1.

Two rocks are dropped simultaneously from the top of a tall building. Rock 1 has mass M_(1), and rock 2 has mass M_(2). If air resistance is negligible, what is the ratio of rock 1's momentum to rock 2's momentum just before they hit the ground?

Answer»

`(sqrt(M_(1)))/(M_(2))`
`(M_(1))/(M_(2))`
`((M_(1))/(M_(2)))^(2)`
1

Solution :Because the rocks are DROPPED from REST SIMULTANEOUSLY and air resistance is negligible, both rocks will accelerate at the same rate and have the same speed, v, at impact. So, the ratio of rock 1's momentum to rock 2's momentum will be `(M_(1)v)//(M_(2)v)=M_(1)//M_(2)`.


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