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I got stuck at the following physics question :

A particle has initial velocity,$\mathbf v = v_x \hat i + v_y \hat j $ m/s and a constant force $\mathbf F = F_x \hat i + F_y \hat j$$ acts on the particle. The path of particle will be?

My approach is that $$ \mathbf x = v_x t+ \frac{F_x}{2m} t^2$$ and $$ \mathbf y = v_y t+ \frac{F_y}{2m} t^2$$. But I don't know how to substitute $t$ from those equations. So what should I do now?

Thanks in advance.

  • You can find several approaches for eliminating $t$ from your parametric equations in https://math.stackexchange.com/q/530310/265466 (which I’m considering a duplicate of your question). If you want to skip all that, you can use the formula in https://math.stackexchange.com/q/530310/265466. – amd Nov 12 '19 at 22:47
  • An alternative approach is to work out the trajectory in a coordinate system in which one of the components of the force vanishes, then rotate that solution. – amd Nov 12 '19 at 22:48

1 Answers1

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Rewrite the equations

$$ x= v_x t+ \frac{F_x}{2m} t^2,\>\>\>\>\>\>\> y= v_y t+ \frac{F_y}{2m} t^2$$

as

$$\left( t + \frac {mv_x}{F_x}\right)^2 = \frac{2mx}{F_x} + \frac {m^2v_x^2}{F_x^2},\>\>\>\>\>\>\> \left( t + \frac {mv_y}{F_y}\right)^2 = \frac{2my}{F_y} + \frac {m^2v_y^2}{F_y^2}$$

Eliminate $t$,

$$\frac {mv_x}{F_x}-\frac {mv_y}{F_y} = \sqrt{\frac{2mx}{F_x} + \frac {m^2v_x^2}{F_x^2}} -\sqrt{ \frac{2my}{F_y} + \frac {m^2v_y^2}{F_y^2}}$$

Remove the square roots,

$$\left(\frac{x}{F_x}+ \frac{y}{F_y}+\frac {mv_xv_y}{F_xF_y}\right)^2 = \left(\frac{2x}{F_x} + \frac {mv_x^2}{F_x^2}\right) \left(\frac{2y}{F_y} + \frac {mv_y^2}{F_y^2}\right)$$

Simplify,

$$\frac{x^2}{F_x^2}-\frac{2xy}{F_xF_y}+\frac{y^2}{F_y^2} +\frac{2m}{F_xF_y}\left(\frac {v_x}{F_x} - \frac{v_y}{F_y}\right)(v_yx-v_xy) =0$$

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