Two stones, having masses in the ratio of $3 : 2$ are dropped from the heights in the ratio of $4 : 9$ . The ratio of magnitudes of their linear momentum just before reaching the ground is (neglect air resistance)

4 : 9

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2 : 3

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3 : 2

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1 : 1

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Solution

The correct option is D $1 : 1$
case 1:

$v = \sqrt{2 a s} = \sqrt{2 \times 10 \times 4} = 4 \sqrt{5}$

$p = v m = 4 \sqrt{5} \times 3 = 12 \sqrt{5}$ .....A

case 2:

$v = \sqrt{2 a s} = \sqrt{2 \times 10 \times 9} = 6 \sqrt{5}$

$p = v m = 6 \sqrt{5} \times 2 = 12 \sqrt{5}$ .....B

from A and B, its clear that, ratio is $1 : 1$

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Two stones, having masses in the ratio of 3:2 are dropped from the heights in the ratio of 4:9. The ratio of magnitudes of their linear momentum just before reaching the ground is (neglect air resistance)

The correct option is D 1:1case 1:v=√2as=√2×10×4=4√5p=vm=4√5×3=12√5.....Acase 2:v=√2as=√2×10×9=6√5p=vm=6√5×2=12√5.....Bfrom A and B, its clear that, ratio is 1:1

Two stones, having masses in the ratio of $3 : 2$ are dropped from the heights in the ratio of $4 : 9$ . The ratio of magnitudes of their linear momentum just before reaching the ground is (neglect air resistance)