Two steel spheres approach each other head-on with the same speed and collide elastically. After the collision one of the spheres of radius r comes to rest,the radius of other sphere is
Answers
Explanation:
r/(3)1/3
Hope it helps you
Answer:
According to the given data, conservation of momentum and conservation of energy is to be applied on the given situation to form the equation and then be compared to rule out the relation.
Thereby,
Initial momentum = Final Momentum
\Rightarrow m_{ 1 }v-m_{ 2 }v=m_{ 1 }v" -m_{ 2 }v'⇒m
1
v−m
2
v=m
1
v"−m
2
v
′
\Rightarrow v"=0⇒v"=0
\Rightarrow v'=\frac { v(m_{ 1 }-m_{ 2 }) }{ m_{ 2 } }⇒v
′
=
m
2
v(m
1
−m
2
)
\Rightarrow \frac { 1 }{ 2 } m_{ 1 }v_{ 2 }+\frac { 1 }{ 2 } m_{ 2 }v_{ 2 }=\frac { 1 }{ 2 } m_{ 2 }{ v }_{ 2 }'⇒
2
1
m
1
v
2
+
2
1
m
2
v
2
=
2
1
m
2
v
2
′
Thereby \frac { 1 }{ 2 } m_{ 1 }v_{ 2 }+\frac { 1 }{ 2 } m_{ 2 }v_{ 2 }=\frac { 1 }{ 2 } m_{ 2 }(\frac { m_{ 1 }-m_{ 2 } }{ m_{ 2 } } )v_{ 2 }
2
1
m
1
v
2
+
2
1
m
2
v
2
=
2
1
m
2
(
m
2
m
1
−m
2
)v
2
\Rightarrow \frac { 1 }{ 2 } v_{ 2 }\left( m_{ 1 }+m_{ 2 } \right) =\frac { 1 }{ 2 } m_{ 2 }(\frac { m_{ 1 }-m_{ 2 } }{ m_{ 2 } } )v_{ 2 }⇒
2
1
v
2
(m
1
+m
2
)=
2
1
m
2
(
m
2
m
1
−m
2
)v
2
\Rightarrow m_{ 1 }+m_{ 2 }=\frac { (m_{ 1 }-m_{ 2 })^{ 2 } }{ m_{ 2 } }⇒m
1
+m
2
=
m
2
(m
1
−m
2
)
2
\Rightarrow m_{ 1 }m_{ 2 }+m_{ 2 }^{ 2 }=m_{ 1 }^{ 2 }+m_{ 2 }^{ 2 }-2m_{ 1 }m_{ 2 }⇒m
1
m
2
+m
2
2
=m
1
2
+m
2
2
−2m
1
m
2
\Rightarrow 3m_{ 1 }m_{ 2 }=m_{ 1 }^{ 2 }⇒3m
1
m
2
=m
1
2
\Rightarrow m_{ 2 }=\frac { m_{ 1 } }{ 3 }⇒m
2
=
3
m
1
Then the Volume of sphere =\frac { 4 }{ 3 } \pi r^{ 3 }=
3
4
πr
3
and so now the V of the second sphere becomes \frac { V }{ 3 }
3
V
\Rightarrow \frac { V }{ 3 } =\frac { 4 }{ 3 } \pi { r' }^{ 3 }⇒
3
V
=
3
4
πr
′
3
\Rightarrow \frac { V }{ \frac { V }{ 3 } } =\frac { \frac { 4 }{ 3 } \pi r^{ 3 } }{ \frac { 4 }{ 3 } \pi { r' }^{ 3 } }⇒
3
V
V
=
3
4
πr
′
3
3
4
πr
3
\Rightarrow \frac { 1 }{ 3 } =\frac { r'^{ 3 } }{ r^{ 3 } }⇒
3
1
=
r
3
r
′3
\Rightarrow r'=\frac { r }{ \sqrt [ 3 ]{ 3 } }⇒r
′
=
3
3
r