fisica1-youn-e-freedman-exercicios-resolvidos

8.76: Just after the collision: ∑ F = ma
1600N −
2
v8
T − m8g
= m8
R
2
v8
1.35 m
v = 16.0m s
2
( 8.00 kg)( 9.80m s ) = ( 8.00 kg)
8
Energy and momentum are conserved during the elastic collision.
( 2.00 kg) v = ( 2.00 kg) v + ( 8.00 kg)( 16.0 m s)
1
2
m v
2
v
m v
0
0
0
2
2 0
1
= m2v
2
+ 64.0 m s
2
2
( 2.00 kg) v = ( 2.00 kg) v + ( 8.00 kg)( 16.0m s)
v
0
2
0
= m v
= v
= v
2
2
2
2
2
+ m v
2
2
+
8
2
1
2
2
8
+ 1024m
m v
2
2
8 8
s
2
2
(1)
(2)
Solve (1) and (2) for
v
0
: v0
=
40.0m
s
8.77: a) The coefficient of friction, from either force or energy consideration, is
2
µ
k
= v 2gs,
where v is the speed of the block after the bullet passes through. The speed
of the block is determined from the momentum lost by the bullet,
3
4.00× 10
− kg 280m s = 1.12 kg ⋅ m and so the coefficient of kinetic friction is
( )( ) s,
µ
(( 1.12 kg ⋅ m s)
(0.80 kg) )
2
2( 9.80 m s )( 0.45m)
2
k
=
=
1 3
2
2
b) ( 4.00 10 kg) ( 400 m s) − ( 120m s)
) = 291 J.
2
0.22.
× − c) From the calculation of the
p (1.12 kg⋅m s)
momentum in part (a), the block’s initial kinetic energy was = = 0.784 J.
2
2m
2(0.80 kg)
2