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8.86: (a) For momentum to be conserved, the two fragments must depart in opposite<br />

directions. We can thus write<br />

M V = −M<br />

V<br />

A<br />

A<br />

B<br />

B<br />

Since M = M − , we have<br />

A<br />

M B<br />

( M − MB)<br />

V<br />

V<br />

V<br />

A<br />

A<br />

B<br />

= −M V<br />

B<br />

B<br />

− M<br />

B<br />

=<br />

M − M<br />

B<br />

Then for the ratio of the kinetic energies<br />

KE<br />

KE<br />

A<br />

B<br />

=<br />

1<br />

2<br />

1<br />

2<br />

M<br />

M<br />

A<br />

B<br />

V<br />

V<br />

2<br />

A<br />

2<br />

B<br />

=<br />

M<br />

M<br />

A<br />

B<br />

M<br />

M<br />

2<br />

B<br />

B<br />

=<br />

2<br />

( M − M<br />

B<br />

) M<br />

A<br />

The ratio of the KE’s is simply the inverse ratio of the masses.<br />

From the two equations<br />

M<br />

B<br />

KE<br />

A<br />

= KEB<br />

and KEA<br />

+ KEB<br />

= Q<br />

M<br />

We can solve for<br />

KEB<br />

to find<br />

A<br />

⎛ M ⎞<br />

A<br />

KEA<br />

= Q − KEB<br />

= Q<br />

⎜1−<br />

M<br />

A<br />

M<br />

⎟<br />

⎝ +<br />

B ⎠<br />

Q QM<br />

A<br />

KEB<br />

= =<br />

1+<br />

M + M<br />

M B<br />

M A<br />

(b) If M<br />

B<br />

= 4M<br />

A<br />

, then M<br />

A<br />

will get 4 times as much KE as M<br />

B,<br />

or 80% of Q for<br />

M and 20% for M .<br />

A<br />

B<br />

A<br />

B

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