Question

A 48.5-kg skater is traveling due east at a speed of 2.65 m/s. A 72.0-kg skater is moving due south at a speed of 7.45 m/s. They collide and hold on to each other after the collision, managing to move off at an angle ? south of east, with a speed of vf. Find the following. (a) the angle ? ° (b) the speed vf, assuming that friction can be ignored

Answer 1

(theta = A)

Using Momentum conservation In x-direction

Pix = Pfx

m1*v1x + m2*v2x = M*Vx

m1 = 48.5 kg & m2 = 72 kg

v1x = v1 = 2.65 m/sec

v2x = 0 m/sec

Vx = V*cos A

M = 48.5 + 72 = 120.5

48.5*2.65 + 0 = 120.5*V*cos A

V*cos A = 128.525/120.5 = 1.07 m/sec

In y-direction

m1*v1y + m2*v2y = M*Vy

v1y = 0 m/sec

v2y = -7.45 m/sec

Vy = -V*sin A

0 + 72*(-7.45) = -120.5*V*sin A

V*sin A = 72*7.45/120.5

V*sin A = 4.45

Also we have

V*cos A = 1.07

divide both equation

tan A = 4.45/1.07

A = arctan (4.45/1.07) = 76.48 deg south of east

Part B

square and add both equation

V^2 = 1.07^2 + 4.45^2

V = sqrt (1.07^2 + 4.45^2)

V = 4.58 m/sec

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