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A 150-g (weight = 5.30 oz) baseball pitched at a speed of 41.6 m/s (= 136 ft/s) is hit straight back to the pitcher at a speed of 61.5 m/s (= 202 ft/s). The bat is in contact with the ball for 4.70 ms. Find the average force exerted by the bat on the ball.

A 150-g (weight = 5.30 oz) baseball pitched at a speed of 41.6 m/s (= 136 ft/s) is hit straight back to the pitcher at a speed of 61.5 m/s (= 202 ft/s). The bat is in contact with the ball for 4.70 ms. Find the average force exerted by the bat on the ball.

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
8 years ago
We have to choose the co-ordinate system in such a way that, the baseball is only moving along the x-axis, with away from the batter as positive.

The initial momentum pix of the baseball will be,
pix = mvix

and the final momentum pfx of the baseball will be,

pfx = mvfx

Here, mass of the baseball is m, initial velocity of the baseball is vix and final velocity of the baseball is vfx.
To obtain the final momentum pfx of the ball, substitute 150-g for the mass of the baseball and 61.5 m/s for vfx in the equation pfx = mvfx,
pfx = mvfx
= (150 g) (61.5 m/s)
= (150 g×10-3 kg/1 g) (61.5 m/s)
= 9.23 kg.m/s
To obtain the initial momentum pix of the ball, substitute 150-g for the mass of the baseball and -41.6 m/s (negative sign for opposite direction of the baseball) for vix in the equation pix = mvix,
pix = mvix
= (150 g) (-41.6 m/s)
= (150 g×10-3 kg/1 g) (-41.6 m/s)
= -6.24 kg.m/s
To find out the impulse Jx imparted to the ball, substitute 2.4 N.s for pix and 0 for pfx in the equation Jx = pfx – pix,
Jx = Δp
= pfx – pix
= (9.23 kg.m/s) – (-6.24 kg.m/s)
= 15.47 kg.m/s
Therefore the impulse J imparted to the baseball would be 15.47 kg.m/s.
Average force Fav,x will be,
Fav,x = Jx/ Δt
To obtain the average force Fav,x exerted by the bat on the ball, substitute 15.47 kg.m/s for Jx and 4.70 ms for Δt in the equation Fav,x = Jx/ Δt, we get,
Fav,x = Jx/ Δt
= (15.47 kg.m/s)/(4.70 ms)
= (15.47 kg.m/s)/(4.70 ms×10-3 s/1 ms)
= 3290 kg.m/s2
= (3290 kg.m/s2) (1 N/1kg.m/s2)
= 3290 N
From the above observation we conclude that, the average force Fav,x exerted by the bat on the ball would be 3290 N.




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