Ballistic Fragmentation

Classical Mechanics Level 5

A massive body is initially at rest at ( x , y ) = ( 0 m , 7 m ) (x,y) = (0m,7m) . The body explodes and fragments into two equal masses such that both x x and y y momentum are conserved.

One fragment flies leftward and impacts a target at ( x , y ) = ( − 2 m , 5 m ) (x,y) = (-2m,5m) . The other fragment flies rightward and impacts a target at ( x , y ) = ( 12 m , 0 m ) (x,y) = (12m,0m) .

With respect to the horizontal, what is the launch angle of the rightward fragment?

Details and Assumptions:

  • Give your answer in degrees, to 1 decimal place
  • The ambient gravitational acceleration is 10 m / s 2 10 m/s^2 downward
  • Note that "leftward" and "rightward" should not be understood as indicating purely horizontal motion
  • In this problem, the x x direction is horizontal and the y y direction is vertical.


The answer is 37.7.

Steven Chase by Steven Chase , 37, USA

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1 solution

Rohith M.Athreya
Feb 16, 2017

The angles A ; B A ;B may be negative.

we will find the equation of trajectory of each particle first.

The sub-script r r (and l l ) is indicative of the quantities related to the rightward(leftward) moving particle.

x 1 = v 1 t c o s A x_{1}=v_{1}tcosA and y 1 = 7 + v 1 t s i n A − g t 2 2 y_{1}=7+v_{1}tsinA -\frac{gt^{2}}{2} and thus,

y 1 = 7 + x 1 t a n A − g x 1 2 2 v 1 2 c o s 2 A \large y_{1}=7+x_{1}tanA - \frac{gx_{1}^{2}}{2v_{1}^{2}cos^{2}A}

similarly, y 2 = 7 + x 2 t a n A − g x 2 2 2 v 2 2 c o s 2 B \large y_{2}=7+x_{2}tanA - \frac{gx_{2}^{2}}{2v_{2}^{2}cos^{2}B} .

We have been given one point on each of these trajectories.

We will now conserve momentum(linear).

horizontal v 1 c o s A = v 2 c o s B v_{1}cosA=v_{2}cosB

vertical v 1 s i n A = − v 2 s i n B v_{1}sinA=-v_{2}sinB

By substituting the given points in the respective equations of trajectories, we obtain a system of linear equations in t a n A tanA and g 2 v 2 2 c o s 2 B \frac{g}{2v_{2}^{2}cos^{2}B}

solving which, t a n A = 65 84 tan A = \frac{65}{84}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I think we will not require the conservation of momentum in the vertical direction.

Rohit Gupta - 4 years, 3 months ago

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then you have either solved the question very very differently or u haven't understood what i did fully

Rohith M.Athreya - 4 years, 3 months ago

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Yeah, my mistake. I was reading through your solution and thought we need to find the value of angle A A as your solution ends at tan ⁡ A = 65 84 \tan A= \frac{65}{84} . But, on second reading I realize that we need to find the angle B B for which we will require the conservation in the vertical direction.

Rohit Gupta - 4 years, 3 months ago

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