Answer:
Step-by-step explanation:
In the absence of external forces
Δ
→
p
=
0
so
m
b
1
→
v
b
1
+
m
b
2
→
v
b
2
=
m
a
1
→
v
a
1
+
m
a
2
→
v
a
2
Here we have
m
b
1
=
1000
,
→
v
b
1
=
25
ˆ
i
m
b
2
=
2000
,
→
v
b
2
=
15
ˆ
j
m
a
1
=
1000
,
→
v
a
1
=
→
v
a
m
a
2
=
2000
,
→
v
a
2
=
→
v
a
so
1000
⋅
25
ˆ
i
+
2000
⋅
15
ˆ
j
=
(
1000
+
2000
)
→
v
a
so
→
v
a
=
(
25
3
ˆ
i
+
10
ˆ
j
)
with
∣
∣
→
v
a
∣
∣
=
5
√
61
3
=
13.02
∠
→
v
a
=
arctan
(
10
/
(
25
3
)
)
=
50.2
∘
Kinetic energy before
K
b
=
1
2
m
b
1
∣
∣
→
v
b
1
∣
∣
2
+
1
2
m
b
2
∣
∣
→
v
b
2
∣
∣
2
Kinetic energy after
1
2
(
m
b
1
+
m
b
2
)
∣
∣
→
v
a
∣
∣
2
so
Δ
K
=
K
a
−
K
b
=
−
283333
.
[J]