Person on cart throws ball of mass $m$ at angle $\theta$, what is the angle perceived by ground observer if $m$ grows without bound?

Question

Consider a person of mass $m_1$ standing on a cart of mass $m_2$, holding a ball of mass $m_3$. The person throws the ball at an angle $\theta$ (relative to $x$-direction) and initial speed $v_0$, both with respect to the reference frame of the person in the cart. There is no friction.

The angle at which an observer from the ground reference frame sees the ball leave the cart (I will show calculations below) is:

$$\alpha=\tan^{-1}{\frac{(m_1+m_2+m_3)\tan{\theta}}{m_1+m_2}}$$

We can see that if $m_3>>m_1+m_2$ then $\alpha$ approaches $\frac{\pi}{2}$.

What is the intuition behind this result?

Note that the other extreme case, if $m_1+m_2>>m_3$, then $\alpha$ approaches $\theta$: the angle is the same for the ground observer and the observer on the cart.

In the first case, where the ball has an enormous mass, the person "throwing" the ball essentially means the person plus cart are pushing off an enormous mass that is essentially immovable. However, in this case, the person plus cart would move away with a very high speed, but I can't see why the ground observer wouldn't see this with an angle of zero rather than $\frac{\pi}{2}$.

A few of the calculations used

To obtain the angle $\alpha$ as observed from a ground reference frame we need to find the velocity vector of the ball as observed from this frame.

For the $x$ direction, we can use conservation of momentum.

I am using subscripts as follows: $v_{b,g,x}$ means speed of ball, from ground reference, in direction x. If I add an extra 0 it means right before the ball is thrown. Without this extra 0 it means right after the ball is thrown.

$x$-direction initial momentum is $p_{sys,g,x,0}=0$

$x$-direction momentum right after the throw is: $$\vec{p}_{sys,g,x}=m_b v_{b,g,x} +(m_1+m_2)v_{c,g,x}\tag{1}$$

We can express the $x$-direction velocity of the ball relative to the ground reference as follows: $v_{b,g,x}=v_{c,g,x}+v_{b,c,x}$. This is simply the Law of Addition of Velocities for relatively inertial reference frames.

We know that $v_{b,c,x}=v_0\cos{\theta}$ so $$v_{b,g,x}=v_{c,g,x}+v_0\cos{\theta}\tag{2}$$

We now have two equations $(1)$ and $(2)$ in two unknowns $v_{b,g,x}$ and $v_{c,g,x}$ and we obtain

$$v_{b,g,x}=\frac{(m_1+m_2)v_0\cos{\theta}}{m_1+m_2+m_3}$$

$$v_{c,g,x}=\frac{-m_3v_0\cos{\theta}}{m_1+m_2+m_3}$$

To obtain the $y$-direction component of the ball's velocity, from the ground frame we have

$$v_{b,g,y}=v_{c,g,y}+v_{b,c,y}=0+v_0\sin{\theta}$$

Therefore,

$$\tan{\alpha}=\frac{v_{b,g,y}}{v_{b,g,x}}=\frac{(m_1+m_2+m_3)\tan{\theta}}{m_1+m_2}$$

Answer 1

The person pushes the ball in the direction $\alpha$ (the direction in which he wants to move the ball): $0 < \alpha < \pi/2$. After pushed the ball, he observes that the ball moves in the direction of $\theta$. It is worth mentioning that he can change $\alpha$ to any value, but not $\theta$.

The relationship among $\theta$ and $\alpha$ is $$ \tan\theta = \frac{(m_1 + m_2)}{m_1 + m_2 + m_3} \tan\alpha . $$

If $\alpha \neq \pi/2$ (he pushes the ball above right), in the limit $m_3/(m_1+m_2) \to \infty$, $\theta \to 0$. This is reasonable since (seen from ground) the ball doesn't move and he moves left.

If $\alpha \to \pi/2$ (he pushes the ball just above), $\theta \to \pi/2$, and this is again reasonable (seen from ground, he doesn't move and the ball just moves above). However, we don't have the value of $\theta$ in the limit $m_3/(m_1+m_2) \to \infty$ since the right hand side of the equation is indeterminate form. In this case, neither ball or person doesn't move and $\theta$ isn't defined (or he can't observe it !), which agrees the fact that $\lim\theta$ doesn't exist.