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A particle with mass $m_1=m$ moves along the x-axis at a velocity of $v_0$ and collides with another particle $m_2=4m$. As a result of the collision $m_1$ travels upwards at an angle of $90 ^\circ$. (see diagram) enter image description here

How can I find the velocities after the collision?

My approach with conservation of momentum and kinetic enregy:

Momentum in the x direction: $m_1v_0=m_1u_1(cos90^\circ)+m_2u_2 => v_0=4u_2cos(\alpha)$

In the y direction: $u_1=4u_2sin(\alpha)$

Kinetic energy... $v_0^2=u_1^2+4u_2^2$

I've been going at it for quite a while now haha for some reason I'm always left with too many unknowns.

answers should be: $u_1=v_0\sqrt{0.6} \\ u_2=v_0\sqrt{0.1}$

Any help would be appreciated!

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Try to square your solutions for $v_0$ and $u_1$, you'll get a simplification by summing them ; then you'll be able to express $u_1$ and $u_2$ as a function of $v_0$ using this new equation and the one obtained via the conservation of kinetic energy.

Hope this helps ;)

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  • $\begingroup$ Why the summation? $\endgroup$
    – Bad Hombre
    Commented Feb 20 at 12:47
  • $\begingroup$ By squaring you have $v_0^2=16u_2^2 cos^2(\alpha)$ and $u_1^2=16u_2^2 sin^2(\alpha)$. To get rid off the angle, you can just compute the sum $v_0^2 + u_1^2=16u_2^2$ $\endgroup$
    – MauvaiseFoi
    Commented Feb 20 at 12:54

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