••21. A 0.30 kg softball has a velocity of 15 m/s at an angle of35° below the ho

Question

••21.   A 0.30 kg softball has a velocity of 15 m/s at an angle of35° below the horizontal just before making contact with thebat. What is the magnitude of the change in momentum of the ballwhile in contact with the bat if the ball leaves with a velocity of(a) 20m/s, vertically downward, and (b) 20 m/s, horizontally backtoward the pitcher? Thank you! ••21.   A 0.30 kg softball has a velocity of 15 m/s at an angle of35° below the horizontal just before making contact with thebat. What is the magnitude of the change in momentum of the ballwhile in contact with the bat if the ball leaves with a velocity of(a) 20m/s, vertically downward, and (b) 20 m/s, horizontally backtoward the pitcher? Thank you! ••21.   A 0.30 kg softball has a velocity of 15 m/s at an angle of35° below the horizontal just before making contact with thebat. What is the magnitude of the change in momentum of the ballwhile in contact with the bat if the ball leaves with a velocity of(a) 20m/s, vertically downward, and (b) 20 m/s, horizontally backtoward the pitcher?

Explanation / Answer

p=    pafter -pbefore Consider x and y components x component px-before = mV1cos(35) x and y components py-before = mV1sin(35) . (a) 20m/s, vertically downward, and px-after = mV2cos(-90) py-after = mV2sin(-90) __________ px=   m [V2cos(-90)- V1cos(35)] py=   m[V2sin(-90)-V1sin(35)] . (b) 20 m/s, horizontally back toward the pitcher? px-after = mV2cos(0) py-after = mV2sin(0) ______________ px=   m [V2cos(0) -V1cos(35)] py=   m[V2sin(0) -V1sin(35)] . _________ And finally dont forget that p = (px2+py2) and = arcTan(py/px) Please let me know if you have any questions. x and y components py-before = mV1sin(35) . (a) 20m/s, vertically downward, and px-after = mV2cos(-90) py-after = mV2sin(-90) __________ px=   m [V2cos(-90)- V1cos(35)] py=   m[V2sin(-90)-V1sin(35)] . (b) 20 m/s, horizontally back toward the pitcher? px-after = mV2cos(0) py-after = mV2sin(0) ______________ px=   m [V2cos(0) -V1cos(35)] py=   m[V2sin(0) -V1sin(35)] . _________ And finally dont forget that p = (px2+py2) and = arcTan(py/px) Please let me know if you have any questions. x and y components py-before = mV1sin(35) . (a) 20m/s, vertically downward, and px-after = mV2cos(-90) py-after = mV2sin(-90) __________ px=   m [V2cos(-90)- V1cos(35)] py=   m[V2sin(-90)-V1sin(35)] . (b) 20 m/s, horizontally back toward the pitcher? px-after = mV2cos(0) py-after = mV2sin(0) ______________ px=   m [V2cos(0) -V1cos(35)] py=   m[V2sin(0) -V1sin(35)] . _________ And finally dont forget that p = (px2+py2) and = arcTan(py/px) Please let me know if you have any questions. px-after = mV2cos(0) py-after = mV2sin(0) ______________ px=   m [V2cos(0) -V1cos(35)] py=   m[V2sin(0) -V1sin(35)] . _________ And finally dont forget that p = (px2+py2) and = arcTan(py/px) Please let me know if you have any questions. px=   m [V2cos(0) -V1cos(35)] py=   m[V2sin(0) -V1sin(35)] . _________ And finally dont forget that p = (px2+py2) and = arcTan(py/px) Please let me know if you have any questions. And finally dont forget that p = (px2+py2) and = arcTan(py/px) Please let me know if you have any questions.

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