Can't catch the ball

Since the ball always has the same speed with respect to the ground its momentum $Q_{ball} = M_{ball} . V_{ball}$ doesn't change.

By releasing the object Mary gains speed in the opposite direction, and we get:

$Q_0 = 0 = Q_{Mary} - Q_{ball}$ $\Leftrightarrow Q_{Mary} = Q_{ball}$

The ball collides with the wall and reaches Mary back again, thus:

$Q_{Mary+ball} = Q_{ball} + Q_{Mary} = 2 Q_{ball}$

When throwing the ball for the second time, we have minimally $V_{Mary,2} = V_{ball}$ , then:

$Q_{Mary+ball} = Q_{Mary,2} - Q_{ball}$ $\Leftrightarrow 2.m_{ball}.v_{ball} = m_{Mary}.v_{ball} - m_{ball}.v_{ball}$ $\Leftrightarrow 2.m_{ball} = m_{Mary} - m_{ball}$ $\Leftrightarrow 3 = \frac{m_{mary}}{m_{ball}}$

Suppose Mary's mass is $M$ and ball's mass in $m$ . After the first throw, the conservation of momentum says that $mV = M V_1$ . The equation for the second catch and throw is $m V + m V = M V_2 - m V$ , $3 m V = M V_2$ . If the ball cannot catch up with Mary, its speed has to be at most $V_2$ (or smaller). Therefore, the maximum value is $\frac{M}{m} = 3$ .

Let m1 = mass of Mary and m2 = mass of ball.

Mary throws the ball the first time.

v1 <---Mary Ball---> v2

Conservation of Momentum: m1 v1 = m2 v2

Ball bounces back elastically, i.e. comes back at exactly the same speed but different direction.

v1 <--- Mary Ball <--- v2

Mary catches the ball. Since she is holding on to the ball, they both have the same speed.

v3 <--- Mary Ball <--- v3

Conservation of Momentum: m1 v3 + m2 v3 = m1 v1 + m2 v2

Mary throws the ball a second time.

v4 <--- Mary Ball ---> v5

Conservation of Momentum: m1 v3 + m2 v3 = m1 v4 – m2 v5

It is given in the question that she always throws the ball at the same speed with respect to the ground: v2 = v5

Ball bounces back elastically.

v4 <--- Mary Ball <--- v5

In order for it never to reach Mary, the ball’s velocity must not exceed Mary’s, hence we are looking at v4 = v5 as the limit.

In summary, our system of equations are:

1) m1 v1 = m2 v2

2) m1 v3 + m2 v3 = m1 v1 + m2 v2

3) m1 v3 + m2 v3 = m1 v4 – m2 v5

4) v2 = v5

5) v4 = v5

Solving simultaneously,

(Combining 2 and 3)

6) m1 v1 + m2 v2 = m1 v4 – m2 v5

(Combining 4, 5 and 6)

7) m1 v1 + m2 v5 = m1 v5 – m2 v5

m1 (v1 – v5) = – 2 m2 v5

(Combining 1 and 4)

8) m1 v1 = m2 v5

v1 = $\frac{\text{m2} \text{v5}}{\text{m1}}$

(Combining 7 and 8)

9) m1 $(\frac{\text{m2} \text{v5}}{\text{m1}}$ – v5) = – 2 m2 v5

m2 v5 – m1 v5 = – 2 m2 v5

3 m2 v5 = m1 v5

$\frac{\text{m1}}{\text{m2}}$ = 3 [Answer]

Let we assume: Mary's mass: M Ball's mass : m The velocity of the ball: V The velocity of Mary after the first throw: V'1 The velocity of Mary after the second throw: V'2

After the first throw, with the law of conservation of momentum, we get:

                                          0 = m(-V)+M(V'1)
                                      mV = M(V'1)........................(1)

Then, after the second throw, we get:

                        m(V)+M(V'1)= m(-V)+M(V'2)
                                     2mV= -M(V'1)+M(V'2).............(2)

Now, we must join those two equations and we can form:

                                     2mV= -mV+M(V'2)
                                     3mV= M(V'2)
                                 3V/V'2 = M/m............................(3)

After the second throw, we know that the ball never reached Mary, that means after the second throw the velocity of the ball is lower or equal to the velocity of Mary. To find the maximum ratio of Mary's mass to ball's mass, we need to see equation (3). Because V is constant, so we need the lowest possible value for V'2 to find the maximum value of M/m. Therefore, we take V'2=V because V'2 must be higher or equal to V. So, we can find:

                                                      V'2=V

                                           3V/V'2=M/m
                                            3V/V  =M/m
                                                   3 =M/m

So, the maximal value for M/m is 3.

Let $M$ be the mass of Mary, $m$ be the mass of ball, $v$ be the velocity of Mary, $V$ be the velocity of the ball which Mary throws initially.

In the first throw, using law of conservation of momentum (since the initial velocities of both the girl and the ball is zero), we infer that $m.V = M.v$ or $\frac {V} {v} = \frac {M} {m}$

In the second throw, in Mary’s frame of reference she would throw the ball with velocity $V$ and she herself would move backward with velocity $v$ . since Mary is already moving with a velocity, $v$ (due to the first throw) in our frame of reference Mary would now move with a velocity $2v$ and the velocity of the ball will be $(V-v)$ .

For the ball to not reach her after the second throw, velocity of the ball reaching Mary must be less than that of Mary. Since the collision with wall is elastic, the ball bounces off the wall after the second collision with a velocity, $(V-v)$ . Which makes $(V-v)\leq 2v$ . So at the extreme condition: $V-v=2v$ $\Rightarrow V=3v \Rightarrow \frac {V} {v} =3\Rightarrow \frac{M} {m} =3$