Respuesta :

gabriellekeesler08

Answer:

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]

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