# What work is done by the wall when a ball hits it?

So, in my physics textbook it's written that the kinetic energy possessed by a body is equal to the amount of work done by an opposing force to stop the body, now that got me thinking that when a ball hits a wall, the wall which is the opposing force does no work because there is no displacement, so to say isn't the kinetic energy possessed by the ball zero?

• COM does not apply here so your physics textbook's assumption cannot be applied. – QuIcKmAtHs Mar 3 '19 at 2:09

The comment, force by the wall that does not deform (the ideal, usual case assumed in the question) does no work, made by @JánLalinský has helped me rewrite my answer.

I will consider what happens when the wall does not move and thus the wall does no work on a ball.
The effect a golf ball moving at $$150\, \rm mph\,(240 \rm kph)$$ hitting a “fixed” wall can be seen in this video shot at $$70,000$$ frames per second.
From this video I have extracted the following stills. The ball hits the wall with the wall exerting an external force on the left-hand side of the ball which is in contact with the wall.
What you can infer from the stills is that the left-hand side of the ball hits the wall and slows down to a stop very rapidly whilst the inside of the ball is still moving to the right.
As time progresses the ball is compressed with its internal parts slowing down in the process and some of the ball's kinetic energy is converted into elastic potential energy within the ball in the same way as when a spring is compressed.
Eventually the centre of mass of the ball must stop moving relative to the wall and at that stage the ball's initially kinetic energy has been converted into elastic potential energy, oscillatory kinetic energy of the ball and has done work permanently deforming the ball - breaking bonds.
That oscillatory motion of the internal parts of the ball is due to compression pulse(s) moving within the ball and this effect is the basis of some experiments used to measure the speed of sound in a rod as described here. There is also heat generated some of which is due to the damping of the oscillatory motion of the internal parts of the ball.

Thus a ball can stopped the ball by exerting an external force on it due to a wall with the wall assumed not to deform and thus doing no work.

. . . the kinetic energy possessed by a body is equal to the amount of work done by an opposing force to stop the body . . .

It is the internal forces within the ball which are the forces doing the work.

The process is then reversed with the elastic potential energy being converted into the kinetic energy of the ball as its centre of mass moves to the right but the collision is not elastic as some of the ball's initial kinetic energy has been converted into heat, (sound), and used to permanently deform the ball - break bonds.

• Force by the wall that does not deform (the ideal, usual case assumed in the question) does no work. Also not all kinetic energy of the system changes into potential energy, a big part of it changes into heat. – Ján Lalinský Mar 2 '19 at 12:32
• @JánLalinský I did state that the collision was elastic and so there is no heating, permanent deformation of the bodies etc. I further assumed that both bodies were capable of deformation, no matter how small. – Farcher Mar 2 '19 at 12:37
• What you are describing is a collision with a tennis racket, not a wall. In a realistic model of impact on a wall, the wall deforms only a little and does only insignificant amount of work. – Ján Lalinský Mar 3 '19 at 11:06
• @JánLalinský I understand what you are saying and I am in the process of rewriting my answer. Thank you for the steer. – Farcher Mar 3 '19 at 11:57
• @JánLalinský I have rewritten my answer. – Farcher Mar 4 '19 at 10:54

The wall doesn't work. Your textbook should have taken the conservative of momentum approach instead. When the ball hits the wall, momentum is conserved but because the collision isn't perfectly elastic, kinetic energy isn't conserved which results in some kinetic energy being lost as heat, sound, and vibration. Most likely the ball will bounce off the wall, but if it it doesn't, then the wall and everything attached to it must be moving at a ridiculously slow speed that can be calculated using the law of conservation of momentum: $$m_{1i}v_{1i} + m_{2i}v_{2i} = m_{1f}v_{1f} + m_{2f}v_{2f}$$

• So what's the resistive force here? – divyam sureka Mar 2 '19 at 7:47
• Inertia I believe. – TechDroid Mar 2 '19 at 7:50
• So basically my textbook is wrong? – divyam sureka Mar 2 '19 at 7:50
• In that regard, yeah. – TechDroid Mar 2 '19 at 7:53