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?

Question

Physics

Josue Tyler

21 June, 11:25

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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Answers (1)

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Maritza · Answer 1

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