What indicates if an object will bounce back? If I throw a small rock (m = 1kg) at a big rock (100kg) the small rock rebounds.
Let's say my weight is 80kg, if I would jump into a big rock instead of bouncing back I would move in the same direction as a big rock. 
The big rock is heavier but it is not reflecting me. Why is that?
 A: The behavior of the rocks and yourself (which I'll refer to in the calculations as a general "human body") in collisions has to do largely with conservation of momentum and conservation of kinetic energy. In basic physics, there are two types of collisions that we can consider to model this situation: a perfectly elastic collision and a perfectly inelastic collision.
Elastic Collision
If we assume for a moment that the collision is elastic, the final direction of both objects (the $100 \,\text{kg}$ rock and the $80 \,\text{kg}$ human body) depends in part on the starting momentum of each object.
Momentum is defined as $p = m v$ - that is, mass times velocity.
So for each object,
$$p_r=m_r v_r$$
$$p_b=m_b v_b$$
where the "subscripts" r and b correspond to the rock and body, respectively. (I apologize for not being able to figure out how to write out subscripts.)
The total momentum before the collision would thus be written as
$$p_r + p_b = m_r v_r + m_b v_b$$
and the total momentum afterward can be written as
$$p_r' + p_b' = m_r v_r' + m_b v_b'$$
where the $'$ indicates that it's after the collision (not a first derivative). The mass is assumed to remain constant before and after the collision, or that any change in mass is negligible.
Because momentum is conserved, we can write
total momentum before = total momentum after
$$p_r + p_b = p_r'+ p_b'$$
$$m_r v_r + m_b v_b = m_r v_r' + m_b v_b'$$
Algebraically rearranging this equation to move terms with the same mass onto the same side gives us
$$m_r v_r - m_r v_r' = m_b v_b' - m_b v_b$$
Factoring out the mass terms gives us
$$m_r \left(v_r - v_r' \right) = m_b \left(v_b' - v_b \right) \qquad  (1)$$
In an ideal collision, kinetic energy is also conserved. Thus we have
$$(\text{total kinetic energy before}) = (\text{total kinetic energy after})$$
$$\frac{1}{2} m_r v_r^2 + \frac{1}{2} m_b v_b^2 = \frac{1}{2} m_r (v_r')^2 + \frac{1}{2} m_b (v_b')^2$$
The $1/2$ can be divided out of the whole equation, leaving
$$m_r v_r^2 + m_b v_b^2 = m_r (v_r')^2 + m_b (v_b')^2$$
Now rearrange the equation to move terms containing the same mass to the same side (as was done for the conservation of momentum equation):
$$m_r v_r^2 - m_r (v_r')^2 = m_b (v_b')^2 - m_b v_b^2$$
Factoring out the mass terms yields
$$m_r \left[v_r^2 - (v_r')^2 \right] = m_b \left[(v_b')^2 - v_b^2 \right]$$
Recall that the difference of two square numbers (or two squared variables) $a^2 - b^2 = (a+b)(a-b)$. The last equation can thus be further factored as
$$m_r (v_r + v_r')(v_r - v_r') = m_b (v_b'+ v_b)(v_b' - v_b) \qquad (2)$$
Now divide Equation $(2)$ by Equation $(1)$ to obtain
$$m_r (v_r + v_r') = m_b (v_b' + v_b)$$
We were wondering earlier about the direction of movement of the human body after the collision. So let's solve for $v_b'$ - the velocity of the body after collision:
$$v_b' =  \frac{m_r}{m_b} (v_r + v_r') - v_b ~.$$
We know that $m_r = 100 \,\text{kg}$ and $m_b = 80 \,\text{kg}$. This gives us
$$v_b' = \frac{100}{80} (v_r + v_r') - v_b$$
$$v_b' = \frac{5}{4} (v_r + v_r') - v_b ~~\text{, or}$$
$$v_b' + v_b = \frac{5}{4} (v_r + v_r') ~.$$
Thus, if the human body's starting velocity is large enough compared to the velocity of the rock, it might actually be possible to reverse the direction of the rock's movement to be the same as the human's. For the rock and body to come to a standstill (or $v_b' = v_r' = 0$),
$$v_b = \frac{5}{4} v_r ~.$$
If the human body manages to collide with the rock at a velocity greater than $5 v_r/4$, then the movement of both the rock and human body will be in the same direction as the initial direction of movement of the person.
Inelastic Collision
Now, what if we assume an inelastic collision, where the rock and human body "stick" together? This may be a more realistic scenario, since people don't tend to "bounce" like rocks do when they collide (or, better yet, billiard balls or steel ball bearings).
In this scenario, conservation of momentum would be described mathematically as follows:
$$(\text{total momentum before}) = (\text{total momentum after})$$
$$m_r v_r + m_b v_b = (m_r + m_b) v'$$
$$v' = \frac{\left[(m_r v_r) + (m_b v_b)\right]}{(m_r + m_b)}$$
Plugging in values of $m_r = 100 \,\text{kg}$ and $m_b = 80 \,\text{kg}$, we see that
$$v' = \frac{100 v_r + 80 v_b}{100 + 80}$$
$$v' = \frac{5}{9} v_r + \frac{4}{9} v_b$$
For the combined human-rock system to come to rest upon collision ($v' = 0$):
$$0 = \frac{5}{9} v_r + \frac{4}{9} v_b$$
$$-\frac{4}{9} v_b = \frac{5}{9} v_r$$
$$v_b = -\frac{5}{4} v_r ~.$$
Once again, we see that for any initial speed under $5/4$ the speed of the rock will result in the final person-rock system moving in the direction of the rock's initial movement.
Conclusions
Therefore, it is actually possible for the rock to be "reflected" in the opposite direction from its initial movement. It just requires the human body to have a much higher initial velocity (heading toward the rock in the opposite direction as the rock's initial movement) than the rock.
However, I would not highly recommend modeling this physically in real life, as other factors (such as the relative fragility of human tissue and bone compared to most rocks) would render an experiment fairly hazardous to any biological test subjects involved, as well as the experimenter him/herself.
A: Two laws govern collisions: conservation of momentum and conservation of energy. Momentum is the product of mass and velocity, so we can write
$$\sum m_i\cdot \vec{v_i} = const$$
Conservation or energy is a little bit trickier, since energy can be converted from one type to another. In an elastic collision, the kinetic energy is conserved, so
$$\sum \frac12 m_i \cdot v_i^2 = const$$
But for inelastic collisions, some of the energy can be lost as heat (or other forms of energy that are not kinetic). Now given that the total momentum is always conserved, you will get the greatest loss of kinetic energy if the two objects have the same velocity after the impact - in other words, they "stick together". You can show this mathematically or you can simply imagine a spring between the objects as they collide: the spring absorbs some of the energy until they are moving at the same speed, at which point the spring starts pushing them apart (in the process "giving back" some energy, so the kinetic energy of the objects increases just as the potential energy of the spring decreases). If the spring kept all the energy, the objects would remain stuck together.
So there it is. Whether something "bounces off" something else only depends on whether the collision is elastic ("the spring gives back the energy") or not (energy is converted to something else during the collision and not returned). When you hit a big rock, there is all kinds of deformation, maybe bones breaking -- but not much bouncing. While one pebble bouncing off another rock is mostly an elastic collision.
A: Only 7 years late, but I think I actually have an answer to this!
In this answer, object $A$ will be the smaller mass, and object $B$ will be the larger mass.
To figure out when object $B$ will recoil or when it will continue in the same direction, it's best to find point between those two cases: when object $B$ will stop after the collision.
From conservation of momentum, this gives us:
$$M_A V_{Ai} + M_B V_{Bi} = M_A V_{Af} + M_B V_{Bf}$$
$$M_A V_{Ai} + M_B V_{Bi} = M_A V_{Af} + 0$$
Now move the terms for object $A$ to the left and the rest to the right
$$ M_A V_{Ai} - M_A V_{Af} = - M_B V_{Bi} $$
$$ M_A (V_{Ai} - V_{Af}) = - M_B V_{Bi} $$
$$ - M_A (V_{Af} - V_{Ai}) = - M_B V_{Bi} $$
$$ M_A (V_{Af} - V_{Ai}) = M_B V_{Bi} <—[EQ1]$$
This equation will be used in a bit to divide both sides of another equation.
But for now, lets shift view to conservation of kinetic energy formula:
$$ 0.5M_A V_{Ai}^2 + 0.5M_B V_{Bi}^2 = 0.5M_A V_{Af}^2 + 0.5M_B V_{Bf}^2$$
$$ 0.5M_A V_{Ai}^2 + 0.5M_B V_{Bi}^2 = 0.5M_A V_{Af}^2 $$
$$ M_A V_{Ai}^2 + M_B V_{Bi}^2 = M_A V_{Af}^2 $$
$$ M_A V_{Ai}^2 - M_A V_{Af}^2  = - M_B V_{Bi}^2 $$
$$ M_A( V_{Ai} - V_{Af} )( V_{Ai} + V_{Af} )   = - M_B V_{Bi}^2 $$
$$ - M_A( V_{Af} - V_{Ai} )( V_{Ai} + V_{Af} )   = - M_B V_{Bi}^2 $$
$$  M_A( V_{Af} - V_{Ai} )( V_{Ai} + V_{Af} )   =  M_B V_{Bi}^2 $$
We now divide this equation by the previously mentioned formula [EQ1]:
$$ \frac{M_A( V_{Af} - V_{Ai} )( V_{Ai} + V_{Af} )}{M_A (V_{Af} - V_{Ai})} = \frac{M_B V_{Bi}^2}{M_B V_{Bi}} $$
That leaves us with
$$ V_{Ai} + V_{Af} = V_{Bi} $$
$$ V_{Af} = V_{Bi} - V_{Ai} <—[EQ2]$$
This equation is important because we'll use it to substitute away $V_{Af}$ in [EQ1]:
$$ M_A (V_{Af} - V_{Ai}) = M_B V_{Bi} $$
$$ M_A ((V_{Bi} - V_{Ai}) - V_{Ai}) = M_B V_{Bi} $$
$$ M_A (V_{Bi} - 2V_{Ai}) = M_B V_{Bi} $$
And now we just finish up by solving for $V_{Ai}$:
$$ V_{Ai} = \frac{- V_{Bi}}{2} (\frac{M_B}{M_A}-1)$$
TA DA!!! A formula that will tell you what the velocity of the smaller mass must be so that it just stops the larger mass in its tracks. Any velocity greater than that will cause the big mass to recoil back the other way, and any velocity lesser than that will cause the big mass to have a final velocity in the same direction as its original velocity :D
You can even verify it works with this simulation: https://ophysics.com/e2.html
