A baseball (m = 149 g) approaches a bat horizontally at a speed of40.2 m/s (90 m

Question

A baseball (m = 149 g) approaches a bat horizontally at a speed of40.2 m/s (90 mph) and is hit straight back at a speed of 45.6 m/s(102 mph). If the ball is in contact with the bat for a time of1.10 ms, what is the average force exerted on the ball by the bat?Neglect the weight of the ball, since it is so much less than theforce of the bat. Choose the direction of the incoming ball as thepositive direction. (NOTE: I used to be a high school baseballcoach, and one fundamental that I stressed to my players wasfollowing through with their swings. A follow-through allows thebat to spend more time in contact with the ball. Plug in a highervalue for contact time than the one listed, and see why followingthrough is important!)
a. -11 600 N
b. -9820 N
c. -5420 N
d. -10 800 N
e. -6180 N

Explanation / Answer

   Initialvelocity   u   =   40.2   m/s    Finalspeed   v   =  - 45.6   m/s   (-ve sign for reversed direction)    mass ofball   m   =   149   g                               =   0.149   kg    time ofcontact   t   =   1.10ms                                    =   1.10* 10-3   s    According to second law of motion    force   =   timerate of change of momentum                =   p/ t                =   (m* v - m * u) /t                                                               (p   =   m* v)                =   0.149* ( - 45.6 - 40.2) / 1.10 * 10-3                =     -1.16 * 104   N                =   -11600   N

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