A Stubborn Physics Problem Solved (A. Foster)
In the student cafeteria the other day, a very interesting and stubborn physics problem came up while
doing calculus (which by the way was invented to do physics) which goes some-what as follows:
Suppose a shell with mass 𝑀𝑀 (in kilograms) is launched (or fired) with initial speed 𝑣𝑣0 = 17 𝑚𝑚/𝑠𝑠 at an
angle of 60∘ with the horizontal. At the top of its trajectory, the shell explodes into two fragments of
1
equal mass, i.e.; 𝑚𝑚 = 2 𝑀𝑀. One of the fragments lands on the ground directly below the explosion point.
How far from the launch (or firing) point did the other fragment land?
shell
Fragment 1
Fragment 2
Why not take a shot at it: After several attempts at a solution; applying the standard methods I learned
when I took physics some years ago, I came to learn that my attempts were incorrect. Somehow, I felt
there was something about the problem that wasn’t clear to me (one of the fragments had no horizontal
speed at the top of the trajectory. Even though this was not explicitly stated in the problem, it was
implicitly implied), thus leading to the incorrect results ( such as 𝑥𝑥 = 25.5 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚) . Now here is my
detailed final attempt.
Solution:
We must analyze, what is going on before the explosion, and at the time of the explosion and after the
explosion in order to solve this problem correctly.
BEFORE THE EXPLOSION: The fired (or launched) shell executes projectile
motion.
Then we have the following projectile-motion data:
mass = 𝑀𝑀,
𝑦𝑦0 = 0; 𝑣𝑣0 = 17𝑚𝑚/𝑠𝑠
𝜃𝜃 = 60∘ ,
𝑥𝑥0 = 0,
∘
𝑣𝑣𝑥𝑥 = 𝑣𝑣0𝑥𝑥 = 𝑣𝑣0 cos 60 , 𝑣𝑣0𝑦𝑦 = 𝑣𝑣0 sin 60∘
The Shell the instant before it Explodes: The shell has traveled horizontal and vertical maximum
distances 𝑥𝑥max (in meters) and 𝑦𝑦max (in meters) from the launch point and these two distances were
traveled simultaneously in some amount of time 𝑡𝑡 (in seconds).
We need to find the 𝑥𝑥max and 𝑦𝑦max coordinates of the point where the shell explodes. Since the shell
explodes at the top of its trajectory, then we have speed 𝑣𝑣 = 0 i.e., 𝑣𝑣𝑥𝑥 = 0; 𝑣𝑣𝑦𝑦 = 𝑣𝑣0𝑦𝑦 − 𝑔𝑔𝑔𝑔 = 0.
Consequently, we find that it took the shell
𝑡𝑡 =
𝑣𝑣0𝑦𝑦 𝑣𝑣0 sin 60∘
17√3
=
=
≈ 1.50228896575 = 𝟏𝟏. 𝟓𝟓 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔.
𝑔𝑔
𝑔𝑔
2(9.8)
to reach the top of its trajectory, the instant it explodes into two fragments (namely, Fragment 1 which
fell straight down to the ground; no horizontal motion from here on out!) and Fragment 2 (which
continued to execute projectile motion). Thus in 𝟏𝟏. 𝟓𝟓 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 the shell traveled a maximum
horizontal distance 𝑥𝑥max from the origin given by
𝑥𝑥max = 𝑥𝑥0 + (𝑣𝑣0 cos 60∘ ) 𝑡𝑡 = 17(0.5)(1.500228896575 ) ≈ 12.7694562089 = 𝟏𝟏𝟏𝟏. 𝟖𝟖 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎
and a maximum vertical distance 𝑦𝑦max above the ground given by
1
𝑦𝑦max = 𝑦𝑦0 + (𝑣𝑣0 sin 60∘ ) 𝑡𝑡 − 2 𝑔𝑔𝑡𝑡 2 =
11.0586734694 = 𝟏𝟏𝟏𝟏. 𝟏𝟏 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎.
17�√3�
2
⋅ (1.50228896575) −
9.8
2
(1.50228896575 )2 ≈
Conservation of Total Horizontal Momentum: The Total momentum of the shell just before it explodes
is given 𝑝𝑝 = 𝑀𝑀𝑣𝑣0 = 17𝑀𝑀 whose total horizontal and vertical amounts are given by 𝑝𝑝𝑥𝑥 = 𝑀𝑀𝑣𝑣𝑥𝑥 =
1
𝑀𝑀𝑣𝑣0 cos 60∘ = 2 𝑀𝑀𝑣𝑣0 = 8.5𝑀𝑀 and 𝑝𝑝𝑦𝑦 = 𝑀𝑀𝑣𝑣𝑦𝑦 = 𝑀𝑀𝑣𝑣0 sin 60∘ = 8.5√3 𝑀𝑀
Ka-BooM!!!!
(The shell breaks into two Fragments)
Fragment 1 and Fragment 2 are both at the explosion coordinates (12.8, 11.1). However, Fragment 1
falls straight down to the ground directly below the point of the explosion and is in free-fall towards
the ground; while Fragment 2 executes projectile motion starting at the explosion point. Here is the
data for the fragments at this instant:
1
FRAGMENT 1: mass = 2 𝑀𝑀, 𝑥𝑥0 = 12.8 𝑚𝑚, 𝑦𝑦0 = 11.1 𝑚𝑚, 𝑉𝑉0𝑦𝑦 = 0, 𝑉𝑉0𝑥𝑥 = 0, 𝑝𝑝 = 0
FRAGMENT 2 (The all important fragment!)
1
1
mass = 2 𝑀𝑀, 𝑉𝑉0𝑥𝑥 = 2 𝑣𝑣0 = 8.5
8.5𝑀𝑀, 𝑝𝑝0𝑦𝑦 = 0
𝑚𝑚
𝑠𝑠
1
, 𝑋𝑋0 = 𝑥𝑥max = 12.8 𝑚𝑚, 𝑌𝑌0 = 𝑦𝑦max = 11.1 𝑚𝑚, 𝜃𝜃 = 0∘ , 𝑝𝑝0𝑥𝑥 = 2 𝑀𝑀𝑣𝑣0 =
Key point: There are no other horizontal forces at work, thus, the Horizontal component of momentum
is conserved, i.e., the total horizontal momentum before and after the explosion is the same. Therefore,
1
(Total H-Momentum Before) 𝑀𝑀𝑣𝑣0 cos 60∘ = 𝑝𝑝0𝑥𝑥 = 𝑝𝑝𝑥𝑥 = 𝑀𝑀𝑉𝑉𝑋𝑋 (Total H-Momentum After)
2
This means that Fragment 2 has horizontal speed 𝑉𝑉𝑥𝑥 after the Explosion given by
2
𝑉𝑉𝑥𝑥 = � � (𝑀𝑀𝑣𝑣0 cos 60∘ ) = 𝑣𝑣0 = 17 𝑚𝑚/𝑠𝑠
𝑀𝑀
FRAGMENT 2 ON THE GROUND AFTER THE EXPLOSION!
1
Free-fall: 𝑌𝑌 = 𝑌𝑌0 + 𝑉𝑉0𝑦𝑦 𝑇𝑇 − 2 𝑔𝑔𝑇𝑇 2 where 𝑇𝑇 is the total time in seconds it takes Fragment 2 to land on
the ground from the time of the explosion. Since 𝑉𝑉0𝑦𝑦 = 0 and 𝑌𝑌 = 0 (when the fragment 2 has landed
on the ground) we have
2 𝑌𝑌0
1 2
𝑔𝑔𝑇𝑇 = 𝑌𝑌0 ⇔ 𝑇𝑇 = �
(This is the time it takes the fragment 2 to land on the ground)
𝑔𝑔
2
The horizontal distance traveled by the fragment during this time 𝑇𝑇 is given by
2 𝑌𝑌0
2(11.1)
= 12.8 + 17 �
= 38.3 𝑚𝑚
𝑋𝑋 = 𝑋𝑋0 + 𝑉𝑉𝑥𝑥 𝑇𝑇 = 𝑋𝑋0 + 𝑉𝑉𝑥𝑥 �
𝑔𝑔
9.8
Thus, Fragment 2 is on the ground at coordinates (38.3, 0) and therefore must have traveled a total
horizontal distance of 𝟑𝟑𝟑𝟑. 𝟑𝟑 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 away from the initial launch point at the origin. I can say that I feel
that this result is correct! More importantly, because I was persistent, I learned something new!
MY POINT IS A SIMPLE ONE: It matters not what your field of study is, only if you are persistent, will you
ever find solutions to stubborn problems. So always keep on trying and don’t give up until you get it
RIGHT!
Understanding is gained in the BATTLE!