Question

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?

Answer 1

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