Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of ((20.0 m/s)i - (4.0 m/s)j). During the 3 ms that the racket and ball are in contact, the net force on the ball is constant and equal to (-(380 N)i + (110 N)j.a) What are the x and y-components of the impulse of the net force applied tothe ball ?b) What are the x and y-components of the final velocity of the ball ?

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aristocles aristocles · Answer 1 · 2019-11-10T06:04:29+01:00

Answer:

[tex]\Delta P = -1.14 \hat i + 0.33\hat j[/tex]

Part b)

[tex]v_f = 0.03 \hat i + 1.79 \hat j[/tex]

Explanation:

Part a)

As we know that impulse is due to the force applied for small time due to which momentum is changed

so we will have

[tex]\Delta P = F\Delta t[/tex]

so we will have

[tex]F = -380 \hat i + 110 \hat j[/tex]

also we know that

[tex]\Delta t = 3 ms[/tex]

now impulse is given as

[tex]\Delta P = (-380\hat i + 110 \hat j)(3 \times 10^{-3})[/tex]

[tex]\Delta P = -1.14 \hat i + 0.33\hat j[/tex]

Part b)

As we know that

[tex]\Delta P = m(v_f - v_i)[/tex]

[tex](-1.14 \hat i + 0.33 \hat j) = (\frac{0.560}{9.81})(v_f - (20\hat i - 4\hat j))[/tex]

[tex]-19.97 \hat i + 5.79\hat i + 20 \hat i - 4\hat j = v_f[/tex]

[tex]v_f = 0.03 \hat i + 1.79 \hat j[/tex]