Chapter 13 Solutions - Mosinee School District

Chapter 13
v
2
588 N m 0.050 0 m
2 9.80 m s2
0.050 0 m 0.700 m s
3.00 kg
13.68 (a) We apply conservation of mechanical energy from just after the collision until the block comes to rest.
KE PE s KE PE f
s
gives 1 2 1
0 k x 2 0
i
f MV
2 2
or the speed of the block just after the collision is
V
kx2
f
M
900 N m 0.050 0 m
1.00 kg
2
1.50 m s
Now, we apply conservation of momentum from just before impact to immediately after the collision. This
gives
m v 0 m v MV
bullet
i
bullet
f
or
M
vbullet
v
f bullet V
i
m
1.00 kg
400 m s 1.5 m s 100 m s
5.00 10
-3
kg
(b)
The mechanical energy converted into internal energy during the collision is
1 2 1 2 1
i f bullet i
bullet f
E KE KE 2 m v 2 m v 2
MV
2
or
E
1 1
2 2
3
2 2 2
5.00 10 kg 400 m s 100 m s 1.00 kg 1.50 m s
E
374 J
13.69 Choose PE g = 0 when the blocks start from rest. Then, using conservation of mechanical energy from when the
blocks are released until the spring returns to its unstretched length gives
KE PE PE KE PE PE , or
g s
f
g s
i
Page 13.26