92. (Second problem in Cluster 2)

As explained in the first solution in this cluster, we take both angles θ

1

and θ

2

to be positive-valued.

(a) We first examine conservation of the y components of momentum.

0

=

−m

1

v

1f

sin θ

1

+ m

2

v

2f

sin θ

2

0

=

−m

1

v

1f

sin 30

◦

+ 2m

1

v

2f

sin θ

2

Next, we examine conservation of the x components of momentum.

m

1

v

1i

=

m

1

v

1f

cos θ

1

+ m

2

v

2f

cos θ

2

m

1

(10.0 m/s)

=

m

1

v

1f

cos 30

◦

+ 2m

1

v

2f

cos θ

2

From the y equation, we obtain v

1f

= 4 v

2f

sin θ

2

; similarly, the x equation yields 20

− v

1f

√

3 =

4v

2f

cos θ

2

with SI units understood (and the fact that cos 30

◦

=

√

3/2 has been used). Squaring

these two relations and adding them leads to

v

2

1f

(1 + 3)

− 40v

1f

√

3 + 400 = 16 v

2

2f

sin

2

θ

2

+ cos

2

θ

2



and thus to

v

2

2f

= v

2

1f

/4

− 5v

1f

√

3/2 + 25 .

(b) The plot (v

2f

versus v

1f

) is shown below. The units for both axes are meters/second.

2.5

3

3.5

4

4.5

5

v2

0

2

4

6

8

10

12

14

v1

(c) Simply from the total kinetic energy requirement that K

i

≥ K

f

we see immediately that v

1f

≤

v

1i

= 10.0 m/s (where the upper bound represents the trivial case where it passes m

2

by completely

with K

i

= K

f

), and with the more stringent requirement that it does strike m

2

and scatters at

θ

1

= 30

◦

we again find that it is bounded by the K

i

= K

f

case. The elastic collision scenario was

worked in the previous problem with the result v

1f

= 9.34 m/s.

(d) And we also found the result v

2f

= 2.52 m/s.

(e) As mentioned, this is an elastic collision.