Question

9. A 60 kg football receiver starts from rest, accelerating at a rate of 0.36 g’s for 1.42 seconds until he runs as fast as he can. A ball of mass 0.25 kg is thrown, and the player catches the ball 3.35 seconds after the play starts. If the ball is travelling at 16 m/s when the player catches it, (a) how far down field did the player catch it, (b) what will the player’s final speed be after catching the ball, (c) what are the kinetic energies of the ball and player separately before the catch, and (d) what is the kinetic energy of the ball – player system after the catch? (e) Compare your results of (c) and (d) and explain.

Answer 1

here,

mass of the player , m = 60 kg

accelration , a = 0.36 g

t = 1.42 s

mass of ball , m' = 0.25 kg

speed of ball , ub = 16 m/s

(a)

the distance travelled by the player , d = 0 + 0.5 * 0.36*9.8 * 1.42^2^2 + ( 0.36*9.8 * 1.42)* 3.35

d = 20.34 m

the distance travelled by the receiver is 20.34 m

(b)

player speed before catch , v' = 0.36 * 9.8 * 1.42

v' = 5.01 m/s

let the final speed of the player

using conservation of momentum

m*v' - mb * ub = ( m + mb) * u'

60 * 5.01 - 0.25 * 16 = ( 60 + 0.25) *u'

u' = 4.92 m/s

the player's final velocity after the catching ball is 4.92 m/s

(c)

the kinetic energies of the ball , KEb = 0.5 * mb * ub^2

KEb = 0.5 * 0.25 * 16^2

KEb = 32 J

the kinetic energy of the reciver before catching the ball , KEr = 0.5 * m * v'^2

KEr = 0.5 * 60 * 5.01^2

KEr = 753 J

(d)

the kinetic energy of the ball-player system , KE = 0.5 * ( m + m' ) * u'^2

KE = 0.5 * ( 60.25) * 4.92^2

KE = 729.22 J

the kinetic energy of the ball-player system is 729.22 J