Question

Two stones, having masses in the ratio of 3:2 are dropped from the heights in the ratio of 4:9. The ratio of magnitude of their linear momentum just before reaching the ground is (neglect air resistance)
Plzzzzz give me it's full solution

Answer 1

Ratio of masses = M/m = 3:2
Ratio of height= H/h = 4:9
u = 0m/s v = ?
g=10m/s
For finding the velocity of the stones with H;
$2(4 \times 10) + 0 = v^{2} \\ \sqrt{80} =v \\ 4 \sqrt{5} = v$
velocity with h height;
$2(9 \times 10) + 0 = v ^{2} \\ \sqrt{180} = v \\ 6 \sqrt{5} = v$
Ratio of linear momentum of the 2 stones;
P/p = MV/mv=1:1
Therefore, the ratio of the linear momentum of the two stones is 1:1

Answer 2

heya, here is your answer

Ratio of masses = M/m = 3:2

Ratio of height= H/h = 4:9

u = 0m/s v = ?

g=10m/s

For finding the velocity of the stones with H;

velocity with h height;

Ratio of linear momentum of the 2 stones;

P/p = MV/mv=1:1

Therefore, the ratio of the linear momentum of the two stones is 1:1