ for $n \ge 2$. So, $v_{n,down} = \sqrt{2gh_{n-1}} = \sqrt{2g \cdot h/4 \cdot (1/2)^{n-2}} = \sqrt{gh/2 \cdot (1/2)^{n-2}} = \sqrt{gh/2^{n-1}}$. And $e_n = 1/\sqrt{2}$ for $n \ge 2$. $J_n = m(1+1/\sqrt{2})\sqrt{gh/2^{n-1}} = m(1+1/\sqrt{2})\sqrt{gh} \frac{1}{(\sqrt{2})^{n-1}}$.

The sum of impulses from $n=2$ to $\infty$: $\sum_{n=2}^{\infty} J_n = m(1+1/\sqrt{2})\sqrt{gh} \sum_{n=2}^{\infty} \frac{1}{(\sqrt{2})^{n-1}}$. Let $k=n-1$. The sum becomes $\sum_{k=1}^{\infty} \frac{1}{(\sqrt{2})^k}$. This is a geometric series with first term $a = 1/\sqrt{2}$ and common ratio $r = 1/\sqrt{2}$. The sum is $\frac{a}{1-r} = \frac{1/\sqrt{2}}{1-1/\sqrt{2}} = \frac{1/\sqrt{2}}{(\sqrt{2}-1)/\sqrt{2}} = \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)} = \sqrt{2}+1$. So, $\sum_{n=2}^{\infty} J_n = m(1+1/\sqrt{2})\sqrt{gh}(\sqrt{2}+1) = m(\frac{\sqrt{2}+1}{\sqrt{2}})\sqrt{gh}(\sqrt{2}+1) = m\sqrt{gh}\frac{(\sqrt{2}+1)^2}{\sqrt{2}} = m\sqrt{gh}\frac{2+1+2\sqrt{2}}{\sqrt{2}} = m\sqrt{gh}\frac{3+2\sqrt{2}}{\sqrt{2}} = m\sqrt{gh}\frac{3\sqrt{2}+4}{2}$.

Total impulse $J_{total} = J_1 + \sum_{n=2}^{\infty} J_n = \frac{3}{2}m\sqrt{2gh} + m\sqrt{gh}\frac{3\sqrt{2}+4}{2}$. $J_{total} = \frac{3}{2}m\sqrt{2}\sqrt{gh} + m\sqrt{gh}(\frac{3\sqrt{2}}{2} + 2) = m\sqrt{gh}(\frac{3\sqrt{2}}{2} + \frac{3\sqrt{2}}{2} + 2) = m\sqrt{gh}(3\sqrt{2} + 2)$. $J_{total} = m\sqrt{gh}(2+3\sqrt{2})$.

c) Total time of flight: The total time of flight is the sum of the time for the initial fall and the time for all subsequent bounces. Time for initial fall: $t_0 = \sqrt{2h/g}$.

Time for the n-th bounce (from $h_{n-1}$ up to $h_n$ and back down): $t_n = 2 \cdot v_{n,up}/g = 2 \cdot e_n v_{n,down}/g = 2 \cdot e_n \sqrt{2gh_{n-1}}/g = 2e_n\sqrt{2h_{n-1}/g}$. $t_n = 2e_n \sqrt{2h_{n-1}/g}$.

For the first bounce cycle (from $h_0$ to $h_1$ and back): $t_1 = 2e_1\sqrt{2h_0/g} = 2(1/2)\sqrt{2h/g} = \sqrt{2h/g}$.

For $n \ge 2$: $h_{n-1} = h/4 \cdot (1/2)^{n-2}$. $e_n = 1/\sqrt{2}$. $t_n = 2(1/\sqrt{2})\sqrt{2 \cdot (h/4 \cdot (1/2)^{n-2})/g} = \sqrt{2}\sqrt{2h/g \cdot (1/4) \cdot (1/2)^{n-2}} = \sqrt{2}\sqrt{2h/g \cdot (1/2)^{n}}$. $t_n = \sqrt{2h/g} \cdot \sqrt{2} \cdot (1/\sqrt{2})^n = \sqrt{2h/g} \cdot (1/\sqrt{2})^{n-1}$.

Total time $T = t_0 + \sum_{n=1}^{\infty} t_n$. $T = \sqrt{2h/g} + \sum_{n=1}^{\infty} \sqrt{2h/g} \cdot (1/\sqrt{2})^{n-1}$. $T = \sqrt{2h/g} + \sqrt{2h/g} \sum_{n=1}^{\infty} (1/\sqrt{2})^{n-1}$. The sum $\sum_{n=1}^{\infty} (1/\sqrt{2})^{n-1} = \sum_{k=0}^{\infty} (1/\sqrt{2})^k$. This is a geometric series with $a=1$ and $r=1/\sqrt{2}$. The sum is $\frac{1}{1-1/\sqrt{2}} = \frac{1}{(\sqrt{2}-1)/\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}-1} = \frac{\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{2+\sqrt{2}}{1} = 2+\sqrt{2}$.

So, $T = \sqrt{2h/g} + \sqrt{2h/g}(2+\sqrt{2}) = \sqrt{2h/g}(1 + 2 + \sqrt{2}) = \sqrt{2h/g}(3+\sqrt{2})$.

Final Answers: a) Impulse provided by ground to particle in first collision: $J_1 = \frac{3}{2}m\sqrt{2gh}$

b) Total impulse provided by ground till particle comes to rest: $J_{total} = m\sqrt{gh}(2+3\sqrt{2})$

c) Total time of flight: $T = \sqrt{2h/g}(3+\sqrt{2})$