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A 0.144 kg baseball is moving towards home plate with a speed of 43.0 m/s when it is bunted. the bat exerts an average force of 6.50 × 103 n on the ball for 1.30 ms. the average force is directed towards the pitcher, which we take to be the positive x direction. what is the final speed of the ball?

Respuesta :

As per impulse momentum theorem we know that

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

now here we will have

[tex]F = 6.50 \times 10^3 N[/tex]

t = 1.30 ms

m = 0.144 kg

[tex]v_i = -43 m/s[/tex]

now we need to find final speed using above formula

[tex](6.50 \times 10^3)(1.30 \times 10^{-3}) = 0.144 ( v_f - (-43))[/tex]

[tex]v_f = 15.7 m/s[/tex]

so final speed is given as above

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