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Classical Mechanics Level 1

Two balls of equal masses are thrown vertically upward at an interval of 2 seconds with the same initial speed of 39.2 m/s. At what height will they collide?

Use $g = 9.8 \si{\meter/\second}^2$ .

The answer is 73.5.

4 solutions

Jordi Bosch
Oct 26, 2014

The equation of the position of the first and second thrown are

$X_{1} = x_{0} + v_{0}t + \frac{1}{2}at^{2}$

$X_{2} = x_{0} + v_{0}(t-2) + \frac{1}{2}a(t-2)^{2}$

Substituting $x_{0} = 0, v_{0} = 39.2m/s, a = -9.8m/s^{2}$ we get:

$X_{1} = 39.2t - 4.9t^{2}$

$X_{2} = 39.2(t-2) - 4.9(t-2)^{2}$

The moment they collide $X_{1} = X_{2}$ so we equate the two equations and get as a solution $t = 2$ .

Substituting for $t = 2$ in the equation of movement of $X_{1}$ or $X_{2}$ we get $X = \boxed{73.5}$ .

you mean you get t = 5 seconds, not 2.

Satyen Nabar - 6 years, 6 months ago

That is definitely correct. Alternatively, you can choose to start you clock when the second ball is thrown so the first ball is thrown two second earlier. It slightly modifies the equations to be:

$X_1=39.2t-4.9t^2$

$X_2=39.2(t+2)-4.9(t+2)^2$

Setting $X_1=X_2$ for the collision, gives t=3, and plugging t=3 into $X_1$ or $X_2$ gives $X_1=X_2=\boxed{73.5}$

Scott Ripperda - 6 years, 2 months ago

My approach was the following :

The first ball (ball no 1) will reach maximum height in 39.2 / 9.8 = 4 seconds. Exploiting the symmetry in upwards and downwards direction, the distance covered by the first ball is 0.5 9.8 (4)^2 =78.4 m. Now at this point, the second ball would have been travelling in the upward direction for 2 seconds. Plugging t = 2 in the s =ut - 1/2 * g * t^2 equation, we get 58.8 m . Now, obviously after this point, the 2 balls would be travelling in opposite directions. If we denote the time until collision from this point as x, the equation for the first ball will be s1 = 1/2 * 9.8 * x^2

for the second ball travelling upward, we get s2 = 39.2 x - 1/2 9.8*x^2

the x^2 terms for s1 and s2 cancel out and s1 + s2 = 39.2*x

But s1 + s2 = distance between the 2 balls at t = 4 secs when the first ball is on top, that is 78.4 - 58.8 = 19.6 m

Therefore, 39.2x = 19.6 x = 19.6/39.2 = 0.5 secs

Plugging this in the equation for the first ball, we get s1 = 1/2 9.8 (0.5)^2 = 1.225

If we subtract this from the height of the first ball, we get 78.4 - 1.225 = 77.175m

Is there anything wrong with this approach ?

Sundar R - 5 years, 1 month ago

First this is a very complicated, but also valid way to approach this problem. You are tracking the problem at the center of the collision. The issue is that s1, the distance of the first ball from the collision point, is defined after 4 seconds of throwing the first ball, but your equation for s2, distance of the second ball from the collision point, is defined after two seconds of throwing the first ball. You did not account for the slow down of the second ball due to gravity during the extra to seconds. Accounting for the slow down you get:

$s2=(39.2-9.8*2)t-4.9t^2=19.6-4.9t^2$ , where $s1=4.9t^2$ . Thus $s1+s2=19.6t$ and the time after the collision after 4 seconds is given by solving:

$19.6t=(h1-h2)=78.4-58.8=19.6$ , where h1 and h2 are the heights of the first and second balls at 4 seconds after the first ball was thrown, which gives t=1, which gives the total time of the collision after the ball was thrown as 5 seconds which is the same as the other answers. Plugging t=1 back into s1 gives s1=4.9m, which means the height of the collision would be:

$max height-s1(t=1)=78.4-4.9=\boxed{73.5}$

In short: your compared distances at two different incompatible times, thus forgetting to account for the slowdown of ball 2 due to gravity.

Scott Ripperda - 5 years ago

Thanks for the lucid explanation. I realize that i had used the initial velocity (39.2) in the equation for s2 instead of 39.2 - 9.8(2) = 19.6 which is the initial velocity from the time the first ball reaches maximum height and begins to change direction , (as pointed out in your equation for s2.)

I realize that the above is a comparatively inelegant approach but seems the most natural one when trying to understand the physics of the problem

Sundar R - 5 years ago

Yes Jordi. , something wrong. Check your steps and thinking carefully. The collision happens after 5 total seconds, not after 4.5 seconds.

Tony Miller - 2 years, 5 months ago

Nelson Mandela
Jan 5, 2015

given, u1 = u2 = 4g = 39.2 m/s.

s1 = s2 = h = ?.

t1 = t and t2 = t + 2.

a1 = a2 = -g.

As, s = ut + 1/2 x a x t^2,

h = 4gt - 1/2 x g x t^2 = 4g(t +2) - 1/2 x g x (t+2)^2.

Solving it, we have, 8g = 2g + 2gt.

implies, 4 = 1 + t.

Thus, t = 3 seconds.

as, h = 4gt - 1/2 x g x t^2, h = 12g - 9/2 x g = 7.5g = 73.5.

Vinay Kumar
Jul 23, 2020

Very simple. At the point where collision happens, ball 2 will reach that point during its return journey exactly after 2 seconds the first ball has reached. So from that point ball 2 takes 1s to reach max height and 1 s to come to that point. Max height is 78.4m. From the top 1s travel is 4.9m. Subtract the two , ans is 73.5. If you are confused, then I am not a good teacher, but the concept is perfect !

Popular Power
Sep 8, 2019

Since their initial positions are same, then their displacement is also same $h=39.2t-0.5gt^2$ $h=39.2(t+2)-0.5gt^2$ $39.2t-0.5gt^2=39.2(t+2)-0.5gt^2 \implies t=3 \ \text{sec}$ $h=73.5 \ \text{m}$

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The answer is 73.5.

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