fisica1-youn-e-freedman-exercicios-resolvidos

8.70: a) After impact, the block-bullet combination has a total mass of 1.00 kg, and the
speed V of the block is found from 1 2 1 2
k
M V = kX , or V = X . The spring constant k
2
is determined from the calibration; k 0.75 N
=
−3
= 300 N m. Combining,
total
2.
50 × 10
m
2
m
−
( 15.0 × 10 m) = 2.60 m s.
300 N/m
2
V =
1.00 kg
b) Although this is not a pendulum, the analysis of the inelastic collision is the same;
v
M
m
1.00 Kg
=
3
8.0 × 10 Kg
total
= V
−
( 2.60 m s) = 325 m s.
8.71: a) Take the original direction of the bullet’s motion to be the x-direction,
and the direction of recoil to be the y-direction. The components of the stone’s velocity
after impact are then
v
x
v
y
−3
⎛ 6.00×
10 Kg ⎞
=
⎜
0.100 Kg
⎟
⎝
⎠
−3
⎛ 6.00 × 10 Kg ⎞
= −
⎜
0.100 Kg
⎟
⎝
⎠
( 350 m s) = 21.0 m s,
( 250 m s) = 15.0 m s,
and the stone’s speed is ( 21.0 m s) ( 15.0 m s) = 25.8 m s ,
15 .0
1
−3
2
( )
21.0
= 35.5°
. b) K
1
= ( 6.00×
10 kg)( 350 m/s) = 368 J
2
1
−
= ( 6.00 × 10
3 kg)( 250 ) 2 1 2 2
0.100 kg 25.8 m / s =
2 2
m/s
not perfectly elastic.
2
2
+ at an angle of arctan
K ( )( ) 221J,
+ so the collision is
2