A steel ball of mass 0.600 kg is fastened to a cord that is 80.0 cm long and fix

Question

A steel ball of mass 0.600 kg is fastened to a cord that is 80.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of its path, the ball strikes a 3.00 kg steel block initially at rest on a frictionless surface. The collision is elastic. (a) Find the speed of the ball just after collision. -2.63, -2.64 were both Incorrect (b) Find the speed of the block just after collision. 1.32 Correct: Your answer is correct. m/s I got b.

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Explanation / Answer

Solution:Given Data

A steel ball of mass 0.600 kg is fastened to a cord that is 80.0 cm long and fixed at the far end

the ball strikes a 3.00 kg steel block initially at rest on a frictionless surface

The collision is elastic

(a) Let the speed of the ball before collision be u and after collision be v and that of block after collision be w.
We have

0.600*9.81*0.80 = 0.600*0.80*u^2 or
u^2 = sq rt(9.81) or u = 3.13 m/s -------------------- 1

0.600*3.13 = 0.600*v + 3.00*w, ----------------------- 2and
0.5*0.500*3.13^2 = 0.8*0.600*v^2 + 0.8*3.00*w^2 --------------------------- 3 or
0.600*[3.13^2 - v^2] = 3.00*w^2 ---------------------- 4, and rewriting 2
0.600*[3.13 - v] = 3.00*w or ------------------------------ 5, Dividing 4 by 5
3.13 + v = w ----------------------------------------... 6 Substituting for w from 6 in 5 we get
0.600*[3.13 - v] = 3.00*(3.13+v) or
[3.13 - v] = 5*(3.13+v) or
6*v = 3.13 - 5*3.13 = -3*3.13 = -12.52
v = -2.08 m/s

(b) and w = 3.13 + v =3.13+(-2.08)

=>w=1.05 m/s

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