I ONLY NEED PART B ANSWERED A 77.0-kg fullback running east with a speed of 5.40

Question

I ONLY NEED PART B ANSWERED

A 77.0-kg fullback running east with a speed of 5.40 m/s is tackled by a 79.0-kg opponent running north with a speed of 3.00 m/s. (a) Explain why the successful tackle constitutes a perfectly inelastic collision. ___________________

(b) Calculate the velocity of the players immediately after the tackle.

magnitude=_____ m/s

direction=______ ° north of east

HINT: Find the total momentum of the two players before the collision and use conservation of momentum to find their momentum and velocity immediately after the tackle. m/s

(c) Determine the mechanical energy that disappears as a result of the collision. _______ J

Explanation / Answer

let Vx and Vy are the x and y components of velocity the players after the collision.

b) Apply conservation of momentum in x-direction

77*5.4 = (77 + 79)*Vx

==> Vx = 77*5.4/(77 + 79)

= 2.67 m/s

Apply conservation of momentum in y-direction

79*3 = (77 + 79)*Vy

==> Vy = 79*3/(77 + 79)

= 1.52 m/s

so, V = sqrt(Vx^2 +Vy^2)

= sqrt(2.67^2 + 1.52^2)

= 3.07 m/s   <<<<<<<<-------------Answer

direction: theta = tan^-1(Vy/Vx)

= tan^-1(1.52/2.67)

= 29.7 degrees North of East   <<<<<<<<-------------Answer

c) the mechanical energy that disappears as a result of the collision = Ki - Kf

= (1/2)*77*5.4^2 + (1/2)*79*3^2 - (1/2)*(77 + 79)*3.07^2

= 743 J <<<<<<<<-------------Answer

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