Question

Help!!!! Please..... Someone. Problem:

Starting with an initial speed of 5.00 m/s at a height of h = 0.260 m, the m1 = 1.55 kg ball swings downward and strikes the m2 = 4.50 kg ball that is at rest, as the drawing shows.

(a) Using the principle of conservation of mechanical energy, find the speed of the 1.55-kg ball just before impact.
(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.
(c) How high does each ball swing after the collision, ignoring air resistance?

Answer 1

And here is the picture from the problem.

Answer 2

for a)
energy=\[E=1/2 MV ^{2}\]=1.55kg*(25m/s)²*1/2=19.375j
gravitational potential energy = mhg =1.55kg*0.26m*10m/s²=4.03j
total energy before impact = 19.375j + 4.03j=23.405j

\[energy = 1/2mv²\]
\[\sqrt[2]{energy * (2/m)}= v\]
v=5.5m/s

Not sure if my answer is a correct one.

Answer 3

(a) \[m _{1}gh+(1/2)m _{1}v ^{2}=0+(1/2)u ^{2}\rightarrow u=\sqrt{2gh+v ^{^2}}=5.49ms ^{-1}\]
(b) Use conservation of momentum in horizontal direction (I don't think we need to worry about vertical component of momentum right after collision):
\[m_1v_i+0=m_1v_f+m_2u_f\]
where i'm using v for m1 and u for m2. Solve this for v_f in terms of u_f and the known v_i (=5.49m/s, from part a). Then write down the conservation of K.E (collision is elastic):
\[(1/2)m_1(v_i)^{2}+0=(1/2)m_1(v_f)^{2}+(1/2)m_2(u_f)^{2}\]
this equation, after substituting for v_f, will have only one unknown, u_f which you should be able to solve for after some fun algebra! After finding u_f, use that to get v_f. The direction is horizontal.
(c) as in part (a), use conservation of M.E for each mass. In this case your initial speeds are the ones you found in (b), namely v_f, and u_f.
I think this will lead you to correct answers. Hope this helps..

Answer 4

Correction: in part a replace \[(1/2)u ^{2}\]
with
\[(1/2) m_1 u ^{2}\]