Two bodies of masses ma and mb have equal kinetic energy. the ratio of their momentum is:
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Answered by
3
Hello !
Since momentum, P = m×v,
where P is momentum,
m is mass of an object, and
v is velocity of an object.
So mass of first object is given i.e.,ma,
therefore, P = ma×v ....... 1 equation,
Now mass of second object is mb,
therefore, P = mb×v .......2 equation,
Now when we divide 1 equation by 2 equation,
we get,
Ratio = a/b
Hope this will be helpful to you, if u like it then plz mark it as brainliest.
Since momentum, P = m×v,
where P is momentum,
m is mass of an object, and
v is velocity of an object.
So mass of first object is given i.e.,ma,
therefore, P = ma×v ....... 1 equation,
Now mass of second object is mb,
therefore, P = mb×v .......2 equation,
Now when we divide 1 equation by 2 equation,
we get,
Ratio = a/b
Hope this will be helpful to you, if u like it then plz mark it as brainliest.
Answered by
0
.........................Given that two bodies of mass m_Am
A
and m_Bm
B
have equal kinetic energy hence,
\dfrac{1}{2}m_A (v_A)^2 = \dfrac{1}{2}m_B (v_B)^2
2
1
m
A
(v
A
)
2
=
2
1
m
B
(v
B
)
2
\dfrac{1}{2}\dfrac{(m_A)^2 (v_A)^2}{m_A} = \dfrac{1}{2}\dfrac{(m_B)^2 (v_B)^2}{m_B}
2
1
m
A
(m
A
)
2
(v
A
)
2
=
2
1
m
B
(m
B
)
2
(v
B
)
2
\dfrac {(p_A)^2}{m_A} = \dfrac {(p_B)^2}{m_B}
m
A
(p
A
)
2
=
m
B
(p
B
)
2
\dfrac {(p_A)^2}{(p_B)^2} = \dfrac {m_A}{m_B}
(p
B
)
2
(p
A
)
2
=
m
B
m
A
\dfrac {(p_A)^2}{(p_B)^2} = \dfrac {\sqrt {m_A}}{\sqrt {m_B}}
(p
B
)
2
(p
A
)
2
=
m
B
m
A
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