Isiranwaridn
Answered

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A 90-kg tight end moving at 9.0 m/s encounters at 400 N•s impulse. Determine the velocity change of the tight end.

Sagot :

Δp=mΔv
400 N.s = (90kg) (vf-9m/s)
400 = 90 vf - 810
400 + 810 = 90 vf
1210 = 90vf 
vf= 13.4 m/s

Answer: The change velocity of the tight end is 4.44 m/s

Explanation:

Mass of the tight end = 90 kg

Initial velocity =[tex]u[/tex] = 9.0 m/s

Final velocity =[tex]v[/tex]

[tex]Impulse=Force\times Time[/tex]

[tex]Impulse=Mass \times acceleration\times Time[/tex]

[tex]a=\frac{dv}{dt}=\frac{[v-u]}{t}[/tex]

[tex]impulse=Mass \times [v-u][/tex]

[tex]400 Nm=90 kg\times[v-9.0 m/s][/tex]

[tex]v=13.44 m/s[/tex]

Change in velocity = 13.44 m/s - 9m/s = 4.44 m/s

The change velocity of the tight end is 4.44 m/s

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