# Collision of Particles

At time $t_0 = 0$, a particle P1 of mass m is projected vertically from the ground with speed $v_0 > 0$. If no collisions occur, this particle would reach its maximum height $h_1$ at time $t = t_1$. At time $t_2$, $0 < t_2 < t_1$, when particle P1 has the speed $\frac{v_0}{3}$, another particle $P_2$ of mass 3m is projected vertically upwards from the same point of the ground with speed $\frac{v_0}{3}$. Assume that the gravitational acceleration g is constant.

(i) Determine $t_1$ and $h_1$.

(ii) Determine $t_2$ and the height $h_2$ of $P_1$ at $t = t_2$.

(iii) Determine the time $t_3$ and height $h_3$ of the collision between particles $P_1$ and $P_2$.

(iv) Determine speeds of these particles right before the collision. What are directions of motion of particles at this moment of time?

(v) Determine the work done by the gravity force on each of the particles from the beginning of motion until the moment of collision.

So far I got $t_1=\frac{v_0}{g}$, $h_1=v_0^2$, $t_2=\frac{2v_0}{3g}$ and $h_2=\frac{4v_0^2}{9g}$ but I'm stuck from iii) onweards.

I know you have to equate the heights of the two particles which gave me $t_3=\frac{3}{2v_0}+\frac{6v_0}{18g}$ but I feel like I went wrong somewhere.

• I recommend using mathematical typesetting (c.f. math.meta.stackexchange.com/questions/107/…) – YukiJ Apr 18 '18 at 15:25
• Your $h_1$ is incorrect. I suggest that you include more details of how you arrived at this and the other values. – amd Apr 20 '18 at 7:26
• Finally, if $0\lt t_2\lt t_1$ and the initial velocity of the second particle are correct, then the particles can never collide, unless you allow the second one to sit on the ground for a while waiting for the first one to come back down. Even if you change the condition to $t_2\gt t_1$, the particles don’t collide while aloft. Most likely the second particle’s launch velocity is wrong. – amd Apr 20 '18 at 7:44

With the stated conditions, the particles never collide. The particle will be momentarily at rest at its peak, so $t_1={v_0\over g}$. You could plug this into the equation of motion to find $h_1$, but a conservation of energy argument works just as well. At the peak of the path, gravity will have done work equal to $mgh_1$ on the particle, converting all of the initial kinetic energy $\frac12mv_0^2$ into potential energy. Equating these and solving for $h_1$ produces ${v_0^2\over2g}$, which is different from the value you’ve given.
Since the particle undergoes constant deceleration on the way up, it’s pretty obvious that $t_2$ is just $2/3$ of the time it takes to reach its peak, and $h_2$ can again be determined via conservation of energy: $$mgh_2 = \frac12mv_0^2-\frac12m\left({v_0\over3}\right)^2,$$ therefore $h_2 = {4v_0^2\over9g}$. So far, so good.
Now we run into problems. If the second particle is launched with a third of the velocity of the first, it will peak in a third of the time, i.e., it will reach its maximum height at time $$t_2+\frac13t_1 = \frac23t_1+\frac13t_1 = t_1.$$ The graphs of the second particle’s motion is thus just the first particle’s graph shifted downwards. The graphs are otherwise identical parabolas, so they have no intersections.
We can try to fix this by waiting to launch until the first particle is heading back down. By symmetry, this alternate launch time is $\frac43t_1$, so the second particle will return to the ground at time $\frac43t_1+\frac23t_1=2t_1$, which is exactly when the first particle returns as well. From the wording of the problem, I suspect that this might not be what the authors had in mind, either.