Question

Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 20.0 m from each other. Being very skilled players, they play without a net. The tennis ball has mass 0.0580 kg. You can ignore gravity and assume that the ball travels parallel to the ground as it travels between the two players. Unless otherwise specified, all measurements are made by the two men. Lorentz serves the ball at 80m/s. Einstein slams a return at 1.8 x 10^8 m/s Question: Presuming the force that Einstein impacted to the ball was constant and done over two centimeters, what was this force?

Answer 1

Given,

mass of the ball m = 0.0580 kg

Initial velocity, u = 80 m/s

Thus,

Kinetic energy before the impact = 1/2*m*u²

                                                      = 0.5 * 0.0580 * (80)² = 185.6 J

Now,

Speed of the ball after the impact, v = 1.8 * 10⁸ m/s

Thus,

Realtivistic momentum, p = mv/

                                            = 0.0580 * 1.8 * 10⁸ /

                                            = 0.0580 * 1.8 * 10⁸ /

                                             = (0.0580 * 1.8/ 0.8) * 10⁸ = 1.305 * 10⁷

Kinetic energy, K.E = p² / 2m

                                = ( 1.305 * 10⁷ )² / ( 2 * 0.0580)

                                = 1.468 * 10¹⁵ J

Now,

distance for which impact lasted, d = 2 cm = 2 * 10^-2 m

Now,

Since Initial kinetic energy is very small compared to final kinetic energy, hence all the final kinetic energy is due to work done by the force applied by Einstein

=> F*d = 1.468 * 10¹⁵

=> F * 2 * 10^-2 = 1.468 * 10¹⁵

=> F = (1.468 * 10¹⁵) / (2 * 10^-2)

         = 7.34 * 10¹⁶ N

Hence, contant force applied is 7.34 * 10¹⁶ N