Question

A 0.157-kg baseball moving at 21.2 m/s strikes the glove of a catcher. The glove recoils a distance of 10.7 cm.The magnitude of the average force applied by the ball on the glove is __.

Answer 1

The magnitude of average force applied by ball on glove is

|F_avg| =330N

Explanation:

Mass of baseball m = 0.157 kg

Initial velocity v_1 = 21.2 m/s

Final velocity v_f = 0 m/s

Distance moved before coming to rest

s = d = 10.7 cm = 0.107 m

V_f ^2 = V_i^2/2 s

= 0^2 - (21.2)^2/2* 0.107

a= 449.44 /0.214

a = -2100 m/s^2

a- acceleration of baseball

Average force applied by the ball on glove |F_avg| = m|a|

=0.157*2100

=329.7 N

|F_avg| =330 N

Answer 2

Answer:329.72 N

Explanation:

Given

mass of ball $m=0.157\ kg$

velocity of ball $v=21.2\ m/s$

glove recoil distance $d=10.7\ cm$

Energy associated with ball E=kinetic energy of ball

$E=\frac{1}{2}mv^2$

$E=\frac{1}{2}\times 0.157\times 21.2^2$

$E=35.28\ J$

Now if an average for is applied to stop the ball then work done by this force is equal to kinetic energy of the ball

$W=F_{avg}\cdot d=E$

$F_{avg}=\frac{E}{d}=\frac{35.28}{0.107}=329.72\ N$