Question

Two stones of masses in the ratio of 3:4 fall from heights in ratio 4:9. The ratio of their momenta on reaching the ground is

Answer 1

Given :

▪ Ratio of masses of two stones = 3:4

▪ Ratio of heights = 4:9

To Find :

▪ The ratio of their momenta on reaching the ground.

Concept :

→ Momentum is defined as the product of mass and velocity.

→ First, we have to calculate final velocity at which stone touches the ground.

→ For free falling motion,

initial velocity is taken equal to zero.

Calculation :

$\implies\sf\:v^2-u^2=2gH\\ \\ \implies\sf\:v^2-(0)^2=2gH\\ \\ \implies\sf\:v^2=2gH\\ \\ \implies\boxed{\bf{\green{v=\sqrt{2gH}}}}\\ \\ \dashrightarrow\sf\:\dfrac{P_A}{P_B}=\dfrac{M_AV_A}{M_BV_B}\\ \\ \dashrightarrow\sf\:\dfrac{P_A}{P_B}=\dfrac{M_A}{M_B}\times \dfrac{\sqrt{2gH_A}}{\sqrt{2gH_B}}\\ \\ \dashrightarrow\sf\:\dfrac{P_A}{P_B}=\dfrac{M_A\sqrt{H_A}}{M_B\sqrt{H_B}}\\ \\ \dashrightarrow\sf\:\dfrac{P_A}{P_B}=\dfrac{3}{4}\times\sqrt{\dfrac{4}{9}}\\ \\ \dashrightarrow\sf\:\dfrac{P_A}{P_B}=\dfrac{3\times 2}{4\times 3}\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{P_A:P_B=1:2}}}}\:\pink{\bigstar}$

Answer 2

Let us assume that the mass of first stone is 3J and mass of second stone is 4J. Aslo, the height of first stone is 4M and height of second stone is 9M.

Since, both the stones fall from heights in ratio 4:9. Means, their initial velocity i.e. u is 0 m/s.

Using the First Equation Of Motion,

v = u + at

→ v = 0 + at1

→ v = at1

Now, using the Second Equation Of Motion,

s = ut + 1/2 at²

→ 4M = (0)t1 + 1/2 a(t1)²

→ 4M = 1/2 a(t1)²

→ 8M = a(t1)²

→ √(8M/a) = t1

So,

→ v = a√(8M/a)

→ v = √[(8Ma²)/a]

→ v = √(8aM)

Now,

Momentum (p1) = mass × velocity

= 3J × √(8aM) ..........(1st equation)

Similarly,

Using the First Equation Of Motion,

v = u + at

→ v = 0 + at1

→ v = at1

Now, using the Second Equation Of Motion,

s = ut + 1/2 at²

→ 9M = (0)t1 + 1/2 a(t2)²

→ 9M = 1/2 a(t2)²

→ 18M = a(t2)²

→ √(18M/a) = t2

So,

→ v = a√(18M/a)

→ v = √[(18Ma²)/a]

→ v = √(18aM)

Now,

Momentum (p2) = mass × velocity

= 4J × √(18aM) ..........(2nd equation)

We have to find the ratio of momenta on reaching the ground.

[3J × √(8aM)] / [4J × √(18aM)]

Squaring both sides

[3J × √(8aM)]² / [4J × √(18aM)]²

(9J × 8aM) / (16J × 18aM)

(9 × 8)/(16 × 18)

(1 × 1)/(2 × 2)

1/4

Now,

(p1)²/(p2)² = 1/4

(p1)²/(p2)² = (1)²/(2)²

p1/p2 = √(1/2)²

p1/p2 = 1/2

Therefore, ratio of their momenta on reaching the ground is 1:2.