A baseball player pitches a fastball toward home plate at aspeed of 47 m/s. The

Question

A baseball player pitches a fastball toward home plate at aspeed of 47 m/s. The batter swings, connects with the ball of mass141 g, and hits it so that the ball leaves the bat with a speed of39 m/s. Assume that the ball is moving horizontally just before andjust after the collision with the bat.

What is the magnitude of the change in momentum of theball?

What is the impulse delivered to the ball by the bat?

  If the bat and ball are in contact for 0.001s, whatis the magnitude of the average force exerted on the ball by thebat?

Explanation / Answer

         Given that theinitial speed of ball is u = 47 m/s         final velocity ofthe ball is v = -39 m/s         mass of ball is m =0.141 kg --------------------------------------------------------------- The change in momentum is P = mu - mv          = 0.141kg ( 47 m/s+ 39 m/s )                                             =--------- kg m/s The impulse is I = change in momentum    = P                         = -------- kg.m/s Force is F = P /dt                   = -------- N ( where time is dt = 0.001 s )

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