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ASHA 777 [7]
3 years ago
6

A 0.30 kg softball has a velocity of 15 m/s at an angle of 35 degrees below the horizontal just before making contact with the b

at. What is the magnitude of the change in momentum of the ball while in contact with the bat if the ball leaves with a velocity of (a) 20 m/s, vertically downward, and (b) 20 m/s, horizontally back toward the pitcher?
Physics
1 answer:
jarptica [38.1K]3 years ago
6 0

Answer:

a) 5.03 kg m/s

b) 10.03 kg m/s

Explanation:

Hi!

Let us consider the origin of coordinates at the pitcher, and pointing directly towards the initial direction of the ball. Therefore, the angle of the velocity with respect to the x axis is -35° (below the horizontal).

The components of the initial momentum are:

px = (0.3 kg)(15 m/s) cos(-35° ) = 4.5 cos(-35° ) kg m/s = 3.69 kg m/s

py = 4.5 sin(-35° ) kg m/s = -2.581 kg m/s

The final momentum will be:

a)

pfx = 0

pfy = - (20 m/s) (0.3 kg) = -6 m/s

And the difference in momentum is:

dpy = pfy - py = -3.419 kg m/s

dpx = pfx - px = -3.69 kg m/s

And its magnitude:

dp = √(dpy^2 + dpx^2) = 5.03 kg m/s

b)

pfx = -6 kg m/s

pfy = 0

And the difference in momentum is:

dpx = pfx - px = -9.69 kg m/s

dpy = pfy - py = 2.581 kg m/s

And its magnitude:

dp = √(dpy^2 + dpx^2) = 10.03 kg m/s

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