Question

A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by 1.0 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son

Answer 1

a. 2.41 m/s

b. 4.82 m/s

Step-by-step explanation:

Father's KE is:

K. E. = ½mv²

Sons k. e. is:

k. e. = ½MV²

Since M = ½m:

k. e. = ¼mV²

We are given that K. E. = ½k.e., therefore:

½mv² = ½(¼mV²)

½mv² = ⅛mV²

=> v² = V²/4

=> v = V/2 and V = 2v

When the father's speed is (v + 1) m/s, K. E. = k. e.:

K. E. = ½m(v + 1)²

=> ½m(v + 1)² = ¼mV²

(v + 1)² = ½V²

Since V = 2v

=> (v + 1)² = ½(2v)²

v² + 2v + 1 = ½(4v²)

v² + 2v + 1 = 2v²

=> v² - 2v - 1 = 0

=> v = 2.41m/s or -0.41m/s

Since speed cannot be negative:

v = 2.41m/s

Recall that V = 2v

=> V = 2 * 2.41 = 4.82m/s

Hence, the father's speed is 2.41m/s and the son's speed is 4.82m/s.