Question

1. Two bodies of masses m, and m, have same
kinetic energy. The ratio of their momentum is
a)√m2/√m1
b)√m1/√m2
c)m1^2/m2^2
d)m2^2/m1^2​

Answer 1

Correct Question:

Two bodies of masses m1 and m2 have same kinetic energy. The ratio of their momentum is?

a)√m2/√m1

b)√m1/√m2

c)m1^2/m2^2

d)m2^2/m1^2

Answer: Option (b)

$\frac{ \sqrt{m1} }{ \sqrt{m2} }$

Solution:

Kinetic Energy (KE) of m1

$= \frac{1}{2} m_{1} {v_{1}}^{2} \\ \\ = \dfrac{1}{2m_{1}} {(m_{1}v_{1})}^{2}\: \: \: \: \: (Equation\:1)$

Kinetic Energy (KE) of m2

$= \frac{1}{2} m_{2} {v_{2}}^{2} \\ \\ =\dfrac{1}{2m_{2}} {(m_{2}v_{2})}^{2} \: \: \: \: \: (Equation\:2)$

Momentum of m1 = $p_{1} = m_{1} v_{1}\: \: \: \: \: (Equation\:3)$

Momentum of m2 = $p_{2} = m_{2} v_{2}\: \: \: \: \: (Equation\:4$

From equation 1 and 3:

Kinetic Energy (KE) of m1

$=\frac{1}{2} m_{1} {v_{1}}^{2} \\ \\ = \dfrac{1}{2m_{1}} \times p_{1} \\ \\ p_{1} = \sqrt{2 \times KE \times m_{1}}$

Similarly from Equation 2 and 4:

$p_{2} = \sqrt{2 \times KE \times m_{2}}$

Ratio:

$= \dfrac{p_{1}}{p_{2}} \\ \\ = \frac{ \sqrt{2 \times KE \times m1} }{ \sqrt{2 \times KE \times m_{2}} } \\ \\ = \frac{ \sqrt{m1} }{ \sqrt{m_{2}}}$

Answer 2

it's simple dude.....

according to given statement

1/2 m1v1^2=1/2m2V2^2....(taking velocity of masses as V1 and V2)

from this we get

V1/V2=√m2/√m1

now

momentum of first object=m1v1

momentum of another object=m2v2

their ration

m1v1/m2v2=m1/m2×√m2/√m1..(V1/V2=√m2/√m1)

solving m1×√m2/m2×√m1

we get

m1v1/m2v2=√m1/√m2

hope it helps

Keep asking ur doubts........