75. We use momentum conservation choosing +x forward and recognizing that the initial momentum is zero.

We analyze this from the point of view of an observer at rest on the ice.

(a) If v

1 and 2

is the speed of the stones, then the speeds are related by v

1 and 2

+ v

boat

= v

rel

. Thus,

with m

1

= 2m

2

and M =12m

2

, we obtain

0

=

(m

1

+ m

2

) (

−v

1 and 2

) + M v

boat

=(2m

2

+ m

2

) (

−v

rel

+ v

boat

) + 12m

2

v

boat

=

−3m

2

v

rel

+ 15m

2

v

boat

which yields v

boat

=

1
5

v

rel

= 0.2000v

rel

.

(b) Using v

1

+ v

boat

= v

rel

, we find – as a result of the first throw – the boat’s speed:

0

=

m

1

(

−v

1

) + (M + m

2

) v

boat

=

2m

2

(

−v

rel

+ v

boat

) + (12m

2

+ m

2

) v

boat

=

−2m

2

v

rel

+ 15m

2

v

boat

which yields v

boat

=

2

15

v

rel

≈ 0.133v

rel

. Then, using v

2

+ v

boat

= v

rel

, we consider the second

throw:

(M + m

2

) v

boat

=

m

2

(

−v

2

) + M v

boat

(12m

2

+ m

2

)

2

15

v

rel



=

m

2

(

−v

rel

+ v

boat

) + 12m

2

v

boat

26

15

m

2

v

rel

=

−m

2

v

rel

+ 13m

2

v

boat

which yields v

boat

=

41

195

v

rel

≈ 0.2103v

rel

.

(c) Finally, using v

2

+ v

boat

= v

rel

, we find – as a result of the first throw – the boat’s speed:

0

=

m

2

(

−v

2

) + (M + m

1

) v

boat

=

m

2

(

−v

rel

+ v

boat

) + (12m

2

+ 2m

2

) v

boat

=

−m

2

v

rel

+ 15m

2

v

boat

which yields v

boat

=

1

15

v

rel

≈ 0.0673v

rel

. Then, using v

1

+ v

boat

= v

rel

, we consider the second

throw:

(M + m

1

) v

boat

=

m

1

(

−v

1

) + M v

boat

(12m

2

+ 2m

2

)

1

15

v

rel



=

2m

2

(

−v

rel

+ v

boat

) + 12m

2

v

boat

14

15

m

2

v

rel

=

−2m

2

v

rel

+ 14m

2

v

boat

which yields v

boat

=

22

105

v

rel

≈ 0.2095v

rel

.