Where is 90% of the energy lost when you throw a clay ball at a door to move it?

Question

Let us say I have a clay ball of mass $1 \ \text{kg}$. I throw it at a door of mass $10 \ \text{kg}$ with a speed of $10 \ \text{m/s}$. Let us say the ball sticks to the door on contact. I am trying to find the final velocity $v$ of the door and clay.

Using conservation of momentum:

$$1 \cdot 10+10 \cdot 0=11 \cdot v$$

$$v= \frac{10}{11}\approx 0.9$$

Using conservation of energy:

$$0.5 \cdot 1 \cdot 100+0.5 \cdot 10 \cdot 0=0.5 \cdot11 \cdot v \cdot v$$

$$v=\sqrt{\frac{100}{11}} \approx \ 3$$

So, as per google, inelastic collisions do not conserve energy, only momentum is conserved. Therefore, $v=10/11$ is the correct answer to choose.

Now, calculating energy lost for other stuff not involved in moving the door:

$$ \text{Lost energy} =0.5 \cdot 1 \cdot 10 \cdot 10-0.5 \cdot 11 \cdot \left(\frac{10}{11}\right)^2 \approx 50-4.5=45.5$$

So, looks like $90 \%$ of initial energy is lost in the process of clay ball sticking to the door! Is my calculation correct? What is the intuitive explanation for this loss of energy?

Answer 1

Yes, your calculations look correct. The energy lost in an inelastic collision are often turned into sound, light, or heat energy. As the clay hits the door, one or both of the objects deform, and it's the original object's kinetic energy that goes into rearranging the molecules. You may notice that an object usually heats up when it is deformed, this a common occurrence where useful kinetic energy gets dissipated as heat.

Answer 2

So, looks like 90% of initial energy is lost in the process of clay ball sticking to the door! Is my calculation correct? What is the intuitive explanation for this loss of energy?

• Permanent deformation of the ball. This requires work,
• Sound is a minor contribution: such collisions usually generate a 'thudding' noise,
• Viscous deformation of the ball generates a little (imperceptible) heat.

Answer 3

We can apply the second law of Newton for the time of the contact of the clay ball with the door. Considering the force that the door does on the ball has the same direction of its inicial velocity (to make things simpler):

$$F = ma \implies Fdx = m\frac{dv}{dt}dx = m dv\frac{dx}{dt} = mvdv = d(\frac{1}{2}mv^2)$$

Where $dx$ is the infinitesimal displacement of portions of the ball under the action of the contact force. That means: each small change of kinetic energy corresponds to a small quantity $Fdx$. That quantity is called work done on the ball.