Two rocks collide in outer space. Before the collision, one rock had mass 11 kg and velocity ‹ 4250, −2950, 2500 › m/s. The other rock had mass 6 kg and velocity ‹ −700, 2150, 3700 › m/s. A 1 kg chunk of the first rock breaks off and sticks to the second rock. After the collision the 10 kg rock has velocity ‹ 1500, 300, 1900 › m/s. After the collision, what is the velocity of the other rock, whose mass is now 7 kg?

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Hashirriaz830 Hashirriaz830 · Answer 1 · 2020-02-21T10:46:51+01:00

Answer:

p_7kg,f=(25100-15000,38,500)m/s

Explanation:

In the collision the total momentum of the system is conserved so we only need to sum up the moments of our two rocks before and after the collision.

p_11kg,i+p_6kg,i=p_10kg,f+p_7kg,f

11kg(4250, −2950, 2500)m/s+6(−700, 2150, 3700)m/s=10kg(1500, 300, 1900)m/s+p_7kg,f

(46,750-32,450,27,500)kgm/s+(4,200‬-12,900‬,22,200‬)kgm/s=(15,000‬-3000‬,19,000‬)kgm/s+p_7kg,f

p_7kg,f=(46,750-32,450,27,500)kgm/s+(4,200‬-12,900‬,22,200‬)kgm/s-(15,000‬-3000‬,19,000‬)kgm/s

p_7kg,f=(25100-15000,38,500)m/s

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-04-02T08:21:07+02:00

The velocity of the other rock, whose mass is now 7 kg is :

‹ 3935.71, -3221.42, 4385.71› m/s

Given information:

before collision,

m₁ = 11 kg

m₂ = 6 kg

v₁ = ‹ 4250, −2950, 2500 › m/s

v₂ = ‹ −700, 2150, 3700 › m/s

after collision,

M₁ = 10kg

M₂ = 7kg

V₁ = ‹ 1500, 300, 1900 › m/s

V₂ = to be determined

According to the law of conservation of momentum:

m₁v₁ + m₂v₂ = M₁V₁ + M₂V₂

11×‹ 4250, −2950, 2500 › + 6×‹ −700, 2150, 3700 › = 10× ‹ 1500, 300, 1900 › + 7V₂

7V₂ = ‹ 27550,-22550,30700› kgm/s

V₂ = ‹ 3935.71, -3221.42, 4385.71› m/s

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