Having real trouble with the question below that is part of my engineering course. I've sussed question 6 but making no headway with Q7.

  1. A ball of mass 0.2 kg collides elastically with a heavier ball (mass 0.5 kg) which was originally stationary. The initial speed of the lighter ball is 12 m s1. The collision was head-on, with both balls moving away along the same line as the incident ball. What is the speed of the heavier ball after the collision?

  2. In the question above, suppose that the collision was not head-on and that the lighter ball was deflected through 90º. Assuming still that the collision is elastic, what is the final speed of the lighter ball?

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    $\begingroup$ Have you considered conservation of momentum in both vertical and horizontal directions plus conservation of kinetic energy? $\endgroup$ – Triatticus Jan 14 '19 at 17:51
  • $\begingroup$ On problem 7, a picture would be very helpful. To me, a 90 deg angle on the smaller ball implies that the large ball only touched the small ball as it went by, imparting essentially zero momentum to it. $\endgroup$ – David White Jan 15 '19 at 15:48

Let call the initial momenta $\vec p$ and $\vec P$ for the little band big balls, respectively. Add a prime for the final state. Define the axes so:

$$ \vec p \propto \hat x$$

with $\hat y$ orthogonal (which means 90 degrees w.r.t $\hat x$).

So, what are the input facts that we know about the initial state:

$$ p_x = mv$$ $$ p_y = 0$$ $$ P_x = M\cdot0 = 0$$ $$ P_y = M\cdot 0=0$$

are all well defined.

What are we told about the final state?

We know the collision was elastic, so that:

$$ E' = E $$

and all the energy is kinetic energy, and kinetic energy in a system of $N$ balls is:

$$ T = \frac 1 2 \sum_{k=1}^N{\frac {p_i^2} m} $$

We also know the little ball scattered at 90 degrees:

$$ p'_x = 0 $$

and of course, by definition:

$$ p'_y = mv' $$

So the question is: "What is $v'$?"

All we need to do is include momentum conservation:

$$ p_x + P_x = p'_x + P'_x $$ $$ p_y + P_y = p'_y + P'_y $$

to solve.

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Use 2d vector math.

You know the velocity of the 0.2kg projectile prior to collision: {12,0}$ms^{-1}$
You know the angle of the velocity of the projectile after the collision; it's 90$^{\mathsf{o}}$. Let's represent that by saying the post-collision velocity is {0,$x$}$ms^{-1}$, where $x$ is the post-collision speed. It's not entirely clear that it'll be the same as it was in problem six, so we'll leave it unknown for now.

Now you can get an expression of the impulse on the projectile, in terms of $x$. The impulse on the target will be the negative of that. From there you can write expressions for the momentum and velocity of the target post-collision.

Once you have the post-collision velocities of the projectile and the target expressed in terms of $x$, write out their kinetic energies $k_{after}^{object}$ also in terms of $x$.
Then we just write out $$k_{before}^{projectile}+0=k_{after}^{projectile}+k_{after}^{target}$$ and solve for $x$.

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