The ball has a mass of 0.40 kg. Initially, it moves horizontally to the left at 20 m/s, but2.then it is kicked and given a velocity with magnitude 30 m/s and direction 45° upward and to theright. Find the impulse of the force and the average force on the ball, assuming a collision timeAt 0.010 s.Uf 30 m/sAFTERavy0.40 kgm.45°wwwwwF.Ui 20 m/sav,xBEFORE(a) Before-and-after diagram(b) Average force on the ball

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Asked Oct 7, 2019
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The ball has a mass of 0.40 kg. Initially, it moves horizontally to the left at 20 m/s, but
2.
then it is kicked and given a velocity with magnitude 30 m/s and direction 45° upward and to the
right. Find the impulse of the force and the average force on the ball, assuming a collision time
At 0.010 s.
Uf 30 m/s
AFTER
avy
0.40 kg
m.
45°
wwwww
F.
Ui 20 m/s
av,x
BEFORE
(a) Before-and-after diagram
(b) Average force on the ball
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The ball has a mass of 0.40 kg. Initially, it moves horizontally to the left at 20 m/s, but 2. then it is kicked and given a velocity with magnitude 30 m/s and direction 45° upward and to the right. Find the impulse of the force and the average force on the ball, assuming a collision time At 0.010 s. Uf 30 m/s AFTER avy 0.40 kg m. 45° wwwww F. Ui 20 m/s av,x BEFORE (a) Before-and-after diagram (b) Average force on the ball

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Expert Answer

Step 1

Given:

Mass of the ball, m = 0.4 kg

Initial velocity = 20 m/s horizontally left

Final velocity = 30 m/s at 45 degrees from horizontal

Time of collision = 0.010 s

Step 2

Calculating the impulse during the collision:

Impulse, I FE XAt ....(1)
cng
Av
x Д
At
I F XAt
спg
I = mAv
Initial velocity of the ball, ū=-20
Final velocity of the ball, v30 cos(45°)f+30 sin(45°)
v= 21.2(+)
Or Av 21.2(+)-(-20f)
AD 41.2+21.2
Hence, Impulse, I = mAỹ = 0.4x(41.2î+21.2
I =(16.48)i+(8.48) ...(2)
Magnitude of impulse is,
(16.48) +(8.48) =18.53 kgm/s
2
|I 18.53 kgm
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Impulse, I FE XAt ....(1) cng Av x Д At I F XAt спg I = mAv Initial velocity of the ball, ū=-20 Final velocity of the ball, v30 cos(45°)f+30 sin(45°) v= 21.2(+) Or Av 21.2(+)-(-20f) AD 41.2+21.2 Hence, Impulse, I = mAỹ = 0.4x(41.2î+21.2 I =(16.48)i+(8.48) ...(2) Magnitude of impulse is, (16.48) +(8.48) =18.53 kgm/s 2 |I 18.53 kgm

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Step 3

Calculating the average for...

Recalling equation 1, Impulse, I = Fng X At . ...(1)
Time taken in collison, At = 0.010s
Impulse, I(16.48)î +(8.48)j (From eq 2)
I
Hence,
ang
At
(16.48)î +(8.48)
ang
0.010
Fax(1648)+(848)
g
1648)2 +(848)2 =1853.4 N
Magnitude of average force, Fn
Fa1853.4 N 1853 N
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Recalling equation 1, Impulse, I = Fng X At . ...(1) Time taken in collison, At = 0.010s Impulse, I(16.48)î +(8.48)j (From eq 2) I Hence, ang At (16.48)î +(8.48) ang 0.010 Fax(1648)+(848) g 1648)2 +(848)2 =1853.4 N Magnitude of average force, Fn Fa1853.4 N 1853 N

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