Question

Two masses, mA — 45 g and mB 65 g, are suspended as shown in the diagram. The length (L) of the massless string is 30 cm. Both masses start from rest with mA starting at 0 60 deg,

B

A. Determine the velocity of mA just before it collides with mB.

B. The collision between mA and         is elastic. Determine the velocity of and      after the collision.

C. Determine the maximum height that and mB attain after the collision.

Answer 1

(a)

m1 falls through a height of 0.30*cos60º

½mv² = mgh

v = √(2gh) = √(2 x9.81 x 0.3 x cos60º) = 1.71 m/s.

(b)
mA*va + mB*vb = 1.71*mA

0.045va + 0.065vb = 0.077

va + 1.44vb = 1.71 .............. (eq1)
and:
½mA* va² + ½mB*vb² = ½mA.(1.7155)²
0.028va² + 0.076.vb² = 0.1315845

va² + 1.44 vb² = 2.9241 ............... (eq2)

solving both equations

(1.71-1.44Vb)² + 1.44 vb² = 2.9241

(1.71-1.44Vb)² + 1.44 vb² = 2.9241

2.9241 – 4.9248 Vb + 2.0736 Vb² + 1.44 vb² = 2.9241

vb = 1.4 m/s

va = 1.71 – 1.44 x 1.4 = - 0.306 m/s.

(c)

ha = recovery height of mA = 0.045 x 0.306²/(2 x 9.8 x 0.045) = 0.004772 m = 0.477 cm

hb = height of mB = 1.4²/(2 x 9.81) = 0.1 m = 10 cm