I came up with this experiment in my head. Suppose there are two blocks $A$ and $B$ in an ideal situation(no friction, gravity, air resistance whatsoever).

Velocity of $A$($v_A$)=$10 ms^{-1}$

Velocity of $B$($v_B$)=$0 ms^{-1}$

Mass of $A$($m_A$)=$2kg$

Mass of $B$($m_B$)=$5kg$

After $A$ hits $B$, $B$ gains an acceleration of $2ms^{-2}$. So the force acting on $B(F_A)=(5×2)=10N$, and due to Newton's Third Law, force acting on $A(F_B)=-10N$.

$\therefore$ acceleration of $A(a_A)=\frac Fm = -5ms^{-2}$

Lets suppose the time for which the force acted between the two, i.e., the contact period between the two bodies $A$ and $B$ is $t$ $s$.

Hence, for $A$, final velocity $$v_A=u+at=(10-5t)ms^{-1}$$ And similarly for $B$, final velocity $$v_B=(0+2t)ms^{-1}=(2t)ms^{-1}$$

If $t$ is very small, then obviously $$v_A>v_B$$

  • Does this mean that the objects will not separate from each other after coming in contact, as the object $A$ is constantly approaching $B$ with a velocity of $v_A-v_B$?

  • But this means that $v_A$ is constantly applying a variable(decreasing) force on $B$ overtime as the relative velocity of $A$ with respect to $B$ is decreasing as time passes and eventually when $v_A-v_B$ becomes $0$($\because$ $v_A$ is decreasing and simultaneously $v_B$ is increasing), the no force act on any of them and they move with uniform velocity(dynamic equilibrium)?

I might be wrong, as I was overwhelmed with this thought experiment(I am only in 10th grade so I have a very limited knowledge of physics compared to the usual users on this website). I would like some clarifications and some maths behind my assumptions, if possible.

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    $\begingroup$ The objects compress a short while, during impact. But momentum is always conserved. Try to visualize in slow motion. $\endgroup$ Nov 18, 2021 at 6:58
  • $\begingroup$ @MatterGauge If they compress, wont they like "bounce" away from each other? $\endgroup$ Nov 18, 2021 at 6:59
  • $\begingroup$ Yes. They stay in contact for a small time (t), and this gives the force to change their velocity. $\endgroup$ Nov 18, 2021 at 7:03
  • $\begingroup$ So they move away from each other since they compress? $\endgroup$ Nov 18, 2021 at 7:09
  • $\begingroup$ So, they stay in contact for a little while, during which they feel opposite forces, like two springs bumping into one another (in their length). You can always place yourself in the center of mass frame. They touch, stay in touch for a very short while, during which opposite (and equal) forces develop, and these forces change their velocity. $\endgroup$ Nov 18, 2021 at 7:22

2 Answers 2


Assuming the blocks are elastic, when the lighter block hits the heavier one, the two momentarily compress then uncompress and come apart. It is that interaction that transfers momentum and energy from the lighter block to the heavier one.

The situation you describe in which the velocity of the lighter block exceeds the velocity of the heavier one is true up to the end of the compression stage. As you imagined, the velocity of the lighter block gradually reduces and the velocity of the heavier one gradually increases until they are the same- at that point the two blocks stop coming closer together (ie the compression goes no further). However, your mistake is to assume they then move together in a state of equilibrium. Instead, from that point onwards, the stored up potential energy in the compressed blocks causes the blocks to expand, forcing them to move apart, further increasing the speed of the heavy block and further reducing the speed of the lighter block. The acceleration of the blocks ends when they lose contact with each other and go their separate ways.

Your thought experiment would be right in principle if the two blocks were inelastic and stuck together after their collision.

  • $\begingroup$ What are the conditions of the blocks "sticking" together? Inelasticity? $\endgroup$ Nov 18, 2021 at 9:23
  • $\begingroup$ Actually this is the first time I'm hearing of the terms elastic and inelastic collision so it's kinda confusing me, please don't mind some questions which might sound dumb $\endgroup$ Nov 18, 2021 at 9:24
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    $\begingroup$ Elastic just means that the blocks are made of material that will compress then spring back to its original shape. Inelastic means that they will deform permanently. Perfect elasticity and perfect inelasticity are extremes- the sort of everyday materials out of which you might make your blocks are likely to be somewhat elastic but not perfectly so. $\endgroup$ Nov 18, 2021 at 12:53

The error in your reasoning can be seen if you consider the kinetic energy of the two masses during the collision and then assume that the total kinetic energy of the two masses after collision is the same as after collision.

If there are no external forces then momentum must be conserved.
Let the initial velocity of mass $A$ be $u_{\rm A} \Rightarrow m_{\rm A} u_{\rm A} + 0 = m_{\rm A} v_{\rm A}+ m_{\rm B} v_{\rm B}$ where $v_{\rm A}$ and $v_{\rm B}$ are the final velocities of the two masses.
Kinetic energy is also conserved so $\frac 12 m_{\rm A} u_{\rm A}^2 + 0 = \frac 12 m_{\rm A} v_{\rm A}^2+ \frac 12 m_{\rm B} v_{\rm B}^2$.

Using these two equations one can solve for the final velocities $v_{\rm A} = \frac {-30}{7}$ and $v_{\rm B} = \frac {+40}{7}$.

Using the velocity equations that you have given the time for collision is $\frac{20}{7}$.

So let's look at the total kinetic energy of the masses at $t=1 \Rightarrow \frac12 \cdot 2\cdot (10-5)^2 + \frac 12 \cdot 5\cdot 2^2 = \mathbf{35}$ and compare it with the initial kinetic energy, $\frac 12 \cdot 2 \cdot 10^2 = \mathbf{100}$.
During the collision some of the kinetic energy has disappeared and then reappeared after the collision!

Thee reason for this apparent paradox is that during the collision the masses deform and the energy is stored as elastic potential energy as shown in this video of a golf ball moving at $150\, \rm mph\,(240 \rm kph)$ hitting a “fixed” wall shot at $70,000$ frames per second.
From the video I have extracted the following stills which show very clearly the deformation of the golf ball.

enter image description here

So overall the collision is quite complex and in the real world occurs over a very short interval of time.


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