A small ball of mass m is aligned above a larger ball of mass M - 2.3ko (with a

Question

A small ball of mass m is aligned above a larger ball of mass M - 2.3ko (with a slight separation, as with the basebalil and basketbali of Figure (a)), and the two are dropped simultaneously from height - 2.1 m.(Assume the radius of each ball is negligible compared to A.) (a) I the larger ball rebounds elastically from the floor and then the smal bail rebounds elastically from the lerger ball, what value of m results in the larger ball stopping when it collides with the small bal? (b) What height does the small bal then reach (ee Figure (8))

Explanation / Answer

(a) ? kin energy of M-ball is the same as its pot energy, that is

0.5*M*V^2 = Mgh, hence V^2=2gh;

since V is not dependent of mass the speed of m-ball is also V;

? now all this bound-rebound stuff is just lyrics to muddle our brain; and we can re-word the problem like this;

“M-ball and m-ball collide head-to-head, both having speed V;

what is mass m if M-ball stops after collision?”

? then momentum conserv law says:

M*V - m*V = M*0+m*u,

where u is new speed of m-ball after collision;

thus u=(M-m)*V/m;

And energy conserv law says:

0.5m*V^2 +0.5M*V^2 = 0.5M*0^2 +0.5m*u^2;

thence m*V^2+M*V^2 = m*{(M-m)*V /m}^2; or deleting V^2;

m^2 +m*M =M^2 –2m*M +m^2, hence m=M/3 =2.3 /3 = 0.767kg;

and u =(2M/3)*V/(M/3) =2V= 2*sqrt(2gh) – (see ?);

(b) ? now forget M-ball;

here m-ball with kin energy

0.5m*u^2 =0.5m*(2V)^2 =4mgh

will stop with pot energy mgH reaching altitude H;

thus mgH=4mgh, hence H=4h = 4*2.1 = 8.4 m;

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